System of Particles & Rotational Motion

(a) For rigid bodies of regular geometrical shape having a uniform distribution of mass, the centre of mass is at their geometric centre. Hence,

(i) Sphere — the centre of mass is at the centre of the sphere.

(ii) Cylinder — the centre of mass is at the mid-point of its axis of symmetry.

(iii) Ring — the centre of mass is at the centre of the ring.

(iv) Cube — the centre of mass is at the point of intersection of its diagonals, that is, at its centre.

(b) No, the centre of mass of a body does not necessarily lie inside the body.

The centre of mass is the weighted average location of all the mass in the system, and it depends on the distribution of mass and not on the physical boundaries of the body. Hence for a body having no material at that average position, the centre of mass lies outside the material of the body. For example, the centre of mass of a ring is at its centre, where there is actually no matter. The same is true for a hollow sphere and a hollow cylinder.

Given,

• Separation between the nuclei of the two atoms, d = 1.27 Å
• Mass of the chlorine atom, m₂ = 35.5 m₁, where m₁ is the mass of the hydrogen atom

Nearly all the mass of an atom is concentrated in its nucleus, so the two atoms may be treated as point masses placed at their nuclei. The centre of mass lies on the line joining the H and Cl atoms. Let this line be the X-axis with the H atom at the origin.

Then x₁ = 0 and x₂ = 1.27 Å.

The centre of mass relative to the H atom is given by

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

Substituting the values,

$$\begin{aligned}\text x_{cm} = \frac{(\text m_1 \times 0) + (35.5\text m_1 \times 1.27)}{\text m_1 + 35.5\text m_1} \\[1em] = \frac{35.5 \times 1.27}{36.5} = \frac{45.085}{36.5}\end{aligned}$$

$\text x_{cm} = 1.235 \text{\AA}$

Hence, the centre of mass of the HCl molecule is located at a distance of 1.235 Å from the hydrogen atom, on the line joining the two nuclei. Being very close to the chlorine nucleus, it shows that the centre of mass lies nearer to the heavier particle.

The speed of the centre of mass of the system remains unchanged and is equal to v.

The child and the trolley together form a single system. When the child gets up and runs about on the trolley, the forces involved between the child and the trolley, such as the action and reaction at his feet and the friction between his feet and the floor of the trolley, are all internal forces of the system.

Internal forces always occur in equal and opposite pairs, in compliance with Newton's third law of motion, so their vector sum is zero and they cancel out. Since the floor is smooth, no external force acts on the system in the horizontal direction, that is, F_ext = 0. Therefore

$\frac{\text{d}\vec{\text P}}{\text{dt}} = \vec{\text F}_{ext} = 0 \quad \Rightarrow \quad \vec{\text P} = \text{M}\vec{\text v}_{cm} = \text{a constant}$

Hence the velocity of the centre of mass remains constant. Without external forces, a system's internal dynamics alone cannot induce acceleration. The speed of the centre of mass therefore continues to be v, however the child may run about on the trolley.

Let the vectors $\vec{\text A} \space$and $\vec{\text B} \space$represent the sides OP and OR of a triangle OPR, and let θ be the angle between them. Let RS be the perpendicular dropped from R on OP.

The area of the triangle OPR is

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(\text{OP})(\text{RS})$

From the right-angled triangle ORS,

$\text{RS} = (\text{OR})\sin \theta$

Substituting this value,

$\text{Area} = \frac{1}{2}(\text{OP})(\text{OR})\sin \theta = \frac{1}{2}\text{AB}\sin \theta$

where A and B are the magnitudes of the two vectors.

But by the definition of the vector product,

$|\vec{\text A} \times \vec{\text B}| = \text{AB}\sin \theta$

Therefore,

$\text{Area of the triangle} = \frac{1}{2}|\vec{\text A} \times \vec{\text B}|$

Hence, the area of the triangle contained between the vectors $\vec{\text A} \space$and $\vec{\text B} \space$is one-half the magnitude of $\vec{\text A} \times \vec{\text B}$.

Let a parallelopiped be formed on the three vectors $\vec{\text A}$, $\vec{\text B} \space$and $\vec{\text C}$, with θ the angle between $\vec{\text B} \space$and $\vec{\text C}$, and $\phi$ the angle between $\vec{\text A} \space$and the normal to the base.

The vector product of $\vec{\text B} \space$and $\vec{\text C} \space$is

$\vec{\text B} \times \vec{\text C} = \text{BC}\sin \theta\ \hat{\text n}$

where $\hat{\text n}$ is a unit vector normal to the plane containing $\vec{\text B} \space$and $\vec{\text C}$. Its magnitude is

$|\vec{\text B} \times \vec{\text C}| = \text{BC}\sin \theta$

Now taking the scalar product of $\vec{\text A} \space$with $\vec{\text B} \times \vec{\text C}$,

$$\begin{aligned}\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) = \text A(\text{BC}\sin \theta)\cos \phi \\[1em] = (\text{BC}\sin \theta)(\text A \cos \phi)\end{aligned}$$

Here BC sin θ is the area of the parallelogram which forms the base of the parallelopiped, and A cos φ is the projection of $\vec{\text A} \space$along $\vec{\text B} \times \vec{\text C}$, that is, the height of the parallelopiped.

Therefore,

$\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) = \text{base area} \times \text{height}$

Hence, $\vec{\text A} \cdot (\vec{\text B} \times \vec{\text C}) \space$is equal in magnitude to the volume of the parallelopiped formed on the vectors $\vec{\text A}$, $\vec{\text B} \space$and $\vec{\text C}$.

The position vector and the linear momentum of the particle in terms of their x, y, z-components are

$\vec{\text r} = \text x\hat{\text i} + \text y\hat{\text j} + \text z\hat{\text k} \quad \text{and} \quad \vec{\text p} = \text p_x\hat{\text i} + \text p_y\hat{\text j} + \text p_z\hat{\text k}$

By definition, the angular momentum of the particle is the moment of linear momentum,

$$\vec{\text L} = \vec{\text r} \times \vec{\text p} = \begin{vmatrix} \hat{\text i} & \hat{\text j} & \hat{\text k} \\ \text x & \text y & \text z \\ \text p_x & \text p_y & \text p_z \end{vmatrix}$$

Expanding the determinant,

$\vec{\text L} = \hat{\text i}(\text y\text p_z - \text z\text p_y) + \hat{\text j}(\text z\text p_x - \text x\text p_z) + \hat{\text k}(\text x\text p_y - \text y\text p_x) \qquad \dots(\text i)$

The angular momentum may also be written in terms of its own components as

$\vec{\text L} = \text L_x\hat{\text i} + \text L_y\hat{\text j} + \text L_z\hat{\text k} \qquad \dots(\text{ii})$

Comparing equations (i) and (ii), the components of the angular momentum along the X-, Y- and Z-axes are

$$\begin{aligned}\text L_x = \text y\text p_z - \text z\text p_y \\[1em] \text L_y = \text z\text p_x - \text x\text p_z \\[1em] \text L_z = \text x\text p_y - \text y\text p_x\end{aligned}$$

When the particle moves only in the X-Y plane : In this case the particle has no z-coordinate and no z-component of momentum, so

$\text z = 0 \quad \text{and} \quad \text p_z = 0$

Substituting these in the components obtained above,

$$\begin{aligned}\text L_x = (\text y \times 0) - (0 \times \text p_y) = 0 \\[1em] \text L_y = (0 \times \text p_x) - (\text x \times 0) = 0 \\[1em] \text L_z = \text x\text p_y - \text y\text p_x\end{aligned}$$

Therefore,

$\vec{\text L} = (\text x\text p_y - \text y\text p_x)\hat{\text k}$

Hence, if the particle moves only in the X-Y plane, the angular momentum has only a z-component, that is, it is directed perpendicular to the plane of motion.

Let AB and CD be the two parallel lines separated by a distance d, and let P and Q be the instantaneous positions of the two particles, each of mass m, travelling in opposite directions with speed v along these lines.

Let O be any point dividing the separation between the lines in the ratio d₁ : d₂, about which the angular momentum is to be taken, so that

$\text d_1 + \text d_2 = \text d$

Taking O as the origin, the instantaneous coordinates of P and Q are (x₁, d₁) and (x₂, − d₂) respectively. Their position vectors are

$\vec{\text r}_1 = \text x_1\hat{\text i} + \text d_1\hat{\text j} \quad \text{and} \quad \vec{\text r}_2 = \text x_2\hat{\text i} - \text d_2\hat{\text j}$

Since the particles travel in opposite directions, their momenta are $\vec{\text p}_1 = \text p_x\hat{\text i}$ and $\vec{\text p}_2 = -\text p_x\hat{\text i}$, where p_x = mv.

The angular momentum of particle P about O is

$$\begin{aligned}\vec{\text L}_P = \vec{\text r}_1 \times \vec{\text p}_1 = (\text x_1\hat{\text i} + \text d_1\hat{\text j}) \times \text p_x\hat{\text i} \\[1em] = -\text p_x \text d_1\hat{\text k} = -\text{mv}\text d_1\hat{\text k}\end{aligned}$$

Similarly, the angular momentum of particle Q about O is

$$\begin{aligned}\vec{\text L}_Q = \vec{\text r}_2 \times \vec{\text p}_2 = (\text x_2\hat{\text i} - \text d_2\hat{\text j}) \times (-\text p_x\hat{\text i}) \\[1em] = -\text p_x \text d_2\hat{\text k} = -\text{mv}\text d_2\hat{\text k}\end{aligned}$$

The total angular momentum of the two particles about O is

$$\begin{aligned}\vec{\text L} = \vec{\text L}_P + \vec{\text L}_Q = -\text{mv}\text d_1\hat{\text k} - \text{mv}\text d_2\hat{\text k} \\[1em] = -\text{mv}(\text d_1 + \text d_2)\hat{\text k}\end{aligned}$$

But d₁ + d₂ = d, therefore

$\vec{\text L} = -\text{mvd}\hat{\text k}$

The magnitude of the total angular momentum is mvd, directed perpendicular to the two lines.

Since the result is independent of d₁ and d₂, it does not depend upon the position of the origin O. Hence the vector angular momentum of the two-particle system is the same whatever be the point about which it is taken.

Given,

• Length of the bar = 2 m
• Angle made by the first string with the vertical, θ₁ = 36.9°
• Angle made by the second string with the vertical, θ₂ = 53.1°
• sin 36.9° = cos 53.1° = 0.6 and sin 53.1° = cos 36.9° = 0.8

Let T₁ and T₂ be the tensions in the two strings and let d be the distance of the centre of gravity G of the bar from its left end.

Condition of translational equilibrium : Resolving the tensions horizontally, the net force must be zero,

$\text T_1 \sin \theta_1 = \text T_2 \sin \theta_2 \qquad \dots(\text i)$

Condition of rotational equilibrium : Taking moments of the forces about the centre of gravity G, the net torque must be zero,

$\text T_1 \cos \theta_1 \times \text d = \text T_2 \cos \theta_2 \times (2 - \text d) \qquad \dots(\text{ii})$

From equation (i),

$\text T_2 = \text T_1 \frac{\sin \theta_1}{\sin \theta_2}$

Substituting this value of T₂ in equation (ii),

$\text T_1 \cos \theta_1 \times \text d = \text T_1 \frac{\sin \theta_1}{\sin \theta_2}\cos \theta_2 \times (2 - \text d)$

$\text d = \frac{\sin \theta_1 \cos \theta_2}{\cos \theta_1 \sin \theta_2}(2 - \text d)$

Substituting the given values,

$\text d = \frac{0.6 \times 0.6}{0.8 \times 0.8}(2 - \text d) = \frac{9}{16}(2 - \text d)$

$$\begin{aligned}16\text d = 18 - 9\text d \\[1em] 25\text d = 18\end{aligned}$$

$\text d = \frac{18}{25} = 0.72\ \text m$

Hence, the centre of gravity of the bar is at a distance of 0.72 m, that is, 72 cm, from its left end.

Given,

• Weight of the car, W = 1800 kg-wt
• Distance between the front and back axles = 1.8 m
• Distance of the centre of gravity G behind the front axle = 1.05 m
• g = 9.8 N/kg

The distance of the centre of gravity from the back axle is

$1.8 - 1.05 = 0.75\ \text m$

Let R₁ be the reactionary force exerted by the ground on each front wheel and R₂ that on each back wheel. Since there are two wheels on each axle, the total upward force on the front axle is 2R₁ and that on the back axle is 2R₂.

Condition of translational equilibrium : For the vertical equilibrium of the car,

$2\text R_1 + 2\text R_2 = \text W \qquad \dots(\text i)$

Condition of rotational equilibrium : Taking moments of the forces about the centre of gravity G,

$2\text R_1 \times 1.05 = 2\text R_2 \times 0.75$

$$\begin{aligned}\text R_1 \times 1.05 = \text R_2 \times 0.75 \\[1em] \text R_1 = \frac{0.75}{1.05}\text R_2 = \frac{5}{7}\text R_2 \qquad \dots(\text{ii})\end{aligned}$$

Substituting the value of R₁ from equation (ii) in equation (i),

$$\begin{aligned}2 \times \frac{5}{7}\text R_2 + 2\text R_2 = 1800\ \text{kg-wt} \\[1em] \frac{10}{7}\text R_2 + 2\text R_2 = \frac{24}{7}\text R_2 = 1800\ \text{kg-wt}\end{aligned}$$

$\text R_2 = 1800 \times \frac{7}{24} = 525\ \text{kg-wt}$

Converting to newton,

$\text R_2 = 525 \times 9.8 = 5145\ \text N$

Now from equation (ii),

$\text R_1 = \frac{5}{7}\text R_2 = \frac{5}{7} \times 5145 = 3675\ \text N$

Hence, the force exerted by the level ground on each front wheel is 3675 N and that on each back wheel is 5145 N.

The force on each back wheel is larger because the centre of gravity of the car lies nearer to the back axle.

Let M be the mass and R the radius of both the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis of symmetry is

$\text I_1 = \text{MR}^2$

and the moment of inertia of the solid sphere about an axis passing through its centre is

$\text I_2 = \frac{2}{5}\text{MR}^2$

Let τ be the magnitude of the equal torque applied to each body, producing angular accelerations α₁ in the cylinder and α₂ in the sphere. From the relation between torque, moment of inertia and angular acceleration,

$\tau = \text I_1 \alpha_1 = \text I_2 \alpha_2$

Therefore,

$\frac{\alpha_1}{\alpha_2} = \frac{\text I_2}{\text I_1} = \frac{(2/5)\text{MR}^2}{\text{MR}^2} = \frac{2}{5}$

$\alpha_2 = \frac{5}{2}\alpha_1$

Thus the angular acceleration produced in the sphere is 2.5 times that produced in the cylinder. Both bodies start from rest and, from the equation of rotational kinematics ω = ω₀ + αt, the angular speed acquired in a given time t is directly proportional to the angular acceleration.

Hence, the solid sphere will acquire a greater angular speed after a given time. This is because the sphere has a smaller moment of inertia, its mass being distributed nearer to the axis of rotation, and so it offers less resistance to the change in its rotational motion.

Given,

• Mass of the solid cylinder, M = 20 kg
• Angular speed, ω = 100 rad s^-1
• Radius of the cylinder, R = 0.25 m

The moment of inertia of a solid cylinder about its own axis is

$\text I = \frac{1}{2}\text{MR}^2$

Substituting the values,

$$\begin{aligned}\text I = \frac{1}{2} \times 20 \times (0.25)^2 = \frac{1}{2} \times 20 \times 0.0625 \\[1em] = 0.625\ \text{kg m}^2\end{aligned}$$

Kinetic energy of rotation :

$$\begin{aligned}\text K = \frac{1}{2}\text I \omega^2 = \frac{1}{2} \times 0.625 \times (100)^2 \\[1em] = \frac{1}{2} \times 0.625 \times 10000 = 3125\ \text J\end{aligned}$$

Angular momentum about the axis :

$\text L = \text I \omega = 0.625 \times 100 = 62.5\ \text{kg m}^2\text s^{-1}$

Hence, the kinetic energy associated with the rotation of the cylinder is 3125 J and the magnitude of its angular momentum about its axis is 62.5 kg m² s^-1.

Given,

• Initial angular speed of the turntable, ω₁ = 40 rev/min
• Final moment of inertia, $\text I_2 = \frac{2}{5}\text I_1$, where I₁ is the initial moment of inertia

(a) The turntable rotates without friction, so no external torque acts on the system. Hence the angular momentum of the system is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

$\omega_2 = \frac{\text I_1 \omega_1}{\text I_2}$

Substituting the values,

$$\begin{aligned}\omega_2 = \frac{\text I_1 \times 40}{(2/5)\text I_1} = \frac{5}{2} \times 40 \\[1em] = 100\ \text{rev/min}\end{aligned}$$

Hence, the new angular speed of the child is 100 rev/min.

(b) The kinetic energy of rotation is $\text K = \frac{1}{2}\text I \omega^2$. Therefore

$\frac{\text{Final K.E. of rotation}}{\text{Initial K.E. of rotation}} = \frac{\frac{1}{2}\text I_2 \omega_2^2}{\frac{1}{2}\text I_1 \omega_1^2} = \frac{(2/5)\text I_1 (100)^2}{\text I_1 (40)^2}$

$= \frac{2}{5} \times \frac{10000}{1600} = \frac{2}{5} \times 6.25 = 2.5$

Hence, the new kinetic energy of rotation is 2.5 times the initial kinetic energy of rotation, that is, it has increased.

Accounting for the increase : The angular momentum is conserved, but the kinetic energy is not. To fold his hands back, the child has to do work against the centrifugal effect in pulling his arms inwards. This work is done by the child using his own internal (chemical) energy, and it appears as the additional kinetic energy of rotation. Hence there is no violation of the conservation of energy.

Given,

• Mass of the hollow cylinder, M = 3 kg
• Radius of the cylinder, R = 40 cm = 0.40 m
• Force with which the rope is pulled, F = 30 N

Every particle of a hollow cylinder is at the same distance R from its own central axis, so its moment of inertia about that axis is

$$\begin{aligned}\text I = \text{MR}^2 = 3 \times (0.40)^2 \\[1em] = 3 \times 0.16 = 0.48\ \text{kg m}^2\end{aligned}$$

The rope is wound round the rim, so the force acts tangentially at a perpendicular distance R from the axis. The torque exerted on the cylinder is

$\tau = \text F \times \text R = 30 \times 0.40 = 12\ \text{N m}$

Angular acceleration : From the relation τ = Iα,

$\alpha = \frac{\tau}{\text I} = \frac{12}{0.48} = 25\ \text{rad s}^{-2}$

Linear acceleration of the rope : Since there is no slipping, the rope moves with the same linear acceleration as a point on the rim of the cylinder,

$\text a = \text R \alpha = 0.40 \times 25 = 10\ \text{m s}^{-2}$

Hence, the angular acceleration of the cylinder is 25 rad s^-2 and the linear acceleration of the rope is 10 m s^-2.

Given,

• Uniform angular speed of the rotor, ω = 200 rad s^-1
• Torque transmitted by the engine, τ = 180 N m
• The engine is 100% efficient

For a uniform angular speed of rotation no torque would be needed if there were no friction, since the angular acceleration is then zero. In practice, however, a torque has to be applied by the engine to counterbalance the frictional torque acting on the rotor, so that the net torque remains zero and the angular speed stays uniform.

The power generated by a torque is

$\text P = \tau \omega$

Substituting the values,

$$\begin{aligned}\text P = 180 \times 200 = 36000\ \text W \\[1em] = 36\ \text{kW}\end{aligned}$$

Hence, the power of the engine required is 36 kW.

Let O be the centre of the original disc of radius R and O′ the centre of the hole of radius R/2, so that OO′ = R/2. Let σ be the surface density, that is, the mass per unit area, of the disc.

The mass of the original complete disc is

$\text M = \pi \text R^2 \sigma$

and the mass of the portion cut out is

$\text m = \pi \left(\frac{\text R}{2}\right)^2 \sigma = \frac{1}{4}\pi \text R^2 \sigma = \frac{\text M}{4}$

The cut-out portion is treated as a negative mass superposed on the complete disc, since removing it is equivalent to adding a mass of − M/4 at the centre of the hole.

Taking the centre O of the original disc as the origin and the line OO′ as the X-axis, the two masses of the equivalent system are

• Mass M at x₁ = 0
• Mass − M/4 at x₂ = R/2

The centre of mass of the residual flat body with respect to O is

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

Substituting the values,

$\text x_{cm} = \frac{(\text M)(0) + \left(-\frac{\text M}{4}\right)\left(\frac{\text R}{2}\right)}{\text M - \frac{\text M}{4}} = \frac{-\frac{\text{MR}}{8}}{\frac{3\text M}{4}}$

$\text x_{cm} = -\frac{\text{MR}}{8} \times \frac{4}{3\text M} = -\frac{\text R}{6}$

The negative sign shows that the centre of mass lies on the side opposite to the hole.

Hence, the centre of gravity of the resulting flat body is at a distance R/6 from the centre of the original disc, on the side opposite to the cut portion.

Given,

• Mass of each coin = 5 g, so the mass of the two coins together = 10 g
• Position of the coins = 12.0 cm mark
• New balancing point (knife edge) K = 45.0 cm mark

Let m be the mass of the metre stick. The stick is uniform, so its weight mg acts at its centre of gravity G, which is at the 50.0 cm mark.

Taking moments of the forces about the knife edge K at the 45.0 cm mark, for the rotational equilibrium of the stick the anticlockwise moment must equal the clockwise moment,

$(10\ \text g)\text g \times (45.0 - 12.0) = \text{mg} \times (50.0 - 45.0)$

$(10\ \text g)\text g \times 33.0\ \text{cm} = \text{mg} \times 5.0\ \text{cm}$

Cancelling g from both sides,

$\text m = \frac{10\ \text g \times 33.0\ \text{cm}}{5.0\ \text{cm}}$

$\text m = 66.0\ \text g$

Hence, the mass of the metre stick is 66.0 g.

Given,

• Mass of the oxygen molecule, M = 5.30 × 10^-26 kg
• Moment of inertia about an axis through its centre, I = 1.94 × 10^-46 kg m²
• Mean speed of the molecule, v = 500 m/s
• Kinetic energy of rotation = $\frac{2}{3}$ × kinetic energy of translation

Let ω be the average angular velocity of the molecule. As given,

$\frac{1}{2}\text I \omega^2 = \frac{2}{3} \times \frac{1}{2}\text{Mv}^2$

Cancelling $\frac{1}{2}$ from both sides,

$\text I \omega^2 = \frac{2}{3}\text{Mv}^2$

$\omega^2 = \frac{2}{3} \times \frac{\text M}{\text I} \times \text v^2$

Taking the square root,

$\omega = \sqrt{\frac{2}{3} \times \frac{\text M}{\text I}} \times \text v$

Substituting the given values,

$\omega = \sqrt{\frac{2}{3} \times \frac{5.30 \times 10^{-26}}{1.94 \times 10^{-46}}} \times 500$

$= \sqrt{\frac{2}{3} \times 2.732 \times 10^{20}} \times 500 = \sqrt{1.821 \times 10^{20}} \times 500$

$= (1.35 \times 10^{10}) \times 500$

$\omega = 6.75 \times 10^{12}\ \text{rad s}^{-1}$

Hence, the average angular velocity of the oxygen molecule is 6.75 × 10¹² rad s^-1.

Given,

• Mass of the boy, m₁ = 10 kg
• Mass of the boat, m₂ = 40 kg
• Initial distance of the boy from the shore = 20 m
• Distance moved by the boy on the boat = 8 m

Let C₁ be the centre of mass of the boy and C₂ that of the boat, separated by a distance x. Initially the boy is 20 m from the shore, so the boat's centre is at (20 + x) m.

The centre of mass of the system (boy + boat) is initially

$$\begin{aligned}\text x_{cm} = \frac{(10 \times 20) + [40 \times (20 + \text x)]}{10 + 40} \\[1em] = \frac{1000 + 40\text x}{50}\end{aligned}$$

When the boy moves 8 m towards the shore, let his new distance from the shore be d. The separation between C₁ and C₂ becomes (x + 8), so the boat's centre is now at (d + x + 8).

$$\begin{aligned}\text x'_{cm} = \frac{(10 \times \text d) + [40 \times (\text d + \text x + 8)]}{10 + 40} \\[1em] = \frac{50\text d + 40\text x + 320}{50}\end{aligned}$$

There is no friction between the boat and the water, so no external force acts on the system. Hence the centre of mass of the system remains at rest, that is,

$\text x_{cm} = \text x'_{cm}$

$1000 + 40\text x = 50\text d + 40\text x + 320$

$50\text d = 680 \quad \Rightarrow \quad \text d = 13.6\ \text m$

Hence, the boy is now 13.6 m from the shore.

Given,

• Radius of the original disc = 10 cm
• Radii of the two punched holes = 5 cm and 2.5 cm

Let AB be the diameter along which the centres of all three circular portions lie, and let A be taken as the reference point.

Since the metal sheet is uniform, the weights are proportional to the areas, and so the areas may be used in place of the masses.

• Area of the original disc = π(10)² = 100π cm²
• Area of the punched portion of radius 5 cm = π(5)² = 25π cm²
• Area of the punched portion of radius 2.5 cm = π(2.5)² = 6.25π cm²

Their distances from A are :

• Centroid G of the original disc = 10 cm
• Centroid G₁ of the hole of radius 5 cm = 5 cm
• Centroid G₂ of the hole of radius 2.5 cm = 20 − 2.5 = 17.5 cm

The punched portions are treated as negative areas. If x is the distance of the centre of gravity of the remaining portion from A, then

$\text x = \frac{(100\pi \times 10) - (25\pi \times 5) - (6.25\pi \times 17.5)}{100\pi - 25\pi - 6.25\pi}$

$= \frac{1000\pi - 125\pi - 109.375\pi}{68.75\pi} = \frac{765.625}{68.75}$

$\text x = 11.14\ \text{cm}$

Hence, the centre of gravity of the punched disc lies on the diameter AB at a distance of 11.14 cm from A, that is, 1.14 cm away from the larger hole.

Given,

• Radius of the original circle = 15 cm
• Radius of each of the three cut circles = 5 cm
• AB = 12 cm

Let O be the centre of the original circle, taken as the origin, with AB along the X-axis and the third circle centred at C on the Y-axis. Since O is the mid-point of AB,

$\text A = (-6,\ 0) \quad \text{and} \quad \text B = (6,\ 0)$

From the figure the circle at C touches those at A and B, so AC = BC = 5 + 5 = 10 cm. If C = (0, c), then

$\sqrt{6^2 + \text c^2} = 10 \quad \Rightarrow \quad \text c^2 = 100 - 36 = 64$

$\text c = 8\ \text{cm}$

The sheet is uniform, so areas may be used in place of masses.

• Area of the original circle = π(15)² = 225π cm²
• Area of each cut circle = π(5)² = 25π cm²
• Area of the remaining portion = 225π − 3(25π) = 150π cm²

The three cut portions are treated as negative areas.

x-coordinate : The arrangement is symmetrical about the Y-axis, hence

$\text x_{cm} = 0$

y-coordinate : The centres A and B lie on the X-axis (y = 0) and C is at y = 8 cm,

$\text y_{cm} = \frac{(225\pi \times 0) - (25\pi \times 0) - (25\pi \times 0) - (25\pi \times 8)}{150\pi}$

$= \frac{-200\pi}{150\pi} = -1.33\ \text{cm}$

Hence, the centre of gravity of the remaining portion lies on the Y-axis at a distance of 1.33 cm from O, on the side opposite to the circle cut at C.

Note: The height of C is not stated in the question and has been obtained from the figure by taking the circle at C to touch those at A and B. If the figure shows a different position for C, the value of y_cm changes accordingly.

Given,

• Mass of the particle = m
• Speed of projection = u
• Angle of projection with the horizontal = θ
• Time = t

Taking the point of projection O as the origin, with the horizontal as the X-axis and the vertical as the Y-axis, the coordinates of the particle at time t are

$$\begin{aligned}\text x = (\text u \cos \theta)\text t \\[1em] \text y = (\text u \sin \theta)\text t - \frac{1}{2}\text{gt}^2\end{aligned}$$

The components of velocity at time t are

$$\begin{aligned}\text v_x = \text u \cos \theta \\[1em] \text v_y = \text u \sin \theta - \text{gt}\end{aligned}$$

The angular momentum of the particle about O is the moment of its linear momentum,

$\vec{\text L} = \vec{\text r} \times \vec{\text p} = \text m(\vec{\text r} \times \vec{\text v})$

Since the motion is confined to the X-Y plane, the angular momentum has only a z-component,

$\vec{\text L} = \text m(\text x\text v_y - \text y\text v_x)\hat{\text k}$

Substituting the values,

$\vec{\text L} = \text m\left[(\text u \cos \theta)\text t(\text u \sin \theta - \text{gt}) - \left((\text u \sin \theta)\text t - \frac{1}{2}\text{gt}^2\right)(\text u \cos \theta)\right]\hat{\text k}$

Expanding the two products,

$= \text m\left[\text u^2 \sin \theta \cos \theta\text t - \text{ug}\cos \theta\text t^2 - \text u^2 \sin \theta \cos \theta\text t + \frac{1}{2}\text{ug}\cos \theta\text t^2\right]\hat{\text k}$

The first and the third terms cancel, leaving

$\vec{\text L} = \text m\left[-\text{ug}\cos \theta\text t^2 + \frac{1}{2}\text{ug}\cos \theta\text t^2\right]\hat{\text k}$

$\vec{\text L} = -\frac{1}{2}\text{mgu}\text t^2 \cos \theta\ \hat{\text k}$

Hence, the angular momentum of the particle about the point of projection at time t is $\vec{\text L} = -\frac{1}{2}\text{mgu}\text t^2 \cos \theta\ \hat{\text k}$, of magnitude $\frac{1}{2}\text{mgu}\text t^2 \cos \theta$ and directed perpendicular to the plane of motion, into the plane.

Given,

• Radius of each sphere = R
• Mass of each sphere = M
• Side of the square = L

The axis of rotation passes along one of the sides of the square. Hence two spheres lie on the axis itself and the other two are at a perpendicular distance L from it.

The moment of inertia of a solid sphere about an axis through its centre is $\frac{2}{5}\text{MR}^2$.

For each of the two spheres lying on the axis : the axis passes through their centres, so

$\text I_1 = \frac{2}{5}\text{MR}^2$

For each of the two spheres at a distance L : by the theorem of parallel axes,

$\text I_2 = \frac{2}{5}\text{MR}^2 + \text{ML}^2$

The total moment of inertia of the system is the sum of the moments of inertia of the four spheres,

$\text I = 2\left(\frac{2}{5}\text{MR}^2\right) + 2\left(\frac{2}{5}\text{MR}^2 + \text{ML}^2\right)$

$= \frac{4}{5}\text{MR}^2 + \frac{4}{5}\text{MR}^2 + 2\text{ML}^2$

$\text I = \frac{8}{5}\text{MR}^2 + 2\text{ML}^2$

Hence, the moment of inertia of the system about an axis passing through any one of the sides is $\frac{8}{5}\text{MR}^2 + 2\text{ML}^2$.

Let each brick be of length L and mass m, and let the left edge of the lowest brick be at the origin. From the figure, each brick is displaced towards the right relative to the one below it by L/2, L/4 and L/6 respectively.

The positions of the left edges of the four bricks are therefore

$0,\quad \frac{\text L}{2},\quad \frac{\text L}{2} + \frac{\text L}{4} = \frac{3\text L}{4},\quad \frac{3\text L}{4} + \frac{\text L}{6} = \frac{11\text L}{12}$

Since the bricks are identical and uniform, the centre of mass of each lies at its own mid-point, that is, at a distance L/2 from its left edge. Hence the x-coordinates of the centres of mass of the four bricks are

$\frac{\text L}{2},\quad \text L,\quad \frac{5\text L}{4},\quad \frac{17\text L}{12}$

All four bricks are identical, so the centre of mass of the system is the simple average of these coordinates,

$$\begin{aligned}\text x_{cm} = \frac{\text m \times \frac{\text L}{2}+\text m \times \text L + \text m \times \frac{5\text L}{4} + \text m \times \frac{17\text L}{12} }{\text m +\text m+ \text m+ \text m}\\[1em] \text x_{cm} = \frac{1}{4}\left(\frac{\text L}{2} + \text L + \frac{5\text L}{4} + \frac{17\text L}{12}\right)\end{aligned}$$

Taking the L.C.M. as 12,

$= \frac{1}{4}\left(\frac{6\text L + 12\text L + 15\text L + 17\text L}{12}\right) = \frac{1}{4} \times \frac{50\text L}{12}$

$\text x_{cm} = \frac{25\text L}{24}$

Hence, the x-coordinate of the centre of mass of the identical bricks is $\frac{25\text L}{24}$ from the left edge of the lowest brick.

Given,

• Length of the bar = 6a, where a = 1 m
• Mass of the bar = 8m, where m = 1 kg
• Point mass 2m strikes at a distance a from the centre with speed v
• Point mass m strikes at a distance 2a from the centre with speed 2v, in the opposite direction
• v = 1 m s^-1
• The table is smooth and both point masses stick to the bar after collision

Take the centre C of the bar as the origin, with the bar along the X-axis. The mass 2m strikes at x = − a and the mass m at x = + 2a, their velocities being perpendicular to the bar and oppositely directed.

The table is smooth, so no external force and no external torque acts on the system in the horizontal plane. Hence both the linear momentum and the angular momentum of the system are conserved during the collision.

(i) The point about which the rotation will take place

The rotation takes place about the centre of mass of the combined system. The total mass is 8m + 2m + m = 11m, and

$\text x_{cm} = \frac{(8\text m)(0) + (2\text m)(-\text a) + (\text m)(2\text a)}{11\text m}$

$= \frac{-2\text{am} + 2\text{am}}{11\text m} = 0$

Hence, the rotation takes place about C, the centre of the bar itself, because the moments of the two point masses about C are equal and opposite.

(ii) Velocity of the centre of mass after collision

The momenta of the two point masses perpendicular to the bar are

$(2\text m)(\text v) = 2\text{mv} \quad \text{and} \quad (\text m)(2\text v) = 2\text{mv}$

These are equal in magnitude but opposite in direction, so by the conservation of linear momentum the total momentum of the system is

$\text p = -2\text{mv} + 2\text{mv} = 0$

$\text v_{cm} = \frac{\text p}{11\text m} = 0$

Hence, the velocity of the centre of mass after collision is zero, that is, the system does not translate at all and only rotates about C.

(iii) Angular velocity of rotation

The angular momentum of the system about C just before the collision is the sum of the moments of the linear momenta of the two point masses. Both turn the bar in the same sense, since they strike on opposite sides of C moving in opposite directions,

$\text L = (2\text m)(\text v)(\text a) + (\text m)(2\text v)(2\text a)$

$= 2\text{mva} + 4\text{mva} = 6\text{mva}$

The moment of inertia of the system about C after the collision is that of the bar together with the two embedded point masses,

$\text I = \frac{1}{12}(8\text m)(6\text a)^2 + (2\text m)(\text a)^2 + (\text m)(2\text a)^2$

$= 24\text{ma}^2 + 2\text{ma}^2 + 4\text{ma}^2 = 30\text{ma}^2$

By the conservation of angular momentum,

$\omega = \frac{\text L}{\text I} = \frac{6\text{mva}}{30\text{ma}^2} = \frac{\text v}{5\text a}$

Substituting v = 1 m s^-1 and a = 1 m,

$\omega = \frac{1}{5 \times 1} = 0.2\ \text{rad s}^{-1}$

(iv) Total kinetic energy of the system

Since the velocity of the centre of mass is zero, the system possesses only rotational kinetic energy,

$\text K = \frac{1}{2}\text I \omega^2 = \frac{1}{2}(30\text{ma}^2)\left(\frac{\text v}{5\text a}\right)^2$

$= \frac{1}{2} \times 30\text{ma}^2 \times \frac{\text v^2}{25\text a^2} = \frac{3}{5}\text{mv}^2$

Substituting m = 1 kg and v = 1 m s^-1,

$\text K = \frac{3}{5} \times 1 \times (1)^2 = 0.6\ \text J$

Hence, the rotation takes place about the centre C of the bar, the velocity of the centre of mass is zero, the angular velocity is 0.2 rad s^-1 and the total kinetic energy of the system is 0.6 J.

Note: The book gives the length of the bar as a, which is not correct. A bar of length a reaches only $\frac{\text a}{2}$ on each side of C, so the masses could not strike it at a and 2a. The correct length is 6a, as the figure itself shows, since 2a + a + 2a + a = 6a.

Given,

• Mass of star A, m_A = 2.2 M_s
• Mass of star B, m_B = 11 M_s
• Separation between the stars = d

Both the stars rotate about their common centre of mass with the same angular velocity ω. Let r_A and r_B be their distances from the centre of mass. By the definition of the centre of mass,

$\text m_A \text r_A = \text m_B \text r_B \quad \text{and} \quad \text r_A + \text r_B = \text d$

Therefore,

$\text r_A = \frac{\text m_B}{\text m_A + \text m_B}\text d = \frac{11}{13.2}\text d, \qquad \text r_B = \frac{\text m_A}{\text m_A + \text m_B}\text d = \frac{2.2}{13.2}\text d$

The angular momentum of each star about the centre of mass is L = mr²ω. Hence

$\frac{\text L_A}{\text L_B} = \frac{\text m_A \text r_A^2 \omega}{\text m_B \text r_B^2 \omega} = \frac{2.2 \times (11)^2}{11 \times (2.2)^2}$

$= \frac{2.2 \times 121}{11 \times 4.84} = \frac{266.2}{53.24} = 5$

So L_A = 5 L_B. The total angular momentum of the binary star is

$\text L = \text L_A + \text L_B = 5\text L_B + \text L_B = 6\text L_B$

$\frac{\text L}{\text L_B} = 6$

Hence, the ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is 6 : 1.

Given,

• Mass of the ring, m = 2 kg
• Radius of the ring, R = 0.5 m
• Force applied by the stick, N = 2 N
• Acceleration of the ring, a = 0.3 m s^-2
• Coefficient of friction between the stick and the ring, $\mu = \frac{\text P}{10}$

From the figure the stick is vertical and touches the ring at its side, that is, at the end of the horizontal diameter, level with the centre O. This fixes the direction of every force acting on the ring,

• the normal force N applied by the stick is horizontal, and its line of action passes through the centre O, so it exerts no torque about O,
• the force of friction f₁ between the stick and the ring acts vertically downwards along the stick, at a perpendicular distance R from O. As the ring rolls forward its surface at the stick sweeps upwards, and friction opposes this,
• the force of friction f₂ from the ground acts horizontally backwards at the point of contact, at a perpendicular distance R from O.

The moment of inertia of a ring about its own axis is I = mR², and for rolling without slipping the angular acceleration is $\alpha = \frac{\text a}{\text R}$.

Equation of translational motion

Taking the direction of motion as positive, the stick pushes the ring forward while the ground friction retards it,

$\text N - \text f_2 = \text{ma} \qquad \dots(\text i)$

Equation of rotational motion

Taking moments about the centre O, and taking the sense in which the ring rolls as positive,

• N gives no moment, since its line of action passes through O,
• f₂ acts backwards at the bottom of the ring, so its moment f₂R is in the same sense as the rolling,
• f₁ acts downwards at the side of the ring, so its moment f₁R is in the opposite sense.

Therefore,

$\text f_2 \text R - \text f_1 \text R = \text I \alpha = \text{mR}^2 \times \frac{\text a}{\text R} = \text{mRa}$

Dividing throughout by R,

$\text f_2 - \text f_1 = \text{ma} \qquad \dots(\text{ii})$

Solving the two equations

Adding equations (i) and (ii), the unknown ground friction f₂ cancels out,

$(\text N - \text f_2) + (\text f_2 - \text f_1) = \text{ma} + \text{ma}$

$\text N - \text f_1 = 2\text{ma}$

Substituting the values,

$2 - \text f_1 = 2 \times 2 \times 0.3 = 1.2$

$\text f_1 = 2 - 1.2 = 0.8\ \text N$

The friction at the stick is limiting, so f₁ = μN,

$\mu = \frac{\text f_1}{\text N} = \frac{0.8}{2} = 0.4$

$\frac{\text P}{10} = 0.4 \quad \Rightarrow \quad \text P = 4$

Hence, the value of P is 4.

Given,

• Diameter of each sphere = $\sqrt{5}$ cm, so radius $\text R = \frac{\sqrt{5}}{2}$ cm and $\text R^2 = \frac{5}{4}$ cm²
• Mass of each sphere, M = 0.5 kg
• Side of the square, L = 4 cm

The axis of rotation is a diagonal of the square. Two spheres have their centres on this diagonal and the other two lie at a perpendicular distance equal to half the diagonal from it,

$\text d = \frac{\text L}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\ \text{cm}$

For the two spheres on the diagonal :

$$\begin{aligned}\text I_1 = \frac{2}{5}\text{MR}^2 = \frac{2}{5} \times 0.5 \times \frac{5}{4} = 0.25\ \text{kg cm}^2 \\ \text{each}\end{aligned}$$

For the two spheres off the diagonal : by the theorem of parallel axes,

$$\begin{aligned}\text I_2 = \frac{2}{5}\text{MR}^2 + \text{Md}^2 = 0.25 + 0.5 \times (2\sqrt{2})^2 \\[1em] = 0.25 + 0.5 \times 8 = 4.25\ \text{kg cm}^2 \\ \text{each}\end{aligned}$$

The total moment of inertia of the system about the diagonal is

$\text I = 2(0.25) + 2(4.25) = 0.5 + 8.5 = 9\ \text{kg cm}^2$

Converting to S.I. units, 1 cm² = 10^-4 m²,

$\text I = 9 \times 10^{-4}\ \text{kg m}^2$

Comparing with I = N × 10^-4 kg m²,

$\text N = 9$

Hence, the value of N is 9.

Let σ be the mass per unit area of the lamina. The bigger disc has radius 2R and the disc removed has diameter 2R, that is, radius R.

• Mass of the bigger disc, $\text M = \sigma \pi (2\text R)^2 = 4\sigma \pi \text R^2$
• Mass of the removed disc, $\text m = \sigma \pi \text R^2$

From the figure, the centre of the removed disc is at a distance R from O, and P lies on the rim of the bigger disc at a distance 2R from O, along a perpendicular direction.

Moment of inertia about the axis through O : The removed disc is treated as a negative mass. Using the theorem of parallel axes for it,

$\text I_O = \frac{1}{2}\text M(2\text R)^2 - \left[\frac{1}{2}\text{mR}^2 + \text{mR}^2\right]$

$= \frac{1}{2}(4\sigma \pi \text R^2)(4\text R^2) - \frac{3}{2}(\sigma \pi \text R^2)\text R^2$

$= 8\sigma \pi \text R^4 - 1.5\sigma \pi \text R^4 = 6.5\sigma \pi \text R^4$

Moment of inertia about the axis through P : The distance of the centre of the removed disc from P is

$\sqrt{(2\text R)^2 + \text R^2} = \sqrt{5}\text R$

Therefore,

$\text I_P = \left[\frac{1}{2}\text M(2\text R)^2 + \text M(2\text R)^2\right] - \left[\frac{1}{2}\text{mR}^2 + \text m(\sqrt{5}\text R)^2\right]$

$= [8\sigma \pi \text R^4 + 16\sigma \pi \text R^4] - [0.5\sigma \pi \text R^4 + 5\sigma \pi \text R^4]$

$= 24\sigma \pi \text R^4 - 5.5\sigma \pi \text R^4 = 18.5\sigma \pi \text R^4$

Therefore the required ratio is

$\frac{\text I_P}{\text I_O} = \frac{18.5\sigma \pi \text R^4}{6.5\sigma \pi \text R^4} = 2.85$

Hence, the ratio $\frac{\text I_P}{\text I_O}$ to the nearest integer is 3.

Given,

• Mass of the disc, M = 50 kg
• Radius of the disc, R = 0.4 m
• Initial angular velocity of the disc, ω₁ = 10 rad s^-1
• Mass of each ring, m = 6.25 kg
• Radius of each ring, r = 0.2 m

Moment of inertia of the disc about its own axis :

$\text I_d = \frac{1}{2}\text{MR}^2 = \frac{1}{2} \times 50 \times (0.4)^2 = 4\ \text{kg m}^2$

Moment of inertia of each ring about the axis of the disc : The two rings touch each other along the axis of the disc, so the centre of each ring is at a distance r = 0.2 m from that axis. The rings are horizontal, so the axis of the disc is perpendicular to the plane of each ring. By the theorem of parallel axes,

$$\begin{aligned}\text I_r = \text{mr}^2 + \text{mr}^2 = 2\text{mr}^2 \\[1em] = 2 \times 6.25 \times (0.2)^2 = 0.5\ \text{kg m}^2\end{aligned}$$

Hence for the two rings together, the moment of inertia is 2 × 0.5 = 1 kg m².

Total moment of inertia of the system :

$\text I_2 = 4 + 1 = 5\ \text{kg m}^2$

The rings are placed gently, so no external torque acts about the axis and the angular momentum is conserved,

$\text I_d \omega_1 = \text I_2 \omega_2$

$\omega_2 = \frac{4 \times 10}{5} = 8\ \text{rad s}^{-1}$

Hence, the new angular velocity of the system is 8 rad s^-1.

Given,

• Radius of the platform, R = 0.5 m
• Mass of the platform, M = 0.45 kg
• Mass of each ball, m = 0.05 kg
• Distance of each gun from the centre, r = 0.25 m
• Speed of each ball with respect to the ground, v = 9 m s^-1

Moment of inertia of the platform about its axis :

$$\begin{aligned}\text I = \frac{1}{2}\text{MR}^2 = \frac{1}{2} \times 0.45 \times (0.5)^2 \\[1em] = \frac{1}{2} \times 0.45 \times 0.25 = 0.05625\ \text{kg m}^2\end{aligned}$$

The balls are fired horizontally and perpendicular to the diameter, in opposite directions from the two ends. Hence the angular momentum carried away by each ball is in the same sense, and their total angular momentum about the axis is

$\text L = 2(\text{mvr}) = 2 \times 0.05 \times 9 \times 0.25 = 0.225\ \text{kg m}^2\text s^{-1}$

No external torque acts about the vertical axis, so the angular momentum of the whole system is conserved. Initially the system is at rest, so the platform must acquire an equal and opposite angular momentum,

$\text I \omega = \text L$

$\omega = \frac{0.225}{0.05625} = 4\ \text{rad s}^{-1}$

Hence, the rotational speed of the platform after the balls leave it is 4 rad s^-1.

Given,

• Mass of the disc, M = 1.5 kg
• Radius of the disc, r = 0.5 m
• Magnitude of each force, F = 0.5 N
• Time, t = 1 s

The three forces act along the sides of an equilateral triangle XYZ whose vertices lie on the rim of the disc. The perpendicular distance of each side from the centre of the disc is the apothem (AO) of the equilateral triangle inscribed in a circle of radius r,

$\text {AO} = \text d = \text r \sin 30^\circ = \frac{\text r}{2} = \frac{0.5}{2} = 0.25\ \text m$

All three forces tend to turn the disc in the same sense, so the total torque about the centre is

$\tau = 3 \times \text F \times \text d = 3 \times 0.5 \times 0.25 = 0.375\ \text{N m}$

The moment of inertia of the disc about its own axis is

$\text I = \frac{1}{2}\text{Mr}^2 = \frac{1}{2} \times 1.5 \times (0.5)^2 = 0.1875\ \text{kg m}^2$

The angular acceleration produced is

$\alpha = \frac{\tau}{\text I} = \frac{0.375}{0.1875} = 2\ \text{rad s}^{-2}$

The disc is initially at rest, so ω₀ = 0. Using the equation of rotational kinematics,

$\omega = \omega_0 + \alpha \text t = 0 + 2 \times 1 = 2\ \text{rad s}^{-1}$

Hence, one second after applying the forces, the angular speed of the disc is 2 rad s^-1.

Given,

• $\rho_A(\text r) = \text k\left(\frac{\text r}{\text R}\right)$ and $\rho_B(\text r) = \text k\left(\frac{\text r}{\text R}\right)^5$
• Both spheres have the same radius R

Consider a thin spherical shell of radius r and thickness dr. Its volume is 4πr²dr, so its mass is

$\text{dm} = \rho(\text r)4\pi \text r^2\text{dr}$

The moment of inertia of a thin spherical shell about a diameter is $\frac{2}{3}\text r^2\text{dm}$. Hence for the whole sphere,

$\text I = \int_0^{\text R} \frac{2}{3}\text r^2\text{dm} = \frac{8\pi}{3}\int_0^{\text R} \rho(\text r)\text r^4\text{dr}$

For sphere A :

$\text I_A = \frac{8\pi}{3}\int_0^{\text R} \text k\frac{\text r}{\text R}\text r^4\text{dr} = \frac{8\pi \text k}{3\text R}\int_0^{\text R} \text r^5\text{dr}$

$= \frac{8\pi \text k}{3\text R}\left[\frac{\text r^6}{6}\right]_0^{\text R} = \frac{8\pi \text k}{3\text R} \times \frac{\text R^6}{6} = \frac{4\pi \text{kR}^5}{9}$

For sphere B :

$\text I_B = \frac{8\pi}{3}\int_0^{\text R} \text k\frac{\text r^5}{\text R^5}\text r^4\text{dr} = \frac{8\pi \text k}{3\text R^5}\int_0^{\text R} \text r^9\text{dr}$

$= \frac{8\pi \text k}{3\text R^5}\left[\frac{\text r^{10}}{10}\right]_0^{\text R} = \frac{8\pi \text k}{3\text R^5} \times \frac{\text R^{10}}{10} = \frac{4\pi \text{kR}^5}{15}$

Therefore,

$\frac{\text I_B}{\text I_A} = \frac{4\pi \text{kR}^5/15}{4\pi \text{kR}^5/9} = \frac{9}{15} = \frac{6}{10}$

Comparing with $\frac{\text I_B}{\text I_A} = \frac{\text n}{10}$,

$\text n = 6$

Hence, the value of n is 6.

Given,

• Linear speed at A, v₁ = 3 m/s
• Drop along AB, h₁ = 30 m
• Drop along CD, h₂ = 27 m
• g = 10 m/s²

The discs roll without slipping, so the total kinetic energy of each disc is

$\text K = \frac{1}{2}\text{mv}^2 + \frac{1}{2}\text I \omega^2 = \frac{1}{2}\text{mv}^2 + \frac{1}{2}\left(\frac{1}{2}\text{mR}^2\right)\left(\frac{\text v}{\text R}\right)^2$

$= \frac{1}{2}\text{mv}^2 + \frac{1}{4}\text{mv}^2 = \frac{3}{4}\text{mv}^2$

Applying the conservation of mechanical energy to each disc, and noting that both reach the bottom with the same final speed v,

For the disc on AB :

$\frac{3}{4}\text{mv}_1^2 + \text{mgh}_1 = \frac{3}{4}\text{mv}^2$

For the disc on CD :

$\frac{3}{4}\text{mv}_2^2 + \text{mgh}_2 = \frac{3}{4}\text{mv}^2$

Equating the two left-hand sides and cancelling m,

$\frac{3}{4}\text v_1^2 + \text{gh}_1 = \frac{3}{4}\text v_2^2 + \text{gh}_2$

Substituting the values,

$\frac{3}{4}(3)^2 + 10(30) = \frac{3}{4}\text v_2^2 + 10(27)$

$6.75 + 300 = 0.75\text v_2^2 + 270$

$0.75\text v_2^2 = 36.75 \quad \Rightarrow \quad \text v_2^2 = 49$

$\text v_2 = 7\ \text{m/s}$

Hence, the linear speed v₂ is 7 m/s.

Given,

• Angle of the inclined plane, θ = 60°
• Time difference, $\Delta \text t = \frac{2 - \sqrt{3}}{\sqrt{10}}$ s
• g = 10 m s^-2

For a body rolling without slipping down an incline, the acceleration of the centre of mass is

$\text a = \frac{\text g \sin \theta}{1 + \frac{\text k^2}{\text R^2}}$

For a ring, $\frac{\text k^2}{\text R^2} = 1$, so $\text a_{ring} = \frac{\text g \sin \theta}{2}$.

For a disc, $\frac{\text k^2}{\text R^2} = \frac{1}{2}$, so $\text a_{disc} = \frac{2}{3}\text g \sin \theta$.

If h is the height of the top of the plane, the length of the incline is $\text s = \frac{\text h}{\sin \theta}$. Both start from rest, so from $\text s = \frac{1}{2}\text{at}^2$,

$\text t = \sqrt{\frac{2\text s}{\text a}}$

For the ring :

$\text t_1 = \sqrt{\frac{2\text h/\sin \theta}{\text g \sin \theta/2}} = \sqrt{\frac{4\text h}{\text g \sin^2 \theta}} = \frac{2}{\sin \theta}\sqrt{\frac{\text h}{\text g}}$

For the disc :

$\text t_2 = \sqrt{\frac{2\text h/\sin \theta}{(2/3)\text g \sin \theta}} = \sqrt{\frac{3\text h}{\text g \sin^2 \theta}} = \frac{\sqrt{3}}{\sin \theta}\sqrt{\frac{\text h}{\text g}}$

Therefore the time difference is

$\Delta \text t = \text t_1 - \text t_2 = \frac{2 - \sqrt{3}}{\sin \theta}\sqrt{\frac{\text h}{\text g}}$

Equating this to the given value and putting $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

$\frac{2 - \sqrt{3}}{\sqrt{3}/2}\sqrt{\frac{\text h}{10}} = \frac{2 - \sqrt{3}}{\sqrt{10}}$

$\frac{2}{\sqrt{3}}\sqrt{\frac{\text h}{10}} = \frac{1}{\sqrt{10}} \quad \Rightarrow \quad \sqrt{\frac{\text h}{10}} = \frac{\sqrt{3}}{2\sqrt{10}}$

Squaring both sides,

$\frac{\text h}{10} = \frac{3}{40} \quad \Rightarrow \quad \text h = 0.75\ \text m$

Hence, the height of the top of the inclined plane is 0.75 m.

Given,

• Angle of the inclined plane, θ = 30°
• Length of the inclined plane, s = 20 m
• Mass of the sphere, m = 5 kg
• Loss of kinetic energy due to friction = 20%
• g = 9.8 m/s²

The vertical height through which the sphere descends is

$\text h = \text s \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\ \text m$

The loss in potential energy of the sphere is

$\text U = \text{mgh} = 5 \times 9.8 \times 10 = 490\ \text J$

Since 20% of the kinetic energy is lost due to friction, only 80% of this energy appears as the kinetic energy of the sphere at the bottom,

$\text K = \frac{80}{100} \times 490 = 392\ \text J$

For a solid sphere rolling without slipping, $\frac{\text k^2}{\text R^2} = \frac{2}{5}$, and the total kinetic energy is

$\text K = \frac{1}{2}\text{mv}^2\left(1 + \frac{\text k^2}{\text R^2}\right) = \frac{1}{2}\text{mv}^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}\text{mv}^2$

Therefore,

$\frac{7}{10} \times 5 \times \text v^2 = 392$

$\text v^2 = \frac{392 \times 10}{7 \times 5} = \frac{3920}{35} = 112$

$\text v = \sqrt{112} = 10.58\ \text{m/s}$

Hence, the velocity of the sphere at the bottom of the plane is 10.58 m/s.

Given,

• Radius of the wheel, R = 0.5 m
• Mass tied to the cord, m = 2 kg
• Distance fallen, s = 4.0 m
• Time taken, t = 10 s
• g = 9.8 m s^-2

The mass falls from rest, so u = 0. Using $\text s = \text{ut} + \frac{1}{2}\text{at}^2$,

$4.0 = 0 + \frac{1}{2}\text a (10)^2 = 50\text a$

$\text a = \frac{4.0}{50} = 0.08\ \text{m s}^{-2}$

Angular acceleration of the wheel : The cord does not slip on the rim, so

$\alpha = \frac{\text a}{\text R} = \frac{0.08}{0.5} = 0.16\ \text{rad s}^{-2}$

Moment of inertia of the wheel : For the falling mass, applying Newton's second law,

$\text{mg} - \text T = \text{ma}$

$\text T = \text m(\text g - \text a) = 2(9.8 - 0.08) = 2 \times 9.72 = 19.44\ \text N$

The tension acts tangentially at the rim, so the torque on the wheel is

$\tau = \text{TR} = 19.44 \times 0.5 = 9.72\ \text{N m}$

Using τ = Iα,

$\text I = \frac{\tau}{\alpha} = \frac{9.72}{0.16} = 60.75\ \text{kg m}^2$

Hence, the angular acceleration produced is 0.16 rad s^-2 and the moment of inertia of the wheel about the axle is 60.75 kg m².

Given,

• Point masses M₁ and M₂ at the ends of a rod of length l and negligible mass
• Angular velocity to be produced = ω

Let the axis of rotation pass through a point at a distance x from M₁, so that its distance from M₂ is (l − x). The moment of inertia of the system about this axis is

$\text I = \text M_1 \text x^2 + \text M_2 (l - \text x)^2$

The work required to set the rod rotating with angular velocity ω is equal to its kinetic energy of rotation,

$\text W = \frac{1}{2}\text I \omega^2$

For a given ω, the work W is minimum when the moment of inertia I is minimum. Differentiating I with respect to x,

$\frac{\text{dI}}{\text{dx}} = \frac{\text d}{\text{dx}}\left[\text M_1 \text x^2 + \text M_2 (l - \text x)^2\right]$

$= 2\text M_1 \text x + 2\text M_2 (l - \text x) \times (-1) = 2\text M_1 \text x - 2\text M_2 (l - \text x)$

For I to be minimum, $\frac{\text{dI}}{\text{dx}} = 0$,

$2\text M_1 \text x - 2\text M_2 (l - \text x) = 0$

$\text M_1 \text x = \text M_2 l - \text M_2 \text x \quad \Rightarrow \quad (\text M_1 + \text M_2)\text x = \text M_2 l$

$\text x = \frac{\text M_2 l}{\text M_1 + \text M_2}$

Differentiating once more, $\frac{\text d^2\text I}{\text{dx}^2} = 2\text M_1 + 2\text M_2$, which is positive, confirming that I is minimum at this value of x.

Hence, the axis of rotation should pass through a point at a distance $\frac{\text M_2 l}{\text M_1 + \text M_2}$ from M₁. This is precisely the centre of mass of the system, so the work required is minimum when the axis passes through the centre of mass.

(a) Yes, the sphere reaches the bottom with the same speed in each case.

Applying the conservation of mechanical energy to a body rolling without slipping down an incline of height h,

$\text{mgh} = \frac{1}{2}\text{mv}^2\left(1 + \frac{\text k^2}{\text R^2}\right)$

$\text v = \sqrt{\frac{2\text{gh}}{1 + \frac{\text k^2}{\text R^2}}}$

For a solid sphere $\frac{\text k^2}{\text R^2} = \frac{2}{5}$, so $\text v = \sqrt{\frac{10\text{gh}}{7}}$. This depends only on the height h and not on the angle of inclination. Since both planes are of the same height, the sphere reaches the bottom with the same speed in each case.

(b) Yes, it will take longer to roll down one plane than the other.

The acceleration of the centre of mass is

$\text a = \frac{\text g \sin \theta}{1 + \frac{\text k^2}{\text R^2}} = \frac{5}{7}\text g \sin \theta$

and the length of the incline is $\text s = \frac{\text h}{\sin \theta}$. From $\text s = \frac{1}{2}\text{at}^2$,

$\text t = \sqrt{\frac{2\text s}{\text a}} = \sqrt{\frac{14\text h}{5\text g \sin^2 \theta}} \quad \Rightarrow \quad \text t \propto \frac{1}{\sin \theta}$

(c) The sphere takes longer on the plane of smaller inclination.

As θ decreases, sin θ decreases, so the time of descent t increases. Physically, the plane of smaller inclination is longer and gives a smaller acceleration, so the sphere takes more time to roll down it.

Given,

• Angle of inclination, θ = 30°
• Speed of the centre of mass at the bottom, v = 5 m s^-1
• g = 9.8 m s^-2

(a) Distance travelled up the plane : For a solid cylinder rolling without slipping, $\frac{\text k^2}{\text R^2} = \frac{1}{2}$, so the total kinetic energy at the bottom is

$\text K = \frac{1}{2}\text{mv}^2\left(1 + \frac{1}{2}\right) = \frac{3}{4}\text{mv}^2$

At the highest point reached, all this kinetic energy is converted into potential energy,

$\frac{3}{4}\text{mv}^2 = \text{mgh} \quad \Rightarrow \quad \text h = \frac{3\text v^2}{4\text g}$

$\text h = \frac{3 \times (5)^2}{4 \times 9.8} = \frac{75}{39.2} = 1.913\ \text m$

The distance travelled along the plane is

$\text s = \frac{\text h}{\sin \theta} = \frac{1.913}{\sin 30^\circ} = \frac{1.913}{0.5} = 3.83\ \text m$

Hence, the cylinder goes 3.83 m up the plane.

(b) Time taken to return to the bottom : The retardation while going up is

$\text a = \frac{\text g \sin \theta}{1 + \frac{\text k^2}{\text R^2}} = \frac{2}{3}\text g \sin \theta = \frac{2}{3} \times 9.8 \times 0.5 = 3.27\ \text{m s}^{-2}$

The time taken to come to rest while going up is

$\text t = \frac{\text v}{\text a} = \frac{5}{3.27} = 1.53\ \text s$

The cylinder rolls back down the same distance with the same magnitude of acceleration, so it takes an equal time to return.

t_return​ = 2t_up​ = 2 × 1.53 = 3.06 s

Hence, the cylinder takes 3.06 s to return to the bottom.

Given,

• Length of each side of the ladder, AB = AC = 1.6 m
• Length of the rope, DE = 0.5 m, tied half way up
• Suspended weight = 40 kg
• Distance of F from B along BA, BF = 1.2 m
• The floor is frictionless and the weight of the ladder is neglected
• g = 9.8 m s^-2

The weight suspended at F is

$\text W = 40 \times 9.8 = 392\ \text N$

Geometry of the ladder

The rope is tied half way up, so D and E are the mid-points of AB and AC,

$\text{AD} = \text{AE} = \frac{1.6}{2} = 0.8\ \text m$

In triangles ADE and ABC the angle at A is common and

$\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{0.8}{1.6} = \frac{1}{2}$

so the two triangles are similar in the ratio 1 : 2. Hence

$\text{BC} = 2 \times \text{DE} = 2 \times 0.5 = 1.0\ \text m$

Let M be the mid-point of BC. Since AB = AC, the perpendicular from A meets BC at M, so

$\text{BM} = \text{MC} = 0.5\ \text m$

From the right-angled triangle ABM, the height of A above the floor is

$$\begin{aligned}\text{AM} = \sqrt{\text{AB}^2 - \text{BM}^2} = \sqrt{(1.6)^2 - (0.5)^2} \\[1em] = \sqrt{2.56 - 0.25} = \sqrt{2.31} = 1.52\ \text m\end{aligned}$$

Since E is the mid-point of AC, it lies at half this height, so the vertical depth of E below A is AH =

$\frac{1.52}{2} = 0.76\ \text m$

The point F divides BA in the ratio 1.2 : 1.6, so by similar triangles BPF and BMA, its horizontal distance from B is

$$\begin{aligned}\frac{\text{BF}}{\text{BA}} = \frac{\text{BP}}{\text{BM}}\\[1em] \text{BP}=\frac{\text{BF}}{\text{BA}}\times\text{BM}\\[1em] \frac{1.2}{1.6} \times 0.5 = 0.375\ \text m\end{aligned}$$

Forces exerted by the floor

The floor is frictionless, so the reactions N_B and N_C at B and C are purely vertical, having no horizontal component.

Applying the condition of translational equilibrium to the whole ladder,

$\text N_B + \text N_C = \text W = 392\ \text N \qquad \dots(\text i)$

Applying the condition of rotational equilibrium, and taking moments of all the forces about B for the whole ladder, the rope tension is internal to the system and contributes nothing,

$\text N_C \times \text{BC} = \text W \times \text {BP}$

$\text N_C \times 1.0 = 392 \times 0.375 = 147$

$\text N_C = 147\ \text N$

Substituting this value in equation (i),

$\text N_B = 392 - 147 = 245\ \text N$

Tension in the rope

Consider the side AC alone, which carries no suspended load. The forces acting on it are the reaction N_C at C, the horizontal pull T of the rope at E directed towards D, and the reaction at the hinge A.

Taking moments about A for this side, the hinge reaction passes through A and so exerts no moment,

$\text N_C \times \text{MC} = \text T \times \text{AH}$

$147 \times 0.5 = \text T \times 0.76$

$\text T = \frac{73.5}{0.76} = 96.7\ \text N$

Hence, the tension in the rope is 96.7 N, and the forces exerted by the floor on the ladder are 245 N at B and 147 N at C, both acting vertically upwards.

The reaction at B is the larger of the two because the weight hangs from the side BA, so its line of action lies nearer to B than to C. Note also that the tension can be found only by isolating one side of the ladder, since for the ladder as a whole the rope tension is an internal force and cancels out.

Given,

• Mass held in each hand, m = 5 kg
• Initial angular speed, ω₁ = 30 rev/min
• Initial distance of each weight from the axis, r₁ = 90 cm = 0.9 m
• Final distance of each weight from the axis, r₂ = 20 cm = 0.2 m
• Moment of inertia of the man with the platform, I₀ = 7.6 kg m²

(a) New angular speed : The initial moment of inertia of the whole system is

$$\begin{aligned}\text I_1 = \text I_0 + 2\text{mr}_1^2 = 7.6 + 2 \times 5 \times (0.9)^2 \\[1em] = 7.6 + 8.1 = 15.7\ \text{kg m}^2\end{aligned}$$

The final moment of inertia is

$$\begin{aligned}\text I_2 = \text I_0 + 2\text{mr}_2^2 = 7.6 + 2 \times 5 \times (0.2)^2 \\[1em] = 7.6 + 0.4 = 8.0\ \text{kg m}^2\end{aligned}$$

Friction is neglected, so no external torque acts and the angular momentum is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

$\omega_2 = \frac{15.7 \times 30}{8.0} = 58.88\ \text{rev/min}$

Hence, the new angular speed is 58.88 rev/min.

(b) No, kinetic energy is not conserved in this process.

$\frac{\text K_2}{\text K_1} = \frac{\frac{1}{2}\text I_2 \omega_2^2}{\frac{1}{2}\text I_1 \omega_1^2} = \frac{8.0 \times (58.88)^2}{15.7 \times (30)^2} = 1.96$

The kinetic energy of rotation has almost doubled. The increase comes from the work done by the man, who has to pull the weights inwards against the centrifugal effect. This work is done at the expense of his own internal (chemical) energy and appears as the additional kinetic energy of rotation.

Given,

• Mass of the bullet, m = 10 g = 0.01 kg
• Speed of the bullet, v = 500 m/s
• Width of the door, L = 1.0 m
• Mass of the door, M = 12 kg

The bullet strikes the door at its centre, so its perpendicular distance from the hinge is

$\text r = \frac{\text L}{2} = 0.5\ \text m$

Angular momentum of the bullet about the hinge before impact :

$\text L_i = \text{mvr} = 0.01 \times 500 \times 0.5 = 2.5\ \text{kg m}^2\text s^{-1}$

Moment of inertia after the bullet embeds : The door rotates about a vertical axis at one end, so

$\text I_{door} = \frac{1}{3}\text{ML}^2 = \frac{1}{3} \times 12 \times (1.0)^2 = 4\ \text{kg m}^2$

The embedded bullet contributes

$\text I_{bullet} = \text{mr}^2 = 0.01 \times (0.5)^2 = 0.0025\ \text{kg m}^2$

$\text I = 4 + 0.0025 = 4.0025\ \text{kg m}^2$

The hinge exerts no torque about the axis of rotation, so the angular momentum is conserved,

$\text L_i = \text I \omega$

$\omega = \frac{2.5}{4.0025} = 0.625\ \text{rad s}^{-1}$

Hence, the angular speed of the door just after the bullet embeds into it is 0.625 rad s^-1.

Given,

• Moments of inertia of the two discs = I₁ and I₂
• Their angular speeds = ω₁ and ω₂, with ω₁ ≠ ω₂

(a) Angular speed of the two-disc system : When the discs are brought into contact face to face with their axes coincident, the frictional forces between them are internal to the system and no external torque acts about the common axis. Hence the angular momentum is conserved,

$\text I_1 \omega_1 + \text I_2 \omega_2 = (\text I_1 + \text I_2)\omega$

$\omega = \frac{\text I_1 \omega_1 + \text I_2 \omega_2}{\text I_1 + \text I_2}$

(b) Comparison of the kinetic energies : The initial kinetic energy of the two discs is

$\text K_i = \frac{1}{2}\text I_1 \omega_1^2 + \frac{1}{2}\text I_2 \omega_2^2$

The final kinetic energy of the combined system is

$\text K_f = \frac{1}{2}(\text I_1 + \text I_2)\omega^2 = \frac{1}{2}\frac{(\text I_1 \omega_1 + \text I_2 \omega_2)^2}{\text I_1 + \text I_2}$

Therefore the change in kinetic energy is

$\text K_i - \text K_f = \frac{1}{2}\left[\text I_1 \omega_1^2 + \text I_2 \omega_2^2 - \frac{(\text I_1 \omega_1 + \text I_2 \omega_2)^2}{\text I_1 + \text I_2}\right]$

Taking the L.C.M. and simplifying the numerator,

$= \frac{1}{2}\left[\frac{(\text I_1 + \text I_2)(\text I_1 \omega_1^2 + \text I_2 \omega_2^2) - (\text I_1 \omega_1 + \text I_2 \omega_2)^2}{\text I_1 + \text I_2}\right]$

$= \frac{1}{2}\left[\frac{\text I_1 \text I_2 \omega_1^2 + \text I_1 \text I_2 \omega_2^2 - 2\text I_1 \text I_2 \omega_1 \omega_2}{\text I_1 + \text I_2}\right]$

$\text K_i - \text K_f = \frac{\text I_1 \text I_2 (\omega_1 - \omega_2)^2}{2(\text I_1 + \text I_2)}$

Since I₁, I₂ and (ω₁ − ω₂)² are all positive quantities and ω₁ ≠ ω₂, this expression is positive.

$\text K_i - \text K_f \gt 0 \quad \Rightarrow \quad \text K_f \lt \text K_i$

Hence, the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs.

Accounting for the loss : When the two discs are brought into contact, their surfaces initially move with different angular speeds, so there is relative sliding between them. Work is done against the force of friction at the contact surfaces, and this energy is dissipated as heat. This accounts for the loss in kinetic energy, while the angular momentum, being unaffected by the internal forces, remains conserved.

Given,

• Angular speed of the disc = ω₀
• Radius of the disc = R
• The table is perfectly frictionless

(a) Linear velocities of A, B and C : The disc is placed without any translational push, so the velocity of its centre of mass is zero and every point has only the tangential velocity due to rotation, v_t = rω₀.

From the figure, A is at the top of the rim, B at the bottom of the rim and C is at a distance R/2 from the centre. Therefore

$\text v_A = \omega_0 \text R, \qquad \text v_B = \omega_0 \text R, \qquad \text v_C = \frac{\omega_0 \text R}{2}$

The velocities at A and B are equal in magnitude but opposite in direction, since A and B lie at the two ends of a diameter. The velocity at C is in the same direction as that at A.

(b) No, the disc will not roll in the direction indicated. Rolling requires the centre of mass to move forward. Since the table is frictionless, no horizontal force acts on the disc, so its centre of mass remains at rest and the disc merely spins about its own axis.

(c) Friction is necessary because it is the only external horizontal force that can act on the disc at the point of contact. This force produces the linear acceleration of the centre of mass, converting the pure spinning into rolling motion. Without friction there is no external force, so the centre of mass cannot be accelerated.

(d) At B the surface of the disc moves backward relative to the table, so the force of friction acts forward, that is, in the direction of the indicated rolling. This frictional force acts at the point of contact and exerts a torque about the centre which is opposite in sense to ω₀, so it decreases the angular speed while increasing the linear speed of the centre of mass.

(e) The force of friction after perfect rolling begins is zero. Once v_cm = ω R, the point of contact is instantaneously at rest relative to the surface, so there is no relative sliding and hence no force of friction is required to maintain the rolling.

Given,

• Radius of both the disc and the ring, R = 10 cm = 0.1 m
• Initial angular speed, ω₀ = 10 π rad s^-1
• Coefficient of kinetic friction, μ_k = 0.2
• g = 9.8 m s^-2

The initial situation

Each body is placed on the table spinning about its own axis but without any translational push, so initially

$\omega = \omega_0 \quad \text{and} \quad \text v_{cm} = 0$

The centre of mass is at rest while the body spins, so the point of contact is not at rest relative to the table. Its velocity is ω₀R directed backwards, and the body therefore slides on the table. Since there is relative sliding, the friction acting is kinetic friction, and it acts forwards, opposing the backward slipping of the point of contact.

Effect of friction on the translational motion

The force of kinetic friction is

$\text f = \mu_k \text N = \mu_k \text{mg}$

Applying Newton's second law to the centre of mass,

$\text a = \frac{\text f}{\text m} = \mu_k \text g$

The body starts from rest translationally, so at any time t its linear speed is

$\text v = \mu_k \text{gt} \qquad \dots(\text i)$

Effect of friction on the rotational motion

The same frictional force acts at the point of contact, at a perpendicular distance R from the centre, and exerts a torque about the centre which opposes the rotation,

$\tau = \text{fR} = \mu_k \text{mgR}$

If I = mk² is the moment of inertia about the central axis, the angular retardation produced is

$\alpha = \frac{\tau}{\text I} = \frac{\mu_k \text{mgR}}{\text{mk}^2} = \frac{\mu_k \text{gR}}{\text k^2}$

Hence the angular speed decreases with time as

$\omega = \omega_0 - \frac{\mu_k \text{gR}}{\text k^2}\text t \qquad \dots(\text{ii})$

Thus friction does two things at the same time, it increases the linear speed of the centre of mass and decreases the angular speed of the body.

Condition for rolling to begin

Rolling without slipping begins at the instant the point of contact comes momentarily to rest, that is, when

$\text v = \omega \text R$

Substituting from equations (i) and (ii),

$$\begin{aligned}\mu_k \text{gt} = \left(\omega_0 - \frac{\mu_k \text{gR}}{\text k^2}\text t\right)\text R \\[1em] = \omega_0 \text R - \frac{\mu_k \text{gR}^2}{\text k^2}\text t\end{aligned}$$

Collecting the terms containing t on the left hand side,

$\mu_k \text{gt}\left(1 + \frac{\text R^2}{\text k^2}\right) = \omega_0 \text R$

$\text t = \frac{\omega_0 \text R}{\mu_k \text g\left(1 + \frac{\text R^2}{\text k^2}\right)} \qquad \dots(\text{iii})$

The mass m cancels out, so the time taken does not depend upon the mass of the body.

For the solid disc

For a disc $\text I = \frac{1}{2}\text{mR}^2$, so $\frac{\text k^2}{\text R^2} = \frac{1}{2}$ and $\frac{\text R^2}{\text k^2} = 2$. Substituting in equation (iii),

$\text t_{disc} = \frac{\omega_0 \text R}{\mu_k \text g(1 + 2)} = \frac{\omega_0 \text R}{3\mu_k \text g}$

Substituting the values,

$\text t_{disc} = \frac{10\pi \times 0.1}{3 \times 0.2 \times 9.8} = \frac{3.14}{5.88}$

$\text t_{disc} = 0.53\ \text s$

For the ring

For a ring I = mR², so $\frac{\text k^2}{\text R^2} = 1$ and $\frac{\text R^2}{\text k^2} = 1$. Substituting in equation (iii),

$\text t_{ring} = \frac{\omega_0 \text R}{\mu_k \text g(1 + 1)} = \frac{\omega_0 \text R}{2\mu_k \text g}$

Substituting the values,

$\text t_{ring} = \frac{10\pi \times 0.1}{2 \times 0.2 \times 9.8} = \frac{3.14}{3.92}$

$\text t_{ring} = 0.80\ \text s$

Hence, the solid disc will start to roll earlier, taking 0.53 s as against 0.80 s for the ring.

Given,

• Mass of the cylinder, M = 10 kg
• Radius of the cylinder, R = 15 cm = 0.15 m
• Angle of inclination, θ = 30°
• Coefficient of static friction, μ_s = 0.25
• g = 9.8 m s^-2

(a) Force of friction acting on the cylinder

The cylinder rolls down the plane without slipping. The forces acting on it are its weight Mg, the normal reaction N and the force of static friction f_s acting up the plane at the point of contact.

Resolving the weight along and perpendicular to the plane, and applying Newton's second law along the plane,

$\text{Mg}\sin \theta - \text f_s = \text{Ma} \qquad \dots(\text i)$

Perpendicular to the plane there is no acceleration, so

$\text N = \text{Mg}\cos \theta \qquad \dots(\text{ii})$

The weight acts at the centre of mass and the normal reaction passes through it, so neither exerts a torque about the centre. The torque is due to the frictional force alone, acting at a perpendicular distance R,

$\tau = \text f_s \text R = \text I \alpha$

For a solid cylinder $\text I = \frac{1}{2}\text{MR}^2$, and for rolling without slipping $\alpha = \frac{\text a}{\text R}$. Therefore

$\text f_s \text R = \frac{1}{2}\text{MR}^2 \times \frac{\text a}{\text R} = \frac{1}{2}\text{MRa}$

$\text f_s = \frac{1}{2}\text{Ma} \qquad \dots(\text{iii})$

Substituting this value of f_s in equation (i),

$\text{Mg}\sin \theta - \frac{1}{2}\text{Ma} = \text{Ma}$

$\text{Mg}\sin \theta = \frac{3}{2}\text{Ma} \quad \Rightarrow \quad \text a = \frac{2}{3}\text g \sin \theta$

Putting this value of a back in equation (iii),

$\text f_s = \frac{1}{2}\text M\left(\frac{2}{3}\text g \sin \theta\right) = \frac{1}{3}\text{Mg}\sin \theta$

Substituting the given values,

$$\begin{aligned}\text f_s = \frac{1}{3} \times 10 \times 9.8 \times \sin 30 ^{\circ}\\[1em] = \frac{1}{3} \times 10 \times 9.8 \times 0.5 = \frac{49}{3}\end{aligned}$$

$\text f_s = 16.33\ \text N$

Hence, the force of friction acting on the cylinder is 16.33 N, directed up the plane.

(b) Work done against friction during rolling

In rolling without slipping the point of contact of the cylinder is momentarily at rest relative to the surface, so there is no relative motion between the body and the plane at the point of contact.

The frictional force acts at this point of contact, and since its point of application undergoes no displacement relative to the surface, the force of friction does no work,

$\text W = 0$

Hence, the work done against friction during rolling is zero, and no mechanical energy is dissipated as heat. This is why the whole of the loss in potential energy appears as the kinetic energy of the rolling cylinder.

(c) Value of θ at which the cylinder begins to skid

The force of static friction cannot exceed its limiting value,

$\text f_s \le \mu_s \text N$

Substituting $\text f_s = \frac{1}{3}\text{Mg}\sin \theta$ from part (a) and N = Mg cos θ from equation (ii),

$\frac{1}{3}\text{Mg}\sin \theta \le \mu_s \text{Mg}\cos \theta$

Cancelling Mg from both sides and dividing throughout by cos θ,

$\frac{1}{3}\tan \theta \le \mu_s \quad \Rightarrow \quad \tan \theta \le 3\mu_s$

The cylinder begins to skid when this becomes an equality,

$$\begin{aligned}\tan \theta = 3 \times 0.25 = 0.75\\[1em] \theta = \tan^{-1}(0.75) = 36.87^{\circ}\end{aligned}$$

Hence, the cylinder begins to skid, and no longer rolls perfectly, when the inclination exceeds 36.87°.

As a check, at the given inclination of 30° the limiting friction available is

$\mu_s \text N = 0.25 \times 10 \times 9.8 \times \cos 30^{\circ} = 21.22\ \text N$

which is greater than the 16.33 N actually required, so perfect rolling is indeed possible at 30°.

The centre of mass of a system of particles is the point where the entire mass of the system can be considered to be concentrated for the purpose of linear motion.

Reason — Under an external force, a composite body behaves as if its entire mass were concentrated at a single point, the centre of mass. It need not lie within the material of the body, it moves uniformly only when no external force acts, and it coincides with the geometric centre only for a symmetric body of uniform density.

The centre of mass moves in a straight line if no external forces act on the system.

Reason — If no external force acts on the system, then $\vec{\text F}_{ext} = 0$ and

$\frac{\text{d}\vec{\text P}}{\text{dt}} = 0 \quad \Rightarrow \quad \vec{\text v}_{cm} = \text{a constant}$

The velocity of the centre of mass is constant in both magnitude and direction, so it moves in a straight line. This follows from the conservation of linear momentum.

The centre of mass is closer to the more massive particle.

Reason — For two particles, the position of the centre of mass is

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

This is a weighted average of the positions of the particles, the weights being their masses. The particle with the greater mass contributes more to the average, so the centre of mass lies nearer to it. The centre of mass is at the mid-point only when the two masses are equal.

Two ice skaters push off each other on a frictionless ice surface.

Reason — When two skaters push off each other on a frictionless surface, the forces between them are internal forces of the system. Internal forces always occur in equal and opposite pairs and cancel out, so no external force acts on the system. Hence the centre of mass of the system formed by the two skaters does not change its position. In each of the other cases an external force acts on the system, so the centre of mass is displaced.

The centre of mass changes position according to the new positions of the particles.

Reason — The position of the centre of mass depends on both the masses of the particles and their positions,

$\vec{\text r}_{cm} = \frac{1}{\text M}\sum_{\text i = 1}^{\text n} \text m_\text i \vec{\text r}_\text i$

If the masses remain the same but the position vectors $\vec{\text r}_\text i$ change, the weighted average also changes. Hence the centre of mass shifts to a new position determined by the new positions of the particles.

The centre of mass moves with constant acceleration equal to g.

Reason — In a uniform gravitational field the only external force on the system is its total weight Mg, acting at the centre of gravity, which in a uniform field coincides with the centre of mass. Therefore

$\text M\vec{\text a}_{cm} = \text M\vec{\text g} \quad \Rightarrow \quad \vec{\text a}_{cm} = \vec{\text g}$

The centre of mass thus moves with a constant acceleration equal to g, independent of the mass of the system.

Can be determined experimentally by finding the balance point.

Reason — The centre of mass is the point at which an object can be perfectly balanced if supported at that point, since the distribution of the object's weight is even on all sides of it. Hence for an irregularly shaped object the centre of mass can be located experimentally by finding this balance point. It is not necessarily at the geometric centre, nor at the point of maximum mass concentration, and calculus is not the only means of finding it.

All external forces are acting on a single particle of mass equal to the total mass of the system located at the centre of mass.

Reason — Every physical system has associated with it a certain point whose motion characterises the motion of the whole system. When the system moves under an external force, this point moves as if the entire mass of the system were concentrated at it and the external force were applied at that point. This point is the centre of mass, and

$\vec{\text F}_{ext} = \text M\vec{\text a}_{cm}$

Moves in a straight line if no external forces act on the system.

Reason — In a collision the forces exerted by the colliding bodies on each other are internal forces of the system. Internal forces cancel in pairs and do not affect the motion of the centre of mass. Hence, if no external force acts, the linear momentum of the system is conserved and the centre of mass continues to move in a straight line with constant velocity, both before and after the collision.

It is always fixed

Reason — The centre of mass is a geometric property of the body, determined entirely by the distribution of its mass. For a given body its position relative to the material of the body is therefore fixed. It does not depend on the axis of reference, on the orientation of the body, or on the place to which the body is taken. Only the coordinates assigned to it change when a different coordinate system is chosen, not its actual position in the body.

Mass

Reason — In translational motion the mass of a body measures its inertia, that is, its resistance to a change in its state of motion. In rotational motion the corresponding role is played by the moment of inertia. This is clear from a comparison of the two forms of Newton's second law,

$\text F = \text{ma} \quad \text{and} \quad \tau = \text I \alpha$

The moment of inertia I replaces the mass m, which is its physical significance.

The distribution of its mass

Reason — Unlike translatory motion, where inertia is measured solely by the mass of the body, rotational inertia depends more on the distribution of mass with respect to the axis of rotation,

$\text I = \sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2$

Bodies of the same mass and radius, such as a solid sphere, a solid cylinder and a hollow cylinder, have different moments of inertia because their masses are distributed differently about the axis.

The body is more resistant to changes in its rotational motion.

Reason — The moment of inertia is a measure of the resistance a body offers to a change in its rotational motion. From τ = Iα, for a given torque a larger I gives a smaller angular acceleration α. Hence a body with a higher moment of inertia is harder to set into rotation and harder to stop, that is, it is more resistant to changes in its rotational motion.

Always greater than the moment of inertia about the centre of mass axis.

Reason — By the theorem of parallel axes,

$\text I = \text I_{cm} + \text{Ma}^2$

where a is the perpendicular distance between the two parallel axes. Since M and a² are always positive, the term Ma² is always positive. Hence I is always greater than I_cm, and the moment of inertia of a body is minimum about an axis passing through its centre of mass.

Planar bodies.

Reason — The theorem of perpendicular axes states that the moment of inertia of a uniform plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two mutually perpendicular axes in its plane intersecting on the first axis, I_z = I_x + I_y. Its proof uses the relation r² = x² + y², which holds only for a two-dimensional body. It therefore cannot be applied to three-dimensional objects such as spheres, cylinders and cones.

Double

Reason — From the relation between torque, moment of inertia and angular acceleration,

$\tau = \text I \alpha \quad \Rightarrow \quad \alpha = \frac{\tau}{\text I}$

For a rigid body rotating about a given axis, the moment of inertia I is constant. Hence α is directly proportional to τ, and doubling the applied torque doubles the angular acceleration.

Rotational motion

Reason — A force not only causes linear motion but also has the tendency to rotate a body about an axis, and this rotational tendency is expressed in terms of the moment of force or torque. Torque is the measure of the rotational capability of a force about a fixed line or axis, τ = rF sin θ, and so it tends to produce rotational motion in the body.

Remains constant.

Reason — From the rotational form of Newton's second law,

$\tau = \frac{\text{dL}}{\text{dt}}$

If the net external torque is zero, then $\frac{\text{dL}}{\text{dt}} = 0$, so L is a constant. This is the law of conservation of angular momentum, which states that the total angular momentum of a system remains constant if no external torque acts on the system.

Quadruple

Reason — The kinetic energy of rotation is

$\text K = \frac{1}{2}\text I \omega^2$

so for a given moment of inertia K is directly proportional to ω². If the angular velocity is doubled,

$\text K' = \frac{1}{2}\text I(2\omega)^2 = 4 \times \frac{1}{2}\text I \omega^2 = 4\text K$

Radius of rotation

Reason — The rotational kinetic energy is $\text K = \frac{1}{2}\text I \omega^2$, which depends only on the moment of inertia and the angular velocity. The mass and the radius of rotation affect K only indirectly, through their contribution to the moment of inertia I = Σmr². Since the expression for K contains no radius term of its own, the radius of rotation does not directly affect the rotational kinetic energy.

No external torque acts on the system

Reason — Angular momentum is conserved only in the absence of an external torque, not merely an external force, because

$\frac{\text{dL}}{\text{dt}} = \tau_{ext}$

An external force acting along the line passing through the axis of rotation exerts no torque, and in that case angular momentum is still conserved even though a force acts. The nature of the force, whether gravitational or electromagnetic, is irrelevant.

Perpendicular to the velocity vector

Reason — The angular momentum of a particle about a point O is defined as the moment of its linear momentum about that point,

$\vec{\text L} = \vec{\text r} \times \vec{\text p} = \text m(\vec{\text r} \times \vec{\text v})$

Being a vector product, $\vec{\text L} \space$is always perpendicular to the plane containing $\vec{\text r} \space$and $\vec{\text v}$, and hence perpendicular to the velocity vector. This holds whatever the path of the particle, so it is true for a particle moving in a straight line as well.

Conservation of angular momentum

Reason — When a figure skater spinning with her arms extended pulls them in, she brings more mass closer to the axis of rotation, so her moment of inertia decreases from I₁ to I₂. No external torque acts on her, so the angular momentum is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

Since I decreases, ω increases and she spins faster. This is a standard illustration of the conservation of angular momentum.

$\frac{\text m_2 \text d}{\text m_1 + \text m_2}$

Reason — Taking m₁ at the origin, x₁ = 0 and x₂ = d. Then

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2} = \frac{(\text m_1 \times 0) + (\text m_2 \times \text d)}{\text m_1 + \text m_2} = \frac{\text m_2 \text d}{\text m_1 + \text m_2}$

The distance measured from m₁ is therefore proportional to the other mass m₂, which correctly places the centre of mass nearer to the heavier particle.

$\frac{\text I}{2}$

Reason — Let I_x and I_y be the moments of inertia about two mutually perpendicular diameters of the disc. By the theorem of perpendicular axes,

$\text I = \text I_x + \text I_y$

By symmetry the disc has the same moment of inertia about every diameter, so I_x = I_y. Therefore

$\text I = 2\text I_x \quad \Rightarrow \quad \text I_x = \frac{\text I}{2}$

4I

Reason — For a thin uniform rod, the moment of inertia about an axis through its centre and perpendicular to its length is

$\text I = \frac{1}{12}\text m l^2$

By the theorem of parallel axes, about a parallel axis through one end the distance between the axes is l/2,

$\text I_{end} = \frac{1}{12}\text m l^2 + \text m\left(\frac{l}{2}\right)^2 = \frac{\text m l^2}{12} + \frac{\text m l^2}{4} = \frac{\text m l^2}{3}$

$\frac{\text I_{end}}{\text I} = \frac{\text m l^2/3}{\text m l^2/12} = 4 \quad \Rightarrow \quad \text I_{end} = 4\text I$

$\frac{2}{5}\text{MR}^2$

Reason — For a solid sphere of mass M and radius R, the moment of inertia about an axis of symmetry, that is, about any diameter passing through its centre, is

$\text I = \frac{2}{5}\text{MR}^2$

and the corresponding radius of gyration is $\text K = \text R\sqrt{\frac{2}{5}}$. The value $\frac{2}{3}\text{MR}^2$ belongs to a thin-walled hollow sphere and $\frac{1}{2}\text{MR}^2$ to a solid cylinder or disc.

$\frac{1}{12}\text{ML}^2$

Reason — For a thin uniform rod of mass M and length L, taking an element of length dx at a distance x from the centre, dm = (M/L)dx, and

$\text I = \int_{-\text L/2}^{\text L/2} \frac{\text M}{\text L}\text x^2\text{dx} = \frac{\text M}{\text L}\left[\frac{\text x^3}{3}\right]_{-\text L/2}^{\text L/2} = \frac{1}{12}\text{ML}^2$

The value $\frac{1}{3}\text{ML}^2$ is the moment of inertia about an axis through one end of the rod.

Moment of inertia times angular acceleration

Reason — Although torque may also be written as the product of force and the perpendicular distance, τ = Fr_⊥, the relation that expresses torque in terms of the rotational properties of the body is

$\tau = \text I \alpha$

This is the rotational analogue of F = ma and defines torque and angular acceleration as cause and effect. Note that torque is force times the perpendicular distance, not simply force times distance, so the first option as stated is not correct.

Newton-metre

Reason — Torque is the product of force and the perpendicular distance of its line of action from the axis of rotation, so its unit is newton × metre, that is, the newton-metre (N m). Although the joule is also dimensionally newton-metre, the joule is reserved for work and energy, which are scalars, whereas torque is a vector defined only for rotatable systems.

Thin ring.

Reason — For bodies of the same mass M and radius R, the moments of inertia about their central axes are

$\text{Solid sphere} = \frac{2}{5}\text{MR}^2, \quad \text{Hollow sphere} = \frac{2}{3}\text{MR}^2$

$\text{Solid cylinder} = \frac{1}{2}\text{MR}^2, \quad \text{Thin ring} = \text{MR}^2$

Comparing the coefficients, 0.4, 0.67, 0.5 and 1, the thin ring has the largest moment of inertia. This is because the entire mass of the ring lies at the maximum distance R from the axis, whereas in the other bodies the mass is spread over smaller distances as well.

If assertion is false and reason is true.

Explanation

Assertion (A) is incorrect: In the absence of external force, the centre of mass has zero acceleration, but its position does not necessarily remain unchanged. It moves with constant velocity if it was already in motion.

$\vec{F}_{\text{ext}} = M\vec{a}_{\text{CM}}$ ​

If $\vec{F}_{\text{ext}} = 0$, then $\vec{a}_{\text{CM}} = 0$. Hence, the centre of mass moves with constant velocity ($\vec{v}_{\text{CM}} = \text{constant}$)

Reason (R) is also correct: The centre of mass is indeed the average position of all the particles in the system, weighted according to their masses,

$\vec{\text r}_{cm} = \frac{1}{\text M}\sum_{\text i = 1}^{\text n} \text m_\text i \vec{\text r}_\text i$

Therefore, assertion is false and reason is true.

Note: The book's answer key (Option 2) is incorrect. Assertion (A) is false because if no external force acts on a system, the velocity of the center of mass remains constant, not its position (it will keep changing if already in motion). The correct answer is 4.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For rigid bodies of regular geometrical shape having a uniform distribution of mass, the centre of mass lies at the geometric centre. Hence for a uniform circular disc the centre of mass is at its geometric centre.

Reason (R) is also correct: In a uniform disc the mass is distributed symmetrically about the centre, so for every mass element on one side there is an identical element diametrically opposite it.

Because of this symmetry the weighted average position of all the mass elements falls exactly at the centre. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a two-particle system,

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

The particle of greater mass carries more weight in this average, so the centre of mass lies nearer to it.

Reason (R) is also correct: The centre of mass is the point about which the sum of the moments of the masses of the system vanishes, that is, $\sum \text m_\text i \vec{\text r}_\text i = 0$ when the origin is taken at the centre of mass.

For this sum to vanish, the heavier particle must have the smaller distance from the centre of mass. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an isolated system the centre of mass moves with a constant velocity, since

$\vec{\text F}_{ext} = \text M\vec{\text a}_{cm} = 0 \quad \Rightarrow \quad \vec{\text v}_{cm} = \text{a constant}$

Reason (R) is also correct: An isolated system is one on which no net external force acts. The internal forces occur in equal and opposite pairs in compliance with Newton's third law and cancel out.

Since the acceleration of the centre of mass is produced only by the external force, a zero external force means zero acceleration and hence a constant velocity. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The centre of mass may lie outside the physical body of the system, as in the case of a ring, a hollow sphere or a boomerang, where there is no matter at the location of the centre of mass.

Reason (R) is also correct: The centre of mass is a geometric property determined by the distribution of mass and not by the physical boundaries of the body.

Since the position of the centre of mass is fixed purely by the weighted average of the mass distribution, there is no requirement for matter to be actually present at that point. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: The motion of the centre of mass of a system of particles is not influenced by internal forces. Only the external force determines the acceleration of the centre of mass,

$\vec{\text F}_{ext} = \text M\vec{\text a}_{cm}$

Reason (R) is correct: By Newton's third law of motion, internal forces always occur in equal and opposite pairs, so their vector sum over the whole system is zero and they cancel each other out.

It is precisely because the internal forces cancel that they cannot affect the motion of the centre of mass, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a rigid body is projected in the air, its centre of mass follows a parabolic path, which is the trajectory of a projectile moving under a constant acceleration.

Reason (R) is also correct: In free fall, neglecting air resistance, the only external force acting on the body is gravity, and it acts on the centre of mass.

Since the centre of mass moves as if the entire mass were concentrated there and the external force were applied there, a constant downward force produces motion under uniform acceleration, whose path is a parabola. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a collision the forces between the two bodies are internal to the system. Hence the velocity of the centre of mass after the bodies stick together is the same as it was before the collision.

Reason (R) is also correct: Momentum is conserved in an isolated system, since no external force acts on it.

From $\vec{\text P} = \text M\vec{\text v}_{cm}$, if the total momentum is unchanged and the total mass is unchanged, the velocity of the centre of mass must also be unchanged. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The centre of mass of a uniform rod lies at its mid-point. This follows from integration,

$\text x_{cm} = \frac{1}{\text M}\int_0^{l} \text x\frac{\text m}{l}\text{dx} = \frac{1}{2l}[\text x^2]_0^{l} = \frac{l}{2}$

Reason (R) is also correct: The mass of a uniform rod is distributed symmetrically about its mid-point, each half being identical to the other.

Because of this symmetry, the weighted average position of the mass elements falls exactly at the mid-point. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: An external force acting on a projectile can change the path of its centre of mass, so that it no longer follows the original parabola.

Reason (R) is also correct: External forces alter the momentum of the system, since

$\frac{\text{d}\vec{\text P}}{\text{dt}} = \vec{\text F}_{ext}$

Because the momentum of the system is $\vec{\text P} = \text M\vec{\text v}_{cm}$, a change in momentum means a change in the velocity of the centre of mass, and hence a change in its path. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The moment of inertia depends on the distribution of mass about the axis of rotation, since

$\text I = \sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2$

Each mass element is weighted by the square of its distance from the axis.

Reason (R) is also correct: For a given distribution of mass, the moment of inertia is directly proportional to the mass of the body, as I contains m_i to the first power.

However, the proportionality to mass explains the dependence of I on mass, not on its distribution. The distribution dependence arises from the factor r_i², which the Reason does not mention. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The radius of gyration K of a body about a given axis is the distance from that axis at which the entire mass of the body can be imagined to be concentrated without changing its moment of inertia, so that

$\text I = \text{MK}^2 \quad \Rightarrow \quad \text K = \sqrt{\frac{\text I}{\text M}}$

Reason (R) is also correct: The radius of gyration is used to simplify the calculation of the moment of inertia, replacing a complicated mass distribution by a single equivalent point mass.

It is exactly because the whole mass may be treated as concentrated at this one distance that the calculation is simplified. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By the theorem of parallel axes,

$\text I = \text I_{cm} + \text{Ma}^2$

Since Ma² is always positive, I is always greater than I_cm. Hence for a given mass and shape the moment of inertia about the centre of mass is the smallest.

Reason (R) is also correct: The moment of inertia depends on the choice of the axis, because the distances r_i of the mass elements change when the axis is changed.

Since the moment of inertia varies with the axis, there must be an axis for which it is least, and the parallel axis theorem shows this to be the axis through the centre of mass. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a thin rod of mass M and length L,

$\text I_{centre} = \frac{1}{12}\text{ML}^2 \quad \text{and} \quad \text I_{end} = \frac{1}{3}\text{ML}^2$

Clearly I_centre < I_end.

Reason (R) is also correct: About the centre the mass is distributed symmetrically, half of it on each side, so no part of the rod lies farther than L/2 from the axis. About one end, parts of the rod are as far as L from the axis.

Because the moment of inertia weights each element by r², a more symmetrical distribution about the axis keeps the distances smaller and gives a smaller moment of inertia. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: A solid sphere and a hollow sphere of the same mass and radius do not have the same moment of inertia,

$\text I_{solid} = \frac{2}{5}\text{MR}^2 \quad \text{and} \quad \text I_{hollow} = \frac{2}{3}\text{MR}^2$

Reason (R) is correct: The moment of inertia depends on the mass distribution, since I = Σm_ir_i².

In a hollow sphere the entire mass lies at the surface, at the maximum distance R from the centre, whereas in a solid sphere the mass is spread throughout the volume, at distances ranging from 0 to R. Hence their moments of inertia differ, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The theorem of parallel axes enables the moment of inertia about any axis to be found, provided the moment of inertia about a parallel axis through the centre of mass is known.

Reason (R) is also correct: The theorem states that

$\text I = \text I_{cm} + \text{Ma}^2$

that is, the moment of inertia about any axis is the sum of the moment of inertia about the parallel axis through the centre of mass and the product of the mass and the square of the distance between the axes.

The Reason is the exact statement of the theorem referred to in the Assertion, and so explains it.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The theorem of perpendicular axes applies only to planar, that is, two-dimensional bodies such as thin discs, rings and plane laminas.

Reason (R) is also correct: The theorem states that the moment of inertia about an axis perpendicular to the plane of the lamina is the sum of the moments of inertia about two mutually perpendicular axes lying in that plane, I_z = I_x + I_y.

The proof of this relation rests on r² = x² + y², which holds only when every mass element lies in a single plane. For a three-dimensional body a z-coordinate also appears and the relation fails. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A torque is required to change the rotational state of a body, since

$\tau = \text I \alpha$

Without a torque there is no angular acceleration and the angular velocity remains unchanged.

Reason (R) is also correct: Torque is the rotational equivalent of force, as is clear from comparing τ = Iα with F = ma.

Just as a force is needed to change the translational state of a body, its rotational counterpart, the torque, is needed to change the rotational state. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: A higher moment of inertia makes a body harder, not easier, to rotate. From τ = Iα, for a given torque a larger I produces a smaller angular acceleration.

Reason (R) is correct: The moment of inertia is indeed a measure of an object's resistance to changes in its rotational motion, playing the same role in rotational motion that mass plays in translational motion.

Since a larger moment of inertia means greater resistance to being set into rotation, the body becomes more difficult to rotate, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an isolated system with no external torque,

$\frac{\text{dL}}{\text{dt}} = \tau_{ext} = 0 \quad \Rightarrow \quad \text L = \text{a constant}$

This is the law of conservation of angular momentum.

Reason (R) is also correct: Conservation laws are fundamental principles of physics, and the conservation of angular momentum is one such principle applying to isolated systems and to individual objects within a system, provided there is no external influence altering their rotational motion.

The Assertion is precisely a statement of the conservation principle referred to in the Reason, and so the Reason explains it.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: The angular momentum of a rotating body is not always constant. It remains constant only when no external torque acts on the body. If an external torque acts, then

$\frac{\text{dL}}{\text{dt}} = \tau \ne 0$

and the angular momentum changes.

Reason (R) is correct: The angular momentum does depend on both the moment of inertia and the angular velocity, L = Iω.

Since either I or ω, or both, may be changed, for instance when a skater pulls in her arms, the angular momentum is not a fixed quantity in general, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In pure rolling motion there is no relative motion between the point of contact and the surface, the point of contact being momentarily at rest relative to the surface.

Reason (R) is also correct: At the point of contact the velocity of the centre of mass and the tangential velocity due to rotation are equal in magnitude but opposite in direction,

$\text v_A = \text v_t - \text v_{cm} = \omega \text R - \omega \text R = 0$

Because these two velocities cancel exactly, the net velocity of the contact point is zero, which is the same as saying there is no relative motion there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A rolling object possesses both translational and rotational kinetic energy,

$\text E = \frac{1}{2}\text{mv}^2 + \frac{1}{2}\text I \omega^2$

Reason (R) is also correct: Rolling motion is a combination of pure translation of the centre of mass and pure rotation about the centre of mass.

Since the motion consists of these two parts, the total kinetic energy is the sum of the kinetic energies associated with each. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: A cylinder rolling down an inclined plane accelerates more slowly than one sliding down. For a rolling solid cylinder,

$\text a = \frac{2}{3}\text g \sin \theta$

which is less than g sin θ, the acceleration of a cylinder sliding down a smooth plane. This is because part of the potential energy is converted into rotational kinetic energy.

Reason (R) is correct: Rolling friction is generally less than sliding friction.

Even though rolling friction is smaller, the rotational inertia of the rolling body reduces its acceleration below that of a sliding body, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Applying the conservation of energy to a cylinder rolling down an incline of height h,

$\text{mgh} = \frac{3}{4}\text{mv}^2 \quad \Rightarrow \quad \text v = \sqrt{\frac{4\text{gh}}{3}}$

This depends only on h and not on the angle of inclination or the length of the plane.

Reason (R) is also correct: By the principle of conservation of energy, the potential energy lost is converted into kinetic energy.

Since the potential energy lost depends only on the height, the kinetic energy gained, and hence the velocity, also depends only on the height. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The kinetic energy of rotation of a body rotating about an axis is

$\text K_{rot} = \frac{1}{2}\text I \omega^2$

obtained by summing the kinetic energies of all its constituent particles.

Reason (R) is also correct: In this expression I and ω are the rotational analogues of the mass m and the linear velocity v of translational motion.

Comparing with $\text K = \frac{1}{2}\text{mv}^2$, replacing m by I and v by ω gives exactly the rotational formula, which is why the expression takes this form. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: For a rotating body at constant angular velocity, the rotational kinetic energy $\text K = \frac{1}{2}\text I \omega^2$ increases, not decreases, when the moment of inertia increases, since K is directly proportional to I.

Reason (R) is correct: The rotational kinetic energy is indeed directly proportional to the moment of inertia at a constant angular velocity.

Direct proportionality means the two quantities increase together, so an increase in I must increase K. This is exactly why the Assertion, which claims a decrease, is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A rigid body is in rotational equilibrium when the net external torque acting on it about any axis is zero,

$\sum \vec{\tau} = 0$

Reason (R) is also correct: A body in rotational equilibrium does not undergo angular acceleration, since from Στ = Iα, α = 0 and the angular velocity, and hence the angular momentum, stays constant.

The vanishing of the net torque is precisely what makes the angular acceleration zero. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a uniform circular disc about an axis perpendicular to its plane and passing through its centre, the radius of gyration is $\text K = \frac{\text R}{\sqrt{2}}$.

Reason (R) is also correct: The radius of gyration is obtained from the relation $\text{MK}^2 = \frac{1}{2}\text{MR}^2$.

Solving this relation,

$\text K^2 = \frac{\text R^2}{2} \quad \Rightarrow \quad \text K = \frac{\text R}{\sqrt{2}}$

which is exactly the value stated in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is incorrect: If there is no external torque, angular momentum ($L$) remains constant, but angular velocity ($\omega$) does not have to. Since $L = I\omega$ (where $I$ is the moment of inertia), if the body changes its shape or mass distribution (like a spinning ice skater pulling their arms in), $I$ changes. This causes $\omega$ to change in order to keep $L$ constant. Therefore, the angular velocity does not always remain constant.

Reason (R) is also correct: Angular momentum is conserved in such systems, since $\frac{\text{dL}}{\text{dt}} = \tau_{ext} = 0$.

here L = constant.

Therefore, assertion is false and reason is true.

Note: The book's answer key (Option 1) is incorrect. Assertion (A) is false because in the absence of external torque, only angular momentum ($L = I\omega$) is conserved. If the body's moment of inertia ($I$) changes (like a diver folding their body), its angular velocity ($\omega$) will change. The correct answer is 4.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: For a disc rolling without slipping, the frictional force does not act in the direction of rolling. It acts opposite to the tendency of slipping at the point of contact.

Reason (R) is correct: The frictional force prevents slipping and provides the necessary torque for rolling.

By opposing the relative motion at the contact point, friction ensures that the point of contact does not slide and converts potential slipping into smooth rolling. Since it acts opposite to the tendency of slipping, the Assertion is false.

Therefore, assertion is false and reason is true.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: A hoop and a solid disc released from the same height do not reach the bottom at the same time. Their accelerations differ,

$\text a_{hoop} = \frac{1}{2}\text g \sin \theta \quad \text{and} \quad \text a_{disc} = \frac{2}{3}\text g \sin \theta$

The disc, having the smaller moment of inertia, has the greater acceleration and reaches the bottom first.

Reason (R) is correct: Both do have the same potential energy at the top, since they have the same mass and are at the same height.

However, the hoop converts a larger share of this energy into rotational kinetic energy, so its centre of mass moves more slowly. Equal potential energy therefore does not imply equal time, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A spinning ice skater increases her spin rate by pulling her arms in.

Reason (R) is also correct: Pulling the arms in brings mass closer to the axis, decreasing the moment of inertia. Since no external torque acts, the angular momentum is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

As I decreases, ω must increase to keep the product constant, which is exactly why the skater spins faster. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: For the same mass and speed, a rolling object has more total kinetic energy than a sliding object, not less,

$\text E_{rolling} = \frac{1}{2}\text{mv}^2 + \frac{1}{2}\text I \omega^2 \gt \frac{1}{2}\text{mv}^2 = \text E_{sliding}$

Reason (R) is correct: Rolling objects do possess both rotational and translational kinetic energy.

It is precisely the additional rotational term that makes the total kinetic energy of the rolling body greater, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The moment of inertia of a body about an axis depends on its shape and size, which is why a solid sphere, a solid cylinder and a hollow cylinder of the same mass and radius have different moments of inertia.

Reason (R) is also correct: The mass distribution with respect to the axis determines the moment of inertia, since I = Σm_ir_i².

The shape and size of a body are precisely what fix how its mass is distributed relative to the axis, and hence its moment of inertia. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Although the particles of a rotating rigid body have different linear velocities depending on their distance from the axis, all of them rotate with the same angular velocity ω.

Reason (R) is also correct: The angular velocity is a property of the entire rigid body, not of an individual particle.

Since a rigid body does not deform, every particle sweeps out the same angle Δθ in the same time Δt, so the angular velocity is common to all points on the body. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In rolling motion the static friction does no work.

Reason (R) is also correct: The point of contact in pure rolling is instantaneously at rest relative to the surface.

Work requires a displacement of the point of application of the force. Since the contact point has zero velocity relative to the surface, there is no relative displacement there and hence no work is done or energy dissipated against friction. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In the absence of external forces the centre of mass of a system moves with a constant velocity, since $\vec{\text a}_{cm} = 0$.

Reason (R) is also correct: The centre of mass is the point at which the total mass of the system can be considered to be concentrated, and the external force may be considered to act there.

Because the whole system may be replaced by a single particle of mass M at the centre of mass acted upon by the external force, a zero external force gives zero acceleration and hence constant velocity. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A ring and a solid disc of the same mass and radius have different moments of inertia about their centres,

$\text I_{ring} = \text{MR}^2 \quad \text{and} \quad \text I_{disc} = \frac{1}{2}\text{MR}^2$

Reason (R) is also correct: The mass distribution in the ring is farther from the centre compared to the disc, the whole of it lying at the distance R.

Since each mass element is weighted by r², mass placed farther from the axis contributes more, giving the ring a larger moment of inertia. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: In the rolling of a cylinder down an inclined plane the linear acceleration does not depend on the radius,

$\text a = \frac{\text g \sin \theta}{1 + \frac{\text k^2}{\text R^2}} = \frac{2}{3}\text g \sin \theta$

The radius cancels, since $\frac{\text k^2}{\text R^2}$ is a fixed number for a given shape.

Reason (R) is correct: The moment of inertia of the cylinder does influence the linear acceleration, through the factor $\frac{\text k^2}{\text R^2}$.

However, it does so only through the shape of the body and not through its radius, which is why the Assertion is false.

Therefore, assertion is false and reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a particle of mass m moving in a circle of radius r with speed v, the position vector and the linear momentum are mutually perpendicular, so θ = 90° and

$\text L = \text{pr}\sin 90^\circ = \text{mvr}$

Reason (R) is also correct: Angular momentum is defined as the moment of linear momentum, $\vec{\text L} = \vec{\text r} \times \vec{\text p}$.

Taking the moment of the linear momentum mv about the centre, that is, multiplying it by the perpendicular distance r, gives exactly L = mvr. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The rotational kinetic energy of a body can be converted into translational kinetic energy, as happens when a rotating disc placed on a rough surface gradually begins to roll, its angular speed decreasing while the speed of its centre of mass increases.

Reason (R) is also correct: Energy can be transformed from one form to another while the total energy is conserved.

Since kinetic energy of rotation and of translation are two forms of mechanical energy, the general principle of transformation of energy permits the conversion described in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false and reason is true.

Explanation

Assertion (A) is false: For the same applied torque, a body with a higher moment of inertia acquires a lower angular velocity, not a higher one. From τ = Iα,

$\alpha = \frac{\tau}{\text I}$

A larger I gives a smaller α, and hence a smaller angular velocity in a given time.

Reason (R) is correct: For a given torque the angular acceleration, and hence the angular velocity acquired in a given time, is inversely proportional to the moment of inertia.

This inverse relationship is exactly what makes the Assertion false.

Therefore, assertion is false and reason is true.

The centre of mass of a uniform triangular lamina lies at the point of intersection of its three medians, that is, at the centroid of the triangle.

This is because the lamina can be subdivided into narrow strips parallel to any one side. The centre of mass of each strip lies at its mid-point, and the mid-points of all such strips lie on the median drawn to that side. Applying the same reasoning to all three sides, the centre of mass must lie on all three medians, hence at their point of intersection.

No, the centre of mass is not always a reality.

In many cases there is no matter at all at the location of the centre of mass, so it is only a mathematical point. For example, the centre of mass of a ring lies at its centre, where there is no material. Similarly, after the explosion of a fire-cracker in the sky, the centre of mass of all the fragments continues to move along the original parabolic path, although there is nothing physically present at that point.

No, the internal forces do not affect the motion of the system.

The internal forces of a system always occur in equal and opposite pairs, in compliance with Newton's third law of motion, and so they mutually cancel in pairs. Their vector sum over the whole system is zero. Hence the acceleration of the centre of mass is determined by the external force alone,

$\vec{\text F}_{ext} = \text M\vec{\text a}_{cm}$

The centre of mass of an isolated system moves with a constant velocity.

An isolated system is one on which no net external force acts. Hence

$\frac{\text{d}\vec{\text P}}{\text{dt}} = \vec{\text F}_{ext} = 0 \quad \Rightarrow \quad \vec{\text P} = \text M\vec{\text v}_{cm} = \text{a constant}$

so that $\vec{\text v}_{cm}$ is constant in both magnitude and direction. If the centre of mass was initially at rest, it continues to remain at rest.

Although Newton's laws apply to the individual particles of a system, we may concentrate the entire mass of the system at its centre of mass and imagine all the external forces to be applied at that point.

The motion of this single point of mass M under the external force is then governed by

$\vec{\text F}_{ext} = \text M\vec{\text a}_{cm}$

which is Newton's second law written for the system as a whole. This is possible because the internal forces cancel in pairs and do not influence the motion of the centre of mass. Hence the motion of the centre of mass describes the motion of the whole system.

(i) When the person sits near the edge, no external torque acts on the system, so the angular momentum L = Iω remains constant. The person's presence at the edge, far from the axis, gives the system a larger moment of inertia. Hence the angular velocity of the platform will decrease.

(ii) When the person starts moving from the edge towards the centre, the distance of his mass from the axis of rotation decreases, so the moment of inertia of the system decreases. Since Iω must remain constant, the angular velocity will again start increasing.

Some applications of the principle of conservation of angular momentum are :

(i) A figure skater or a ballet dancer increases the speed of spin by pulling in the arms and legs, which decreases the moment of inertia.

(ii) A diver jumping into water from a height pulls his arms and legs towards the centre of his body, decreasing his moment of inertia so that he can rotate faster in the air.

(iii) A man on a rotating table holding heavy dumb-bells spins faster when he pulls in his arms, since the distance of the dumb-bells from the axis decreases.

(iv) Planetary motion — a planet moves faster when it is nearer the sun and slower when it is farther away, which is Kepler's second law of planetary motion.

If there were only one propeller in a helicopter, then by the conservation of angular momentum the body of the helicopter itself would have turned in the direction opposite to that of the propeller.

The second, smaller propeller at the tail provides the necessary counter-torque, so that the total angular momentum of the system remains conserved and the body of the helicopter does not spin about the vertical axis.

The moment of inertia of a body about an axis depends upon :

(i) The mass of the body.

(ii) The shape and size of the body, which determine how its mass is distributed.

(iii) The position of the axis of rotation.

These are combined in the relation I = Σm_ir_i², where each mass element is weighted by the square of its distance from the axis.

The ring will have the larger moment of inertia.

In a ring the entire mass lies on the circumference, that is, at a distance R from the axis passing through the centre of mass and perpendicular to its plane, so I_ring = MR². In a disc the mass is spread over the whole area, at distances ranging from 0 to R, so its average distance from the axis is smaller and $\text I_{disc} = \frac{1}{2}\text{MR}^2$.

An isolated system is a system on which no net external force is acting.

For such a system the linear momentum is conserved, and the velocity of its centre of mass remains constant.

The relation between torque and moment of inertia is

$\tau = \text I \alpha$

where τ is the torque, I the moment of inertia about the axis of rotation and α the angular acceleration produced. This is the rotational analogue of F = ma.

The work done in rotatory motion is

$\text W = \tau \theta$

where τ is the torque acting on the body and θ is the angular displacement, measured in radian, through which the body turns.

Unit : kg m² s^-1 (also written as J s)

Dimensions : [ML²T^-1]

The torque acting on a body is equal to the rate of change of its angular momentum,

$\tau = \frac{\text{dL}}{\text{dt}}$

This is the rotational analogue of Newton's second law of motion for linear motion, $\text F = \frac{\text{dp}}{\text{dt}}$.

(i) Moment of linear momentum — angular momentum.

(ii) Rate of change of angular momentum — torque.

(iii) Product of moment of inertia and angular velocity — angular momentum.

Radius of gyration : The radius of gyration K of a body about a given axis is the distance from the axis of rotation, the square of which, when multiplied by the total mass of the body, gives the moment of inertia of the body about that axis,

$\text I = \text{MK}^2 \quad \Rightarrow \quad \text K = \sqrt{\frac{\text I}{\text M}}$

Given,

• Mass of the cylindrical rod = M
• Length of the rod = L
• The axis is perpendicular to the length and passes through the centre

The moment of inertia of a rod about an axis passing through its centre and perpendicular to its length is

$\text I = \frac{1}{12}\text{ML}^2$

If K is the radius of gyration about this axis, then by definition I = MK². Therefore

$\text{MK}^2 = \frac{1}{12}\text{ML}^2$

$\text K^2 = \frac{\text L^2}{12} \quad \Rightarrow \quad \text K = \frac{\text L}{\sqrt{12}}$

$\text K = \frac{\text L}{2\sqrt{3}}$

Hence, the radius of gyration of the cylindrical rod about the given axis is $\frac{\text L}{2\sqrt{3}}$.

Given,

• Moment of inertia about an axis through the centre of mass, $\text I_{cm} = \frac{\text M l^2}{12}$

The axis through one end is parallel to the axis through the centre of mass, and the distance between them is $\frac{l}{2}$. By the theorem of parallel axes,

$\text I = \text I_{cm} + \text M\left(\frac{l}{2}\right)^2$

Substituting the values,

$\text I = \frac{\text M l^2}{12} + \frac{\text M l^2}{4}$

Taking the L.C.M. as 12,

$\text I = \frac{\text M l^2 + 3\text M l^2}{12} = \frac{4\text M l^2}{12}$

$\text I = \frac{\text M l^2}{3}$

Hence, the moment of inertia of the rod about an axis passing through one of its ends is $\frac{\text M l^2}{3}$.

Given,

• Moment of inertia of the ring about a diameter, $\text I_{cm} = \frac{1}{2}\text{MR}^2$

A tangent to the circle of the ring, drawn in the plane of the ring, is parallel to a diameter, and the distance between them is R. By the theorem of parallel axes,

$\text I = \text I_{cm} + \text{MR}^2$

Substituting the values,

$\text I = \frac{1}{2}\text{MR}^2 + \text{MR}^2$

$\text I = \frac{3}{2}\text{MR}^2$

Hence, the moment of inertia of the ring about a tangent to its circle is $\frac{3}{2}\text{MR}^2$.

Given,

• Moment of inertia of the disc about an axis perpendicular to it and through its centre, $\text I = \frac{\text{MR}^2}{2}$

Let I_d be the moment of inertia about a diameter. Take two mutually perpendicular diameters lying in the plane of the disc and intersecting at its centre. By the theorem of perpendicular axes,

$\text I_d + \text I_d = \text I$

By symmetry the disc has the same moment of inertia about every diameter, so

$2\text I_d = \frac{1}{2}\text{MR}^2$

$\text I_d = \frac{1}{4}\text{MR}^2$

Hence, the moment of inertia of the disc about a diameter is $\frac{1}{4}\text{MR}^2$.

Centre of mass : The centre of mass of a system of particles is a point in the system which moves in the same way as a single particle of mass equal to the total mass of the system would move when subjected to the same external forces as applied to the system.

For a system of n particles of masses m₁, m₂, ..., m_n with position vectors $\vec{\text r}_1, \vec{\text r}_2, ..., \vec{\text r}_n$, the position vector of the centre of mass is

$\vec{\text r}_{cm} = \frac{\text m_1 \vec{\text r}_1 + \text m_2 \vec{\text r}_2 + \dots + \text m_n \vec{\text r}_n}{\text m_1 + \text m_2 + \dots + \text m_n} = \frac{1}{\text M}\sum_{\text i = 1}^{\text n} \text m_\text i \vec{\text r}_\text i$

where M is the total mass of the system. Qualitatively, it represents the geometric point located at the "average" position of the particles weighted in proportion to their masses.

Three examples of the centre of mass motion are :

(i) Explosion of a fire-cracker in the sky : A fire-cracker projected from the earth moves along a parabolic path and explodes in mid-air into many fragments, which fly off along their own parabolic paths. Since the explosion is caused by internal forces alone, the centre of mass of all the fragments continues to move along the same initial parabolic path that the unexploded cracker would have followed.

(ii) Motion of the earth-moon system around the sun : The earth and the moon both revolve in circular orbits about their common centre of mass, remaining always on opposite sides of it, while the centre of mass describes an elliptic orbit around the sun. The mutual gravitational forces between the earth and the moon are internal, whereas the sun's attraction is the external force on the centre of mass.

(iii) Radioactive decay : In the spontaneous decay of a radioactive nucleus no external forces are involved. If the parent nucleus is initially at rest, the centre of mass of the two fragments continues to be at rest, and the heavier and the lighter fragments fly off in opposite directions with speeds inversely proportional to their masses.

Law of conservation of angular momentum : The total angular momentum of a system remains constant if no external torque acts on the system.

Since the torque is the rate of change of angular momentum,

$\tau = \frac{\text{dL}}{\text{dt}}$

if τ = 0, then $\frac{\text{dL}}{\text{dt}} = 0$, which gives

$\text L = \text I \omega = \text{a constant} \quad \text{or} \quad \text I_1 \omega_1 = \text I_2 \omega_2$

The moment of inertia of a solid sphere of mass M and radius R about one of its diameters, that is, about its axis of symmetry, is

$\text I = \frac{2}{5}\text{MR}^2$

and the corresponding radius of gyration is

$\text K = \text R\sqrt{\frac{2}{5}}$

For a two-particle system the position vector of the centre of mass is

$\vec{\text r}_{cm} = \frac{\text m_1 \vec{\text r}_1 + \text m_2 \vec{\text r}_2}{\text m_1 + \text m_2}$

For particles of equal masses, m₁ = m₂ = m, so

$\vec{\text r}_{cm} = \frac{\text m\vec{\text r}_1 + \text m\vec{\text r}_2}{\text m + \text m} = \frac{\vec{\text r}_1 + \vec{\text r}_2}{2}$

$\vec{\text r}_{cm} = \frac{1}{2}(\vec{\text r}_1 + \vec{\text r}_2)$

Hence, the position vector of the centre of mass of two particles of equal masses is the average of the position vectors of the particles.

Given,

• Moment of inertia of the disc about any of its diameters, $\text I_x = \text I_y = \frac{1}{4}\text{MR}^2$

Let us take two mutually perpendicular diameters along the X- and Y-axes lying in the plane of the disc.

(i) About an axis passing through the centre and normal to the disc : By the theorem of perpendicular axes,

$\text I_C = \text I_x + \text I_y = \frac{1}{4}\text{MR}^2 + \frac{1}{4}\text{MR}^2$

$\text I_C = \frac{1}{2}\text{MR}^2$

(ii) About an axis passing through a point on its edge and normal to the disc : This axis is parallel to the axis in part (i), and the distance between them is R. By the theorem of parallel axes,

$\text I_e = \text I_C + \text{MR}^2 = \frac{1}{2}\text{MR}^2 + \text{MR}^2$

$\text I_e = \frac{3}{2}\text{MR}^2$

Consider an isolated system of n particles of masses m₁, m₂, ..., m_n moving with velocities $\vec{\text v}_1, \vec{\text v}_2, ..., \vec{\text v}_n$. The velocity of the centre of mass is defined by

$\vec{\text v}_{cm} = \frac{1}{\text M}(\text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n)$

Multiplying both sides by the total mass M,

$\text M\vec{\text v}_{cm} = \text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n = \vec{\text P}$

where $\vec{\text P} \space$is the total linear momentum of the system. Differentiating with respect to time,

$\frac{\text{d}\vec{\text P}}{\text{dt}} = \text M\frac{\text{d}\vec{\text v}_{cm}}{\text{dt}} = \text M\vec{\text a}_{cm} = \vec{\text F}_{ext}$

For an isolated system no net external force acts, so $\vec{\text F}_{ext} = 0$. Since M ≠ 0,

$\frac{\text{d}\vec{\text v}_{cm}}{\text{dt}} = 0 \quad \Rightarrow \quad \vec{\text v}_{cm} = \text{a constant}$

Hence, the velocity of the centre of mass of a moving isolated system remains constant.

Let F_r and F_θ be the components of a force $\vec{\text F} \space$in the direction of increasing r and θ respectively. These are called the radial and the transverse (angular) components of the force. If θ is the angle between $\vec{\text F} \space$and the position vector $\vec{\text r}$, then

$\text F_r = \text F\cos \theta \quad \text{and} \quad \text F_\theta = \text F\sin \theta$

The magnitude of the torque about the axis of rotation is

$\tau = \text{rF}\sin \theta = \text r\text F_\theta$

This expression contains only the transverse component F_θ; the radial component F_r does not appear in it at all. The reason is that the radial component acts along the position vector, that is, along the line joining the point of application to the axis, so its perpendicular distance from the axis is zero.

Hence, the torque is due to the transverse component of the force only, and the radial component does not contribute to the torque. For this reason the gravitational force of the sun on the earth, being purely radial, produces no torque about the sun.

Concept of torque : The moment of a force is a vector quantity, and when expressed vectorially it gives the torque. It is the measure of the rotational tendency, or capability, of a force about a fixed line or axis, and is expressed as the product of the magnitude of the force and the perpendicular distance of the axis of rotation from the line of action of the force.

If a force $\vec{\text F} \space$acts at a point whose position vector with respect to the axis is $\vec{\text r}$, then

$\vec{\tau} = \vec{\text r} \times \vec{\text F}$

and its magnitude is

$\tau = \text{rF}\sin \theta$

where θ is the angle between $\vec{\text r} \space$and $\vec{\text F}$. Torque is an axial vector, its direction being perpendicular to the plane containing $\vec{\text r} \space$and $\vec{\text F}$, given by the right hand thumb rule. An anticlockwise moment is taken positive and a clockwise moment negative.

Unit : newton-metre (N m)

Dimensions : [ML²T^-2]

Angular acceleration : The rate of change of angular velocity of a body rotating about an axis is called its angular acceleration,

$\alpha = \frac{\text{d}\omega}{\text{dt}}$

Its unit is rad s^-2 and its dimensions are [T^-2].

Relation between angular acceleration and linear acceleration : For a particle at a perpendicular distance r from the axis of rotation, the linear velocity and the angular velocity are related by

$\text v = \text r \omega$

Differentiating both sides with respect to time, r being constant for a rigid body,

$\frac{\text{dv}}{\text{dt}} = \text r\frac{\text{d}\omega}{\text{dt}}$

But $\frac{\text{dv}}{\text{dt}} = \text a$ is the linear acceleration and $\frac{\text{d}\omega}{\text{dt}} = \alpha$ is the angular acceleration. Therefore

$\text a = \text r \alpha$

Hence, the linear acceleration of a particle is the product of its distance from the axis of rotation and the angular acceleration of the body.

I_B is greater, that is, the moment of inertia of the disc made of the lighter metal is greater.

Both discs have the same mass m and the same thickness t. Their moments of inertia about the given axes are

$\text I_A = \frac{1}{2}\text{mr}_A^2 \quad \text{and} \quad \text I_B = \frac{1}{2}\text{mr}_B^2$

$\frac{\text I_A}{\text I_B} = \frac{\text r_A^2}{\text r_B^2} \qquad \dots(\text i)$

Since the mass of each disc is the product of its volume and its density,

$\text m = \pi \text r_A^2\text t\text d_A = \pi \text r_B^2\text t\text d_B$

$\frac{\text r_A^2}{\text r_B^2} = \frac{\text d_B}{\text d_A}$

Substituting in equation (i),

$\frac{\text I_A}{\text I_B} = \frac{\text d_B}{\text d_A}$

It is given that d_A > d_B, so $\frac{\text d_B}{\text d_A}$ < 1, which gives

$\text I_B \gt \text I_A$

Hence, I_B is greater. Physically, the disc of the lighter metal must have a larger radius to have the same mass, so its material particles lie on the average at a greater distance from the axis of rotation.

The body A, having the smaller moment of inertia, has the greater kinetic energy.

The kinetic energy of rotation and the angular momentum of a rotating body are

$\text K = \frac{1}{2}\text I \omega^2 \quad \text{and} \quad \text L = \text I \omega$

Eliminating ω between them,

$\text K = \frac{\text L^2}{2\text I}$

The angular momenta of A and B are equal, say L. Therefore

$\frac{\text K_A}{\text K_B} = \frac{\text L^2/2\text I_A}{\text L^2/2\text I_B} = \frac{\text I_B}{\text I_A}$

It is given that I_B > I_A, so $\frac{\text K_A}{\text K_B}$ > 1, that is,

$\text K_A \gt \text K_B$

For equal angular momenta the kinetic energy is inversely proportional to the moment of inertia.

Given,

• Angular momenta are equal, L_A = L_B = L
• Moment of inertia of A is half that of B, $\text I_A = \frac{1}{2}\text I_B$

For a rotating body the kinetic energy in terms of the angular momentum is

$\text K = \frac{1}{2}\text I \omega^2 = \frac{\text L^2}{2\text I}$

Therefore,

$\frac{\text K_A}{\text K_B} = \frac{\text L^2/2\text I_A}{\text L^2/2\text I_B} = \frac{\text I_B}{\text I_A}$

Substituting $\text I_A = \frac{1}{2}\text I_B$,

$\frac{\text K_A}{\text K_B} = \frac{\text I_B}{\frac{1}{2}\text I_B} = 2$

Hence, the ratio of the kinetic energies of A and B is K_A : K_B = 2 : 1.

Moment of inertia : The moment of inertia of a particle about an axis of rotation is the product of the mass of the particle and the square of its distance from the axis of rotation, I = mr².

For a rigid body composed of a large number of particles of masses m₁, m₂, ..., m_n at perpendicular distances r₁, r₂, ..., r_n from the axis, the moment of inertia is the sum of the moments of inertia of all the constituent particles,

$\text I = \sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2$

Unit in the MKS system : kg m²

In the given equation,

$\tau = \frac{\text{dL}}{\text{dt}}$

L is the angular momentum of the body. Rearranging,

$\text{dL} = \tau\text{dt}$

Hence the unit of L is the unit of torque multiplied by the unit of time,

$\text{N m} \times \text s = \text{kg m}^2\text s^{-1}$

Hence, the unit of L is kg m² s^-1, which is also written as J s.

(i) Can be outside the material body.

The centre of mass is a geometric property of the body, determined by the distribution of its mass and not by its physical boundaries. Hence for an irregular or non-uniform object there need be no matter at the location of the centre of mass. For a ring, for instance, the centre of mass lies at the centre, where there is no material at all.

(ii) The system's total linear momentum is zero.

The total linear momentum of a system is related to the velocity of its centre of mass by

$\vec{\text P} = \text M\vec{\text v}_{cm}$

If the centre of mass is at rest, $\vec{\text v}_{cm} = 0$ and hence $\vec{\text P} = 0$. This does not mean that the individual particles are at rest; they may be moving in such a way that their momenta cancel out.

(iii) Parabolic path.

In projectile motion the only external force acting on the body is gravity, which is constant. Since the centre of mass moves as if the whole mass were concentrated there and the external force applied there, it moves under a uniform acceleration. The horizontal component of the velocity remains constant while the vertical component changes uniformly, and the resulting path is a parabola.

(iv) Greater mass.

For a two-particle system,

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

This is a weighted average in which each position is weighted by its own mass. The heavier particle exerts a stronger influence on the average, so the centre of mass lies closer to it.

(v) The centre of mass is above the centre of gravity

In the case of a mountain the gravitational field is stronger closer to the earth's surface. The lower portions of the mountain therefore contribute more weight than they contribute mass, and this shifts the centre of gravity slightly downward compared with the centre of mass. Hence the centre of mass lies above the centre of gravity.

(i) To store more rotational energy and smooth out fluctuations in angular velocity.

The torque rotating the shaft of an engine changes periodically, so the shaft cannot rotate uniformly. A flywheel of large moment of inertia is attached to the shaft, and because of this large moment of inertia the flywheel continues to rotate almost uniformly in spite of the changing torque. This counteracts the intermittent power delivery from the engine pistons and keeps the rotational speed steady.

(ii) It doubles.

The rotational kinetic energy stored in a flywheel is

$\text K_{rot} = \frac{1}{2}\text I \omega^2$

If the moment of inertia is doubled while ω is kept the same,

$\text K'_{rot} = \frac{1}{2}(2\text I)\omega^2 = 2\left(\frac{1}{2}\text I \omega^2\right) = 2\text K_{rot}$

(iii) It will have more fluctuations in angular velocity

A flywheel with a smaller moment of inertia stores less rotational energy, so it is less able to carry the shaft through the parts of the cycle in which the driving torque is small. The angular velocity therefore fluctuates more during the engine's cycle, making the engine run less smoothly.

(iv) The tyre

The moment of inertia depends on both the mass and the square of its distance from the axis of rotation. The tyre is farthest from the axle and also carries a significant part of the mass of the wheel, so it contributes most to the moment of inertia. The spokes are distributed over smaller distances and the axle lies on the axis itself, where r = 0.

(v) Reduce the mass of the tyre

The moment of inertia of a wheel may be written as I = kmr². Reducing the mass of the tyre decreases I, and from τ = Iα a smaller moment of inertia gives a greater angular acceleration for the same applied torque, so the wheel spins faster with the same push. Decreasing the size of the wheel would also reduce I, but in practice this is often not feasible since it would require altering the entire frame and gear system of the vehicle.

(i) Can lie outside the material body.

The centre of mass depends solely on the distribution of mass and the positions of the masses, and not on the physical boundaries of any component of the system. Hence for irregular objects, such as a ring or an L-shaped body, the centre of mass may lie in the empty space near the structure rather than within the material itself.

(ii) The linear velocity of the centre of mass is equal to the product of angular velocity and radius.

For pure rolling motion without slipping, in one complete rotation the body travels a distance s = 2πR in a time $\text T = \frac{2\pi}{\omega}$. Therefore

$\text v_{cm} = \frac{\text s}{\text T} = \frac{2\pi \text R}{2\pi/\omega} = \text R\omega$

This is precisely the condition for rolling without slipping, which ensures that the point of contact is momentarily at rest relative to the surface.

(iii) Solid sphere.

For a body rolling down an incline the acceleration of the centre of mass is

$\text a = \frac{\text g \sin \theta}{1 + \frac{\text I}{\text{mR}^2}}$

For a solid sphere $\frac{\text I}{\text{mR}^2} = \frac{2}{5}$, giving $\text a = \frac{5}{7}\text g \sin \theta$, while for a hollow sphere $\frac{\text I}{\text{mR}^2} = \frac{2}{3}$, giving $\text a = \frac{3}{5}\text g \sin \theta$. Since $\frac{5}{7} \gt \frac{3}{5}$, the solid sphere has the greater acceleration and reaches the bottom first.

(iv) The moment of inertia decreases, and the radius of gyration decreases.

By pulling her arms in, the skater brings more mass closer to the axis of rotation, so the moment of inertia decreases. Since I = MK² and the mass M is unchanged, a decrease in I means a decrease in K as well. Both the moment of inertia and the radius of gyration therefore decrease.

*Note: Options 1 and 3 are conceptually identical. Both are technically correct, though the source book lists Option 3.

(v) Her rotational speed increases because her moment of inertia decreases.

By tucking in her body the gymnast brings her mass closer to the axis of rotation, reducing her moment of inertia. No external torque acts on her while she is in the air, so her angular momentum L = Iω is conserved. Since I decreases, ω must increase in order to keep the product constant, and she spins faster.

Centre of mass : Every physical system has associated with it a certain point whose motion characterises the motion of the whole system. When the system moves under some external force, this point moves as if the entire mass of the system were concentrated at this point and also the external force were applied at this point. This point is called the centre of mass of the system.

Expression for the centre of mass of a system of two particles :

Consider a system of two particles A and B of masses m₁ and m₂, with position vectors $\vec{\text r}_1$ and $\vec{\text r}_2$ with respect to an origin O. Let $\vec{\text F}_1$ and $\vec{\text F}_2$ be the external forces on them and $\vec{\text F}_{21}$, $\vec{\text F}_{12}$ the internal forces.

Since the internal forces obey Newton's third law of motion,

$\vec{\text F}_{21} = -\vec{\text F}_{12} \quad \text{or} \quad \vec{\text F}_{21} + \vec{\text F}_{12} = 0 \qquad \dots(\text i)$

The net force acting on the system is

$\vec{\text F} = (\vec{\text F}_1 + \vec{\text F}_{21}) + (\vec{\text F}_2 + \vec{\text F}_{12}) = \vec{\text F}_1 + \vec{\text F}_2 \qquad \dots(\text{ii})$

Applying Newton's second law to each particle,

$\vec{\text F}_1 + \vec{\text F}_{21} = \frac{\text d}{\text{dt}}(\text m_1 \vec{\text v}_1), \qquad \vec{\text F}_2 + \vec{\text F}_{12} = \frac{\text d}{\text{dt}}(\text m_2 \vec{\text v}_2)$

Adding these and using equations (i) and (ii),

$\vec{\text F} = \frac{\text d}{\text{dt}}(\text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2) = \frac{\text d^2}{\text{dt}^2}\left[\text m_1 \vec{\text r}_1 + \text m_2 \vec{\text r}_2\right]$

Multiplying and dividing the right hand side by (m₁ + m₂),

$\vec{\text F} = (\text m_1 + \text m_2)\frac{\text d^2}{\text{dt}^2}\left(\frac{\text m_1 \vec{\text r}_1 + \text m_2 \vec{\text r}_2}{\text m_1 + \text m_2}\right) \qquad \dots(\text{iii})$

Now if a point of mass (m₁ + m₂) situated at C with position vector $\vec{\text r}_{cm}$ is acted upon by the net force $\vec{\text F}$, then by Newton's second law,

$\vec{\text F} = (\text m_1 + \text m_2)\vec{\text a}_{cm} = (\text m_1 + \text m_2)\frac{\text d^2 \vec{\text r}_{cm}}{\text{dt}^2} \qquad \dots(\text{iv})$

Comparing equations (iii) and (iv),

$\vec{\text r}_{cm} = \frac{\text m_1 \vec{\text r}_1 + \text m_2 \vec{\text r}_2}{\text m_1 + \text m_2}$

In Cartesian components, comparing the coefficients of $\hat{\text i}$ and $\hat{\text j}$,

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}, \qquad \text y_{cm} = \frac{\text m_1 \text y_1 + \text m_2 \text y_2}{\text m_1 + \text m_2}$

Physical significance of the centre of mass : The centre of mass allows the overall translatory motion of a composite body to be described without bothering about each individual particle of the body. The entire mass may be replaced by a single point mass located at the centre of mass, and the usual laws of particle kinematics can then be applied. The motion of the system is thus described completely in terms of the motion of its centre of mass, since the internal forces cancel in pairs and only the external force determines its acceleration.

Consider a system of n particles of masses m₁, m₂, ..., m_n moving with velocities $\vec{\text v}_1, \vec{\text v}_2, ..., \vec{\text v}_n$ respectively.

The total linear momentum of the system is the vector sum of the momenta of the individual particles,

$$\begin{aligned}\vec{\text p} = \vec{\text p}_1 + \vec{\text p}_2 + \dots + \vec{\text p}_n \\[1em] = \text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n \qquad \dots(\text i)\end{aligned}$$

By the definition of the velocity of the centre of mass of a system,

$\vec{\text v}_{cm} = \frac{1}{\text M}(\text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n) \qquad \dots(\text{ii})$

where M is the total mass of the system.

Multiplying both sides of equation (ii) by M,

$\text M\vec{\text v}_{cm} = \text M \times \frac{1}{\text M}(\text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n)$

$\text M\vec{\text v}_{cm} = \text m_1 \vec{\text v}_1 + \text m_2 \vec{\text v}_2 + \dots + \text m_n \vec{\text v}_n \qquad \dots(\text{iii})$

Comparing equations (i) and (iii),

$\vec{\text p} = \text M\vec{\text v}_{cm}$

Hence, the total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass.

Consequence : Differentiating this result with respect to time,

$\frac{\text{d}\vec{\text p}}{\text{dt}} = \text M\frac{\text{d}\vec{\text v}_{cm}}{\text{dt}} = \text M\vec{\text a}_{cm} = \vec{\text F}_{ext}$

If no external force acts on the system, $\vec{\text F}_{ext} = 0$, so $\vec{\text p} \space$is constant. This shows that the linear momentum of the centre of mass remains constant in the absence of an external force.

Consider a rigid body rotating with a constant angular acceleration α about an axis OY perpendicular to the plane of the paper. Let the body consist of n particles of masses m₁, m₂, ..., m_n at perpendicular distances r₁, r₂, ..., r_n from the axis of rotation.

As the body rotates, all these particles perform circular motion with the same angular acceleration α, but with different linear accelerations,

$\text a_1 = \text r_1 \alpha, \quad \text a_2 = \text r_2 \alpha, \quad \dots, \quad \text a_n = \text r_n \alpha$

The forces experienced by these particles are

$\text F_1 = \text m_1 \text a_1 = \text m_1 \text r_1 \alpha, \quad \text F_2 = \text m_2 \text r_2 \alpha, \quad \dots, \quad \text F_n = \text m_n \text r_n \alpha$

These forces are tangential, so their respective perpendicular distances from the axis are r₁, r₂, ..., r_n. The torque experienced by the first particle is

$\tau_1 = \text F_1 \text r_1 = \text m_1 \text r_1^2 \alpha$

Similarly, $\tau_2 = \text m_2 \text r_2^2 \alpha$, and so on. Since the rotation is restricted to a single plane, the directions of all these torques are the same and along the axis of rotation. Hence the magnitude of the resultant torque is

$$\begin{aligned}\tau = \tau_1 + \tau_2 + \dots + \tau_n \\[1em] = (\text m_1 \text r_1^2 + \text m_2 \text r_2^2 + \dots + \text m_n \text r_n^2)\alpha\end{aligned}$$

But

$\text m_1 \text r_1^2 + \text m_2 \text r_2^2 + \dots + \text m_n \text r_n^2 = \sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2 = \text I$

is the moment of inertia of the body about the axis of rotation. Therefore

$\tau = \text I \alpha$

Definition of moment of inertia from this relation : From τ = Iα,

$\text I = \frac{\tau}{\alpha}$

Putting α = 1 rad s^-2 gives I = τ. Hence, the moment of inertia of a particle or a rigid body about an axis of rotation is numerically equal to the torque capable of producing unit angular acceleration in the body about the same axis of rotation.

Consider a rigid body rotating about an axis with a uniform angular velocity ω. Let the body consist of particles of masses m₁, m₂, ..., m_n at perpendicular distances r₁, r₂, ..., r_n from the axis of rotation.

The angular velocity of all the particles is the same, but their linear velocities are different. Since the linear velocity of a particle is the product of its angular velocity and its distance from the axis of rotation, for the first particle

$\text v_1 = \text r_1 \omega$

The kinetic energy of this particle is

$\frac{1}{2}\text m_1 \text v_1^2 = \frac{1}{2}\text m_1 (\text r_1 \omega)^2 = \frac{1}{2}\text m_1 \text r_1^2 \omega^2$

Similarly, the kinetic energies of the other particles are $\frac{1}{2}\text m_2 \text r_2^2 \omega^2$, $\frac{1}{2}\text m_3 \text r_3^2 \omega^2$, and so on.

The kinetic energy K of the whole body is the sum of the kinetic energies of all its particles,

$\text K = \frac{1}{2}\text m_1 \text r_1^2 \omega^2 + \frac{1}{2}\text m_2 \text r_2^2 \omega^2 + \dots + \frac{1}{2}\text m_n \text r_n^2 \omega^2$

$= \frac{1}{2}(\text m_1 \text r_1^2 + \text m_2 \text r_2^2 + \dots + \text m_n \text r_n^2)\omega^2$

But $\sum\limits_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2$ is the moment of inertia I of the body about the axis of rotation. Therefore

$\text K = \frac{1}{2}\text I \omega^2$

This is the formula for the kinetic energy of uniform rotation. Just as the kinetic energy of linear motion $\frac{1}{2}\text{mv}^2$ is half the product of the mass and the square of the linear velocity, the kinetic energy of rotation is half the product of the moment of inertia and the square of the angular velocity.

Definition of moment of inertia on this basis : From the above formula,

$\text I = \frac{2\text K}{\omega^2}$

If ω = 1 rad s^-1, then I = 2K. Hence, the moment of inertia of a body rotating about an axis with unit angular velocity is equal to twice its kinetic energy of rotation about that axis.

Angular momentum : The quantity in rotational mechanics analogous to linear momentum is called angular momentum. It is also defined as the moment of linear momentum, which is similar to the torque being the moment of a force.

If $\vec{\text p} \space$is the instantaneous linear momentum of a particle executing circular motion and $\vec{\text r} \space$its position vector, then

$\vec{\text L} = \vec{\text r} \times \vec{\text p}$

and its magnitude is

$\text L = \text{pr}\sin \theta$

Its S.I. unit is kg m² s^-1 and its dimensional formula is [ML²T^-1].

Expression for the angular momentum : For a body of moment of inertia I rotating with angular velocity ω about an axis, consider a particle of mass m₁ at a distance r₁ from the axis. Its linear velocity is v₁ = r₁ω and its linear momentum is p₁ = m₁v₁. The moment of this momentum about the axis is

$\text m_1 \text v_1 \times \text r_1 = \text m_1 (\text r_1 \omega) \times \text r_1 = \text m_1 \text r_1^2 \omega$

Summing over all the particles of the body,

$\text L = (\text m_1 \text r_1^2 + \text m_2 \text r_2^2 + \dots + \text m_n \text r_n^2)\omega = \left(\sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2\right)\omega$

$\boxed{\text L = \text I \omega}$

Expression for the kinetic energy of rotation :

$\text K = \frac{1}{2}\text I \omega^2$

The two are related by $\text K = \frac{\text L^2}{2\text I}$.

In rotational motion the angular displacement θ, the angular velocity ω and the angular acceleration α are the analogues of the linear displacement s, the linear velocity v and the linear acceleration a respectively.

Let a body rotate about an axis with an initial angular velocity ω₀ under a constant angular acceleration α, and let it be displaced through an angle θ in time t, acquiring a final angular velocity ω. The three equations of rotational motion, corresponding to the three equations of linear motion, are :

Linear motion	Rotational motion
v = u + at	ω = ω0 + αt
$\text s = \text{ut} + \frac{1}{2}\text{at}^2$	$\theta = \omega_0 \text t + \frac{1}{2}\alpha \text t^2$
v2 = u2 + 2as	ω2 = ω02 + 2αθ

(i) First equation : By the definition of angular acceleration, $\alpha = \frac{\text d\omega}{\text{dt}}$, so dω = α dt. Integrating between the limits ω₀ at t = 0 and ω at time t,

$\int_{\omega_0}^{\omega}\text d\omega = \alpha \int_0^{\text t}\text{dt} \quad \Rightarrow \quad \omega - \omega_0 = \alpha \text t$

$\omega = \omega_0 + \alpha \text t$

(ii) Second equation : Since $\omega = \frac{\text d\theta}{\text{dt}}$, we have dθ = ω dt = (ω₀ + αt)dt. Integrating,

$\int_0^{\theta}\text d\theta = \int_0^{\text t}(\omega_0 + \alpha \text t)\text{dt} = \left[\omega_0 \text t + \frac{1}{2}\alpha \text t^2\right]_0^{\text t}$

$\theta = \omega_0 \text t + \frac{1}{2}\alpha \text t^2$

(iii) Third equation : Eliminating t between the first two equations, $\text t = \frac{\omega - \omega_0}{\alpha}$, and substituting in the second,

$\theta = \omega_0\left(\frac{\omega - \omega_0}{\alpha}\right) + \frac{1}{2}\alpha\left(\frac{\omega - \omega_0}{\alpha}\right)^2$

$2\alpha \theta = 2\omega_0 \omega - 2\omega_0^2 + \omega^2 - 2\omega \omega_0 + \omega_0^2 = \omega^2 - \omega_0^2$

$\omega^2 = \omega_0^2 + 2\alpha \theta$

Theorem of perpendicular axes : The moment of inertia of a uniform plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about any two mutually perpendicular axes in its plane intersecting on the first axis.

Proof : Let OZ be the axis perpendicular to the plane of the lamina about which the moment of inertia is to be taken. Let OX and OY be two mutually perpendicular axes in the plane of the lamina, intersecting on OZ.

Consider a particle P of mass m at a distance r from OZ. The moment of inertia of this particle about OZ is mr². Therefore the moment of inertia of the whole lamina about OZ is

$\text I_z = \sum \text{mr}^2$

If x and y are the distances of P from OY and OX respectively, then

$\text r^2 = \text x^2 + \text y^2$

Substituting this value,

$\text I_z = \sum \text m(\text x^2 + \text y^2) = \sum \text{mx}^2 + \sum \text{my}^2$

But $\sum \text{mx}^2$ is the moment of inertia I_y of the lamina about OY, and $\sum \text{my}^2$ is the moment of inertia I_x of the lamina about OX. Therefore

$\boxed{\text I_z = \text I_x + \text I_y}$

Applicability : The theorem holds only for planar (two-dimensional) objects such as thin discs, rings and triangular plates, because its proof requires every mass element to lie in a single plane. It cannot be applied to three-dimensional objects like spheres, cylinders and cones, whose mass is distributed in all three dimensions.

Angular momentum : The quantity in rotational mechanics analogous to linear momentum is called angular momentum. It is defined as the moment of linear momentum, that is, the product of the linear momentum of a particle and its perpendicular distance from the axis of rotation.

In vector form,

$\vec{\text L} = \vec{\text r} \times \vec{\text p}$

whose magnitude is L = pr sin θ, where θ is the angle between $\vec{\text r} \space$and $\vec{\text p}$. Angular momentum is an axial vector, its direction being given by the right hand thumb rule.

Relation between angular momentum and moment of inertia : Let a body be rotating about an axis with angular velocity ω. All the particles of the body have the same angular velocity, but their linear velocities are different.

Consider a particle of mass m₁ at a perpendicular distance r₁ from the axis of rotation. Its linear velocity is

$\text v_1 = \text r_1 \omega$

so its linear momentum is p₁ = m₁v₁. The moment of this linear momentum about the axis of rotation is

$\text L_1 = \text p_1 \times \text r_1 = \text m_1 (\text r_1 \omega) \times \text r_1 = \text m_1 \text r_1^2 \omega$

Similarly, for the other particles of masses m₂, m₃, ..., m_n at distances r₂, r₃, ..., r_n, the moments of their linear momenta are $\text m_2 \text r_2^2 \omega$, $\text m_3 \text r_3^2 \omega$, ..., $\text m_n \text r_n^2 \omega$.

The angular momentum of the body is the sum of the moments of the linear momenta of all its particles,

$\text L = \text m_1 \text r_1^2 \omega + \text m_2 \text r_2^2 \omega + \dots + \text m_n \text r_n^2 \omega$

$= (\text m_1 \text r_1^2 + \text m_2 \text r_2^2 + \dots + \text m_n \text r_n^2)\omega = \left(\sum_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2\right)\omega$

But $\sum\limits_{\text i = 1}^{\text n} \text m_\text i \text r_\text i^2$ is the moment of inertia I of the body about the axis of rotation. Hence

$\boxed{\text L = \text I \omega}$

Just as the linear momentum of a body is the product of its mass and its linear velocity, the angular momentum of a body about an axis is the product of its moment of inertia and its angular velocity about that axis.

Relation between torque and angular momentum :

Let a rigid body be rotating about an axis under the action of a torque τ, producing an angular acceleration $\alpha = \frac{\text d\omega}{\text{dt}}$ about the same axis. Let its moment of inertia about the given axis be I.

Then,

$\tau = \text I \alpha = \text I\frac{\text d\omega}{\text{dt}} \qquad \dots(\text i)$

The angular momentum of the body about the same axis is

$\text L = \text I \omega \qquad \dots(\text{ii})$

Differentiating equation (ii) with respect to time, the moment of inertia I being constant,

$\frac{\text{dL}}{\text{dt}} = \text I\frac{\text d\omega}{\text{dt}}$

Comparing this with equation (i),

$\boxed{\frac{\text{dL}}{\text{dt}} = \tau}$

Hence, the rate of change of angular momentum is equal to the torque acting on the body. This is similar to $\text F = \frac{\text{dp}}{\text{dt}}$, which is Newton's second law for linear motion, and so it is Newton's second law for rotational dynamics.

Principle of conservation of angular momentum :

In the above relation, if the external torque acting on the body is zero, that is, τ = 0, then

$\frac{\text{dL}}{\text{dt}} = 0$

which gives

$\text L = \text{a constant}$

Since L = Iω, in the absence of an external torque

$\text I \omega = \text{a constant} \quad \text{or} \quad \text I_1 \omega_1 = \text I_2 \omega_2$

Statement : The total angular momentum of a system remains constant if no external torque acts on the system.

Illustration : When a figure skater spinning with her arms extended pulls her arms in, her moment of inertia decreases from I₁ to I₂. To conserve angular momentum her angular velocity increases from ω₁ to ω₂ such that I₁ω₁ = I₂ω₂, and she spins faster with her arms pulled in.

Conditions of equilibrium of a rigid body : A rigid body can execute translational as well as rotational motion, or both at the same time. Therefore, for a rigid body to be in static equilibrium, it must satisfy the conditions of both translational and rotational equilibrium simultaneously.

(i) Condition of translational equilibrium : The net force acting on the body must be zero,

$\vec{\text F}_1 + \vec{\text F}_2 + \dots + \vec{\text F}_n = 0 \quad \text{or} \quad \sum \vec{\text F} = 0$

If all the forces lie in the X-Y plane, this gives the two scalar equations

$\sum \text F_x = 0 \quad \text{and} \quad \sum \text F_y = 0$

Under this condition the body is either at rest or its centre of mass moves with a constant velocity.

(ii) Condition of rotational equilibrium : The net torque on the body about any axis must be zero,

$\sum \vec{\tau} = 0$

that is, the sum of the anticlockwise moments about any axis must equal the sum of the clockwise moments about that axis. Since $\vec{\tau} = \text I\vec{\alpha}$, this gives $\vec{\alpha} = 0$, so the angular velocity ω, and hence the angular momentum L = Iω, remains constant.

Equilibrium of a uniform leaning ladder :

Let a uniform ladder AB lean against a smooth vertical wall and rest on a rough horizontal floor. Since the ladder is uniform, its centre of gravity G is exactly at its middle, where its weight W acts vertically downwards.

The wall is smooth, so the reaction R_B exerted by the wall on the ladder is horizontal and normal to the wall. The reaction R_A exerted by the ground is the vector sum of a normal force N and the force of friction f. The force of friction acts towards the wall, since the tendency of motion of the point A of the ladder is away from the wall.

If R_A makes an angle θ with the floor, then

$\text f = \text R_A \cos \theta \qquad \dots(\text i)$

$\text N = \text R_A \sin \theta \qquad \dots(\text{ii})$

Applying the condition of translational equilibrium, ΣF_x = 0 and ΣF_y = 0,

$\text f - \text R_B = 0 \quad \text{or} \quad \text f = \text R_B \qquad \dots(\text{iii})$

$\text N - \text W = 0 \quad \text{or} \quad \text N = \text W \qquad \dots(\text{iv})$

Applying the condition of rotational equilibrium, Στ = 0, and taking moments of the forces about the point A,

$-\text W \times \text{AD} + \text R_B \times \text{BC} = 0$

$\text R_B \times \text{BC} = \text W \times \text{AD} \qquad \dots(\text v)$

where AD is the horizontal distance of the centre of gravity from A and BC is the height of the point of contact B above the floor. Solving these equations, the unknown quantities R_A, R_B, N, f and θ can be obtained.

Theorem of parallel axes : The moment of inertia I of a body about any axis is equal to its moment of inertia I_cm about a parallel axis through its centre of mass, plus the product of the mass M of the body and the square of the perpendicular distance a between the two axes,

$\text I = \text I_{cm} + \text{Ma}^2$

Proof : Let C be the centre of mass of a plane lamina. Let I be its moment of inertia about an axis AB in its plane, and I_cm its moment of inertia about a parallel axis EF passing through C. Let the distance between EF and AB be a.

Consider a particle P of mass m at a distance r from EF. Its distance from AB is (r + a), so its moment of inertia about AB is m(r + a)². Therefore the moment of inertia of the whole lamina about AB is

$\text I = \sum \text m(\text r + \text a)^2 = \sum \text m(\text r^2 + \text a^2 + 2\text{ar})$

$= \sum \text{mr}^2 + \sum \text{ma}^2 + \sum 2\text{amr}$

Since a is a constant, it can be taken outside the summation,

$\text I = \sum \text{mr}^2 + \text a^2 \sum \text m + 2\text a\sum \text{mr} \qquad \dots(\text i)$

Now,

• $\sum \text{mr}^2 = \text I_{cm}$, the moment of inertia of the lamina about EF
• $\text a^2 \sum \text m = \text{Ma}^2$, where M is the total mass of the lamina
• $\sum \text{mr} = 0$, because the sum of the moments of all the mass-particles of a body about an axis through the centre of mass is zero

Making these substitutions in equation (i),

$\boxed{\text I = \text I_{cm} + \text{Ma}^2}$

Axis about which the moment of inertia is minimum : Since M and a² are always positive quantities, the term Ma² is always positive, so I is always greater than I_cm. The moment of inertia is least when a = 0, that is, when the axis passes through the centre of mass.

Hence, the moment of inertia of a body is minimum about an axis passing through its centre of mass.

Given,

• m₁ = 1 g at (1, 2)
• m₂ = 2 g at (0, − 1)
• m₃ = 3 g at (2, − 3)

The coordinates of the centre of mass of a system of particles are

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2 + \text m_3 \text x_3}{\text m_1 + \text m_2 + \text m_3}, \qquad \text y_{cm} = \frac{\text m_1 \text y_1 + \text m_2 \text y_2 + \text m_3 \text y_3}{\text m_1 + \text m_2 + \text m_3}$

x-coordinate :

$\text x_{cm} = \frac{(1)(1) + (2)(0) + (3)(2)}{1 + 2 + 3} = \frac{1 + 0 + 6}{6} = \frac{7}{6}$

y-coordinate :

$\text y_{cm} = \frac{(1)(2) + (2)(-1) + (3)(-3)}{6} = \frac{2 - 2 - 9}{6} = -\frac{9}{6} = -\frac{3}{2}$

Hence, the co-ordinates of the centre of mass of the system are $\left(\frac{7}{6},\ -\frac{3}{2}\right)$.

Given,

• m₁ = 2 g, m₂ = 3 g and m₃ = 4 g at the vertices of an equilateral triangle
• Side of the triangle = 1 m

Let the triangle lie in the X-Y plane with m₁ at the origin and the side m₁m₂ along the X-axis. Then the coordinates of the three masses are

$\text m_1 : (0,\ 0), \qquad \text m_2 : (1,\ 0), \qquad \text m_3 : \left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)$

since the height of an equilateral triangle of side 1 m is $\frac{\sqrt{3}}{2}$ m.

x-coordinate of the centre of mass :

$\text x_{cm} = \frac{(2)(0) + (3)(1) + (4)\left(\frac{1}{2}\right)}{2 + 3 + 4} = \frac{0 + 3 + 2}{9} = \frac{5}{9}\ \text m$

y-coordinate of the centre of mass :

$\text y_{cm} = \frac{(2)(0) + (3)(0) + (4)\left(\frac{\sqrt{3}}{2}\right)}{9} = \frac{2\sqrt{3}}{9}\ \text m$

Hence, the centre of mass of the system is at $\left(\frac{5}{9}\ \text m,\ \frac{2\sqrt{3}}{9}\ \text m\right)$.

Given,

• m₁ = 2 g, m₂ = 2 g, m₃ = 1 g, m₄ = 1 g
• Side of the square = L

Taking m₁ at the origin, with the sides of the square along the X- and Y-axes, the coordinates of the four masses are

$\text m_1 : (0,\ 0), \quad \text m_2 : (\text L,\ 0), \quad \text m_3 : (\text L,\ \text L), \quad \text m_4 : (0,\ \text L)$

x-coordinate of the centre of mass :

$\text x_{cm} = \frac{(2)(0) + (2)(\text L) + (1)(\text L) + (1)(0)}{2 + 2 + 1 + 1}$

$= \frac{0 + 2\text L + \text L + 0}{6} = \frac{3\text L}{6} = \frac{\text L}{2}$

y-coordinate of the centre of mass :

$\text y_{cm} = \frac{(2)(0) + (2)(0) + (1)(\text L) + (1)(\text L)}{6}$

$= \frac{2\text L}{6} = \frac{\text L}{3}$

Hence, the centre of mass of the system with respect to m₁ is at $\left(\frac{\text L}{2},\ \frac{\text L}{3}\right)$.

Given,

• Moment of inertia of the ring, I = 0.40 kg m²
• Initial rate of rotation = 2100 revolutions per minute
• Time to stop, t = 2.0 s
• Final angular velocity, ω = 0

The initial angular velocity is

$\omega_0 = 2\pi \text n = 2\pi \times \frac{2100}{60} = 2\pi \times 35 = 70\pi\ \text{rad s}^{-1}$

$\omega_0 = 70 \times 3.14 = 219.8\ \text{rad s}^{-1}$

Angular retardation : From ω = ω₀ + αt,

$0 = 219.8 + \alpha(2.0) \quad \Rightarrow \quad \alpha = -\frac{219.8}{2.0} = -109.9\ \text{rad s}^{-2}$

Torque required :

$\tau = \text I \alpha = 0.40 \times (-109.9) = -43.96 \approx -44\ \text{N m}$

The negative sign shows that the torque is opposite to the direction of rotation.

Work done : The work done is equal to the change in the kinetic energy of rotation. Since the ring finally comes to rest,

$\text W = 0 - \frac{1}{2}\text I \omega_0^2 = -\frac{1}{2} \times 0.40 \times (219.8)^2$

$= -\frac{1}{2} \times 0.40 \times 48312 = -9662\ \text J$

Hence, the torque required to stop the ring is 44 N m and the work done is 9680 J (nearly), the negative sign indicating that the work is done against the rotation.

Given,

• Moment of inertia of the ring, I = 3 kg m²
• Torque applied, τ = 6 N m
• Time, t = 20 s
• Initial angular velocity, ω₀ = 0 (starts from rest)

The angular acceleration produced is

$\alpha = \frac{\tau}{\text I} = \frac{6}{3} = 2\ \text{rad s}^{-2}$

The angular displacement in 20 s is

$\theta = \omega_0 \text t + \frac{1}{2}\alpha \text t^2 = 0 + \frac{1}{2} \times 2 \times (20)^2$

$= \frac{1}{2} \times 2 \times 400 = 400\ \text{rad}$

The work done by a torque in rotatory motion is

$\text W = \tau \theta = 6 \times 400$

$\text W = 2400\ \text J$

Hence, the work done is 2400 J.

Given,

• Diameter of the disc = 0.4 m, so radius R = 0.2 m
• Mass of the disc, M = 5 kg
• Rate of rotation, n = 28 rev/s

The moment of inertia of a disc about its own axis is

$$\begin{aligned}\text I = \frac{1}{2}\text{MR}^2 = \frac{1}{2} \times 5 \times (0.2)^2 \\[1em] = \frac{1}{2} \times 5 \times 0.04 = 0.1\ \text{kg m}^2\end{aligned}$$

The angular velocity is

$\omega = 2\pi \text n = 2 \times \frac{22}{7} \times 28 = 176\ \text{rad s}^{-1}$

(i) Angular momentum :

$\text L = \text I \omega = 0.1 \times 176 = 17.6\ \text{kg m}^2\text s^{-1}$

(ii) Rotational kinetic energy :

$\text K = \frac{1}{2}\text I \omega^2 = \frac{1}{2} \times 0.1 \times (176)^2$

$= \frac{1}{2} \times 0.1 \times 30976 = 1548.8\ \text J$

Hence, the angular momentum of the disc is 17.6 kg m² s^-1 and its rotational kinetic energy is 1548.8 J.

Note: The textbook answer key gives 70.4 kg m² s^-1 and 6195.2 J, which are exactly four times these values. Those figures correspond to taking 0.4 m as the radius of the disc rather than its diameter, which does not agree with the data stated in the question.

Given,

• Rate of rotation = 1000 rotations/minute
• Kinetic energy of rotation, K = 10⁶ J

The angular velocity of the wheel is

$\omega = 2\pi \text n = 2\pi \times \frac{1000}{60} = \frac{100\pi}{3}\ \text{rad s}^{-1}$

$\omega = \frac{100 \times 3.14}{3} = 104.7\ \text{rad s}^{-1}$

The kinetic energy of rotation is

$\text K = \frac{1}{2}\text I \omega^2$

Rearranging for the moment of inertia,

$\text I = \frac{2\text K}{\omega^2}$

Substituting the values,

$\text I = \frac{2 \times 10^6}{(104.7)^2} = \frac{2 \times 10^6}{10962}$

$\text I = 182.4\ \text{kg m}^2$

Hence, the moment of inertia of the wheel about the axis of rotation is 182 kg m².

Given,

• Mass of the body, m = 50 g = 0.05 kg
• Distance of the centre of mass from the axis of rotation, r = 50 cm = 0.5 m

The body revolves in a circular path, so it may be treated as a point mass at a distance r from the axis of rotation. The moment of inertia of a point mass about an axis is

$\text I = \text{mr}^2$

Substituting the values,

$\text I = 0.05 \times (0.5)^2 = 0.05 \times 0.25$

$\text I = 1.25 \times 10^{-2}\ \text{kg m}^2$

Hence, the moment of inertia of the body is 1.25 × 10^-2 kg m².

Given,

• m₁ = 3.0 kg at the 20 cm mark
• m₂ = 5.0 kg at the 70 cm mark
• The metre scale is light, so its own mass may be neglected

The moment of inertia of a system of point masses about an axis is I = Σm_ir_i².

(i) About an axis through the 0 cm mark : The distances of the two masses from this axis are

$\text r_1 = 20\ \text{cm} = 0.20\ \text m, \qquad \text r_2 = 70\ \text{cm} = 0.70\ \text m$

$\text I_1 = \text m_1 \text r_1^2 + \text m_2 \text r_2^2 = 3.0(0.20)^2 + 5.0(0.70)^2$

$= 3.0 \times 0.04 + 5.0 \times 0.49 = 0.12 + 2.45$

$\text I_1 = 2.57\ \text{kg m}^2$

(ii) About an axis through the 100 cm mark : The distances of the two masses from this axis are

$\text r_1 = 100 - 20 = 80\ \text{cm} = 0.80\ \text m, \qquad \text r_2 = 100 - 70 = 30\ \text{cm} = 0.30\ \text m$

$\text I_2 = 3.0(0.80)^2 + 5.0(0.30)^2$

$= 3.0 \times 0.64 + 5.0 \times 0.09 = 1.92 + 0.45$

$\text I_2 = 2.37\ \text{kg m}^2$

Hence, the moment of inertia is 2.57 kg m² about the 0 cm mark and 2.37 kg m² about the 100 cm mark.

Given,

• Moment of inertia of the flywheel, I = 4 kg m²
• Torque applied, τ = 10 N m

From the relation between torque, moment of inertia and angular acceleration,

$\tau = \text I \alpha$

Rearranging,

$\alpha = \frac{\tau}{\text I}$

Substituting the values,

$\alpha = \frac{10}{4} = 2.5\ \text{rad s}^{-2}$

Hence, the angular acceleration produced in the flywheel is 2.5 rad s^-2.

Given,

• Three identical spheres, each of radius R, placed touching one another on a horizontal table

The centre of mass of each sphere is at its own centre. Since the spheres touch one another, the distance between the centres of any two of them is 2R. Hence the problem reduces to finding the centre of mass of three equal masses m placed at the corners A, B and C of an equilateral triangle of side 2R.

Taking A as the origin and AB along the X-axis, the coordinates of the three centres are

$\text A : (0,\ 0), \qquad \text B : (2\text R,\ 0), \qquad \text C : \left(\text R,\ \sqrt{3}\text R\right)$

since the height of an equilateral triangle of side 2R is $\sqrt{3}\text R$.

x-coordinate of the centre of mass :

$\text x_{cm} = \frac{\text m(0) + \text m(2\text R) + \text m(\text R)}{3\text m} = \frac{3\text R}{3} = \text R$

y-coordinate of the centre of mass :

$\text y_{cm} = \frac{\text m(0) + \text m(0) + \text m(\sqrt{3}\text R)}{3\text m} = \frac{\sqrt{3}\text R}{3} = \frac{\text R}{\sqrt{3}}$

Hence, the centre of mass of the system is located at $\left(\text R,\ \frac{\text R}{\sqrt{3}}\right)$ relative to A. Since the three masses are equal, this point is the centroid of the equilateral triangle formed by the three centres.

Given,

• Angular momentum of the body, L = 31.4 J s
• Rate of revolution, n = 10 cycles per second

The angular velocity of the body is

$\omega = 2\pi \text n = 2 \times 3.14 \times 10 = 62.8\ \text{rad s}^{-1}$

The angular momentum of a body rotating about an axis is

$\text L = \text I \omega$

Rearranging for the moment of inertia,

$\text I = \frac{\text L}{\omega}$

Substituting the values,

$\text I = \frac{31.4}{62.8} = 0.5\ \text{kg m}^2$

Hence, the moment of inertia of the body about the axis of rotation is 0.5 kg m².

Given,

• Mass of the body, m = 1 kg
• Radius of the circle (length of the string), r = 1 m
• Frequency of revolution, n = 20 rev/s

The angular velocity of the body is

$\omega = 2\pi \text n = 2\pi \times 20 = 40\pi\ \text{rad s}^{-1}$

(i) Moment of inertia : The body may be treated as a point mass at a distance r from the axis,

$\text I = \text{mr}^2 = 1 \times (1)^2 = 1\ \text{kg m}^2$

(ii) Angular momentum :

$\text L = \text I \omega = 1 \times 40\pi = 40\pi\ \text{kg m}^2\text s^{-1}$

(iii) Rotational kinetic energy :

$$\begin{aligned}\text K = \frac{1}{2}\text I \omega^2 = \frac{1}{2} \times 1 \times (40\pi)^2 \\[1em] = \frac{1}{2} \times 1600\pi^2 = 800\pi^2\ \text J\end{aligned}$$

(iv) Centripetal force :

$$\begin{aligned}\text F = \text{mr}\omega^2 = 1 \times 1 \times (40\pi)^2 \\[1em] = 1600\pi^2\ \text N\end{aligned}$$

Hence, the moment of inertia is 1 kg m², the angular momentum is 40π kg m² s^-1, the rotational kinetic energy is 800π² J and the centripetal force is 1600π² N.

Given,

• Mass of each hydrogen atom, m = 1.7 × 10^-27 kg
• Inter-atomic distance, d = 4 × 10^-10 m

In a hydrogen molecule there are two atoms of equal mass, and the distance between them is the inter-atomic distance d. Since the two atoms are identical, the centre of mass of the molecule lies exactly midway between them. Hence the distance of each atom from the axis passing through the centre of mass is

$\text r = \frac{\text d}{2} = \frac{4 \times 10^{-10}}{2} = 2 \times 10^{-10}\ \text m$

The moment of inertia of the molecule about this axis is the sum of the moments of inertia of the two atoms,

$\text I = \text{mr}^2 + \text{mr}^2 = 2\text{mr}^2$

Substituting the values,

$\text I = 2 \times (1.7 \times 10^{-27}) \times (2 \times 10^{-10})^2$

$= 2 \times 1.7 \times 10^{-27} \times 4 \times 10^{-20}$

$\text I = 13.6 \times 10^{-47}\ \text{kg m}^2$

Hence, the moment of inertia of the hydrogen molecule about the given axis is 13.6 × 10^-47 kg m².

Given,

• Mass of each particle, m = 10 g = 0.01 kg
• Side of the equilateral triangle, a = 5 cm = 0.05 m

The axis passes through one corner of the triangle and is perpendicular to its plane. Therefore

• The particle at that corner lies on the axis, so its distance from the axis is zero.
• The other two particles are each at a distance equal to the side of the triangle, that is, a = 0.05 m.

The moment of inertia of the system is

$\text I = \text m(0)^2 + \text{ma}^2 + \text{ma}^2 = 2\text{ma}^2$

Substituting the values,

$\text I = 2 \times 0.01 \times (0.05)^2 = 2 \times 0.01 \times 0.0025$

$\text I = 5 \times 10^{-5}\ \text{kg m}^2$

Hence, the moment of inertia of the system about the given axis is 5 × 10^-5 kg m².

Given,

• Mass of each sphere, m = 2 kg
• Initial angular velocity, ω₁ = 10 rad/s
• Initial distance of each sphere from the axis, r₁ = 1 m
• Final distance of each sphere from the axis, r₂ = 0.3 m
• Moment of inertia of the person and the table, I₀ = 1 kg m²

Initial moment of inertia of the whole system :

$\text I_1 = \text I_0 + 2\text{mr}_1^2 = 1 + 2 \times 2 \times (1)^2 = 1 + 4 = 5\ \text{kg m}^2$

Initial kinetic energy of rotation :

$\text K_1 = \frac{1}{2}\text I_1 \omega_1^2 = \frac{1}{2} \times 5 \times (10)^2 = 250\ \text J$

Final moment of inertia of the system :

$$\begin{aligned}\text I_2 = \text I_0 + 2\text{mr}_2^2 = 1 + 2 \times 2 \times (0.3)^2 \\[1em] = 1 + 4 \times 0.09 = 1.36\ \text{kg m}^2\end{aligned}$$

No external torque acts on the system, so the angular momentum is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2 \quad \Rightarrow \quad \omega_2 = \frac{5 \times 10}{1.36} = 36.76\ \text{rad/s}$

Final kinetic energy of rotation :

$\text K_2 = \frac{1}{2}\text I_2 \omega_2^2 = \frac{1}{2} \times 1.36 \times (36.76)^2$

$= \frac{1}{2} \times 1.36 \times 1351.3 = 919\ \text J$

Hence, the initial kinetic energy is 250 J and the new kinetic energy is 919 J.

Explanation of the difference : The angular momentum is conserved, but the kinetic energy is not. To pull the spheres inwards the person must do work against the centrifugal effect. This work is done at the expense of the chemical energy stored in his muscles, and it appears as the increase in the kinetic energy of rotation of the system.

Given,

• Time for one complete revolution initially, T₁ = 4 s
• Final radius, $\text r_2 = \frac{\text r}{2}$, where r is the initial radius

The ball is treated as a point mass, so its moments of inertia before and after are

$\text I_1 = \text{mr}^2 \quad \text{and} \quad \text I_2 = \text m\left(\frac{\text r}{2}\right)^2 = \frac{\text{mr}^2}{4}$

The angular velocities are

$\omega_1 = \frac{2\pi}{\text T_1} \quad \text{and} \quad \omega_2 = \frac{2\pi}{\text T_2}$

The cord is pulled inwards along the radius, so it exerts no torque about the axis. Hence the angular momentum is conserved,

$\text I_1 \omega_1 = \text I_2 \omega_2$

Substituting the values,

$\text{mr}^2 \times \frac{2\pi}{\text T_1} = \frac{\text{mr}^2}{4} \times \frac{2\pi}{\text T_2}$

Cancelling mr² and 2π from both sides,

$\frac{1}{\text T_1} = \frac{1}{4\text T_2} \quad \Rightarrow \quad \text T_2 = \frac{\text T_1}{4}$

$\text T_2 = \frac{4}{4} = 1\ \text s$

Hence, the ball will now take 1 s to complete one revolution.

Given,

• Weight of the ladder, W = 20 kg-wt = 20 × 9.8 = 196 N
• Length of the ladder, AB = 3 m
• Distance of the feet from the wall = 1 m
• The wall is frictionless and the floor is rough
• g = 9.8 N/kg

Let the ladder rest with its foot at A on the floor and its top at B against the wall. The height at which the ladder touches the wall is

$\text h = \sqrt{(3)^2 - (1)^2} = \sqrt{9 - 1} = \sqrt{8} = 2.83\ \text m$

The ladder is uniform, so its weight W acts at its mid-point G, whose horizontal distance from A is

$\frac{1}{2} \times 1 = 0.5\ \text m$

The wall is frictionless, so the reaction R_B exerted by it is horizontal. The reaction at the floor has a normal component N and a frictional component f.

Condition of translational equilibrium :

$\sum \text F_y = 0 \quad \Rightarrow \quad \text N = \text W = 196\ \text N$

$\sum \text F_x = 0 \quad \Rightarrow \quad \text f = \text R_B$

Condition of rotational equilibrium : Taking moments of all the forces about the foot A,

$\text R_B \times \text h = \text W \times 0.5$

$\text R_B = \frac{196 \times 0.5}{2.83} = \frac{98}{2.83} = 34.6\ \text N$

Hence the frictional force at the floor is also f = 34.6 N.

Total reaction at the floor :

$\text R_A = \sqrt{\text N^2 + \text f^2} = \sqrt{(196)^2 + (34.6)^2}$

$= \sqrt{38416 + 1197} = \sqrt{39613} = 199.0\ \text N$

Hence, the reaction force on the wall is 34.6 N and the reaction force on the floor is 199.0 N.

$\left(\frac{5\text a}{12}, \frac{5\text b}{12}\right)$

Reason —

Given, the sheet ABCD has mass M, length a and breadth b. The shaded portion HBGO is one quarter of the sheet, so its mass is $\frac{\text M}{4}$.

The centre of mass of the complete lamina is at its geometric centre,

$(\text x_1,\ \text y_1) = \left(\frac{\text a}{2},\ \frac{\text b}{2}\right)$

and that of the shaded quarter is at

$(\text x_2,\ \text y_2) = \left(\frac{3\text a}{4},\ \frac{3\text b}{4}\right)$

Treating the removed portion as a negative mass,

$\text X_{cm} = \frac{\text M\left(\frac{\text a}{2}\right) - \frac{\text M}{4}\left(\frac{3\text a}{4}\right)}{\text M - \frac{\text M}{4}} = \frac{\frac{\text a}{2} - \frac{3\text a}{16}}{\frac{3}{4}} = \frac{\frac{5\text a}{16}}{\frac{3}{4}} = \frac{5\text a}{12}$

Similarly, by symmetry,

$\text Y_{cm} = \frac{5\text b}{12}$

$\frac{\text a}{5}(\hat{i} - \hat{j})$

Reason —

Given, m_A = m, m_B = 2m, m_C = 3m and m_D = 4m, each having an acceleration of magnitude a in the directions shown.

From the figure the accelerations are $-\text a\hat{i}$ for A, $+\text a\hat{j}$ for B, $+\text a\hat{i}$ for C and $-\text a\hat{j}$ for D.

The acceleration of the centre of mass is

$\vec{\text a}_{cm} = \frac{\text m_A \vec{\text a}_A + \text m_B \vec{\text a}_B + \text m_C \vec{\text a}_C + \text m_D \vec{\text a}_D}{\text m_A + \text m_B + \text m_C + \text m_D}$

Substituting the values,

$= \frac{\text m(-\text a\hat{i}) + 2\text m(\text a\hat{j}) + 3\text m(\text a\hat{i}) + 4\text m(-\text a\hat{j})}{\text m + 2\text m + 3\text m + 4\text m}$

$= \frac{-\text{ma}\hat{i} + 2\text{ma}\hat{j} + 3\text{ma}\hat{i} - 4\text{ma}\hat{j}}{10\text m} = \frac{2\text{ma}\hat{i} - 2\text{ma}\hat{j}}{10\text m}$

$\vec{\text a}_{cm} = \frac{\text a}{5}(\hat{i} - \hat{j})$

$\vec{\text r} = \frac{13}{8}\text L \hat{x} + \frac{5}{8}\text L \hat{y}$

Reason —

From the figure the bar consists of three uniform segments. Taking the mass per unit length as uniform, the horizontal segment of length 2L has mass 2m, and the two segments of length L have mass m each, giving a total mass of 4m.

x-coordinate : The three segments have their own centres of mass at x = L, x = 2L and $\text x = \frac{5\text L}{2}$ respectively,

$\text X_{cm} = \frac{2\text m(\text L) + \text m(2\text L) + \text m\left(\frac{5\text L}{2}\right)}{2\text m + \text m + \text m}$

$= \frac{2\text{mL} + 2\text{mL} + 2.5\text{mL}}{4\text m} = \frac{6.5\text{mL}}{4\text m} = \frac{13}{8}\text L$

y-coordinate : Working in the same way for the vertical positions of the three segments gives

y-coordinate : The three segments have their own centres of mass at y = L, y = $\frac{\text L}{2}$ and y = 0 respectively,

$$\begin{aligned}\text Y_{cm} = \frac{2\text m(\text L) + \text m\left(\frac{\text L}{2}\right) + \text m(0)}{2\text m + \text m + \text m} \\[1em] = \frac{2\text{mL} + \frac{1}{2}\text{mL}}{4\text m} \\[1em] = \frac{\frac{5}{2}\text{mL}}{4\text m} \\[1em] \text Y_{cm} = \frac{5}{8}\text L\end{aligned}$$

So,

$\vec{\text r}_{cm} = \frac{13}{8}\text L\hat{x} + \frac{5}{8}\text L\hat{y}$

$\left(\frac{7}{12}\text{ m}, \frac{\sqrt{3}}{4}\text{ m}\right)$

Reason —

Given, m₁ = 50 g, m₂ = 100 g and m₃ = 150 g at the vertices of an equilateral triangle of side 1 m.

From the figure, m₁ is at the origin, m₂ at (1, 0) and m₃ at the apex, whose coordinates are

$\left(\frac{1}{2},\ \frac{\sqrt{3}}{2}\right)$

x-coordinate of the centre of mass :

$\text x_{cm} = \frac{50(0) + 100(1) + 150\left(\frac{1}{2}\right)}{50 + 100 + 150} = \frac{0 + 100 + 75}{300} = \frac{175}{300} = \frac{7}{12}\ \text m$

y-coordinate of the centre of mass :

$\text y_{cm} = \frac{50(0) + 100(0) + 150\left(\frac{\sqrt{3}}{2}\right)}{300} = \frac{75\sqrt{3}}{300} = \frac{\sqrt{3}}{4}\ \text m$

5

Reason —

Given, disc B of mass M and radius R rolls without slipping on the circumference of the fixed disc A of radius R.

The centre of B moves in a circle of radius 2R about the centre of A with angular speed ω, so the speed of its centre of mass is

$\text v_{cm} = \omega(2\text R)$

Since B rolls without slipping, its own angular speed ω_B about its centre satisfies v_cm = ω_BR, which gives ω_B = 2ω.

The angular momentum of B about the centre of A is the sum of the orbital part and the spin part,

$\text L = \text M \text v_{cm}(2\text R) + \text I_{cm}\omega_B$

$= \text M(2\omega \text R)(2\text R) + \frac{1}{2}\text{MR}^2(2\omega)$

$= 4\text M\omega \text R^2 + \text M\omega \text R^2 = 5\text M\omega \text R^2$

Comparing with L = nMωR², we get n = 5.

$\frac{20}{3}$ m

Reason — Given, m₁ = 10 kg and m₂ = 20 kg at the two ends of a rod of length d = 10 m of negligible mass.

Taking the 10 kg mass at the origin, x₁ = 0 and x₂ = 10 m. The distance of the centre of mass from the 10 kg mass is

$\text x_{cm} = \frac{\text m_1 \text x_1 + \text m_2 \text x_2}{\text m_1 + \text m_2}$

Substituting the values,

$\text x_{cm} = \frac{(10 \times 0) + (20 \times 10)}{10 + 20} = \frac{200}{30}$

$\text x_{cm} = \frac{20}{3}\ \text m$

The centre of mass lies nearer to the heavier 20 kg mass, as expected.

140

Reason — Given, a solid sphere of mass M and radius R is divided into two parts.

First part : mass $\frac{7\text M}{8}$, converted into a uniform disc of radius 2R,

$\text I_1 = \frac{1}{2}\left(\frac{7\text M}{8}\right)(2\text R)^2 = \frac{1}{2} \times \frac{7\text M}{8} \times 4\text R^2 = \frac{7}{4}\text{MR}^2$

Second part : mass $\text M - \frac{7\text M}{8} = \frac{\text M}{8}$, converted into a uniform solid sphere. Since the density is unchanged, the volume is $\frac{1}{8}$ of the original, so its radius is

$\text r = \frac{\text R}{2}$

$\text I_2 = \frac{2}{5}\left(\frac{\text M}{8}\right)\left(\frac{\text R}{2}\right)^2 = \frac{2}{5} \times \frac{\text M}{8} \times \frac{\text R^2}{4} = \frac{\text{MR}^2}{80}$

Therefore,

$\frac{\text I_1}{\text I_2} = \frac{\frac{7}{4}\text{MR}^2}{\frac{1}{80}\text{MR}^2} = \frac{7}{4} \times 80 = 140$

$\frac{2\text K}{\text I}\theta$

Reason —

Given, the kinetic energy of the disc as a function of θ is K = Kθ², and its moment of inertia is I.

The kinetic energy of rotation is

$\frac{1}{2}\text I \omega^2 = \text K\theta^2$

$\omega^2 = \frac{2\text K\theta^2}{\text I}$

Differentiating both sides with respect to time,

$2\omega\frac{\text d\omega}{\text{dt}} = \frac{2\text K}{\text I} \times 2\theta\frac{\text d\theta}{\text{dt}}$

But $\frac{\text d\omega}{\text{dt}} = \alpha$ and $\frac{\text d\theta}{\text{dt}} = \omega$. Therefore

$2\omega \alpha = \frac{4\text K}{\text I}\theta \omega$

Cancelling ω from both sides,

$\alpha = \frac{2\text K}{\text I}\theta$

16 cm

Reason — Given,

• Inner radius of the hollow cylinder, R₁ = 10 cm
• Outer radius of the hollow cylinder, R₂ = 20 cm

The moment of inertia of a hollow cylinder about its own axis is

$\text I = \frac{1}{2}\text M(\text R_1^2 + \text R_2^2)$

Substituting the values,

$\text I = \frac{1}{2}\text M\left[(10)^2 + (20)^2\right] = \frac{1}{2}\text M(100 + 400) = 250\text M$

For a thin cylinder of the same mass M and radius R, the moment of inertia about its axis is I = MR². Equating the two,

$\text{MR}^2 = 250\text M \quad \Rightarrow \quad \text R^2 = 250$

$\text R = \sqrt{250} = 15.8 \approx 16\ \text{cm}$

3MR²

Reason —

Given, the central disc D₁ of mass M and radius R carries two identical discs D₂ and D₃, each of mass M and radius R, attached rigidly at the opposite ends of one of its diameters.

From the figure the axis OO′ passes through the centre of D₁ and is perpendicular to the plane of D₁. The discs D₂ and D₃ stand in planes containing this axis, with their centres on the rim of D₁, that is, at a perpendicular distance R from OO′.

For the central disc D₁ : the axis is perpendicular to its plane and passes through its centre,

$\text I_1 = \frac{1}{2}\text{MR}^2$

For each of the discs D₂ and D₃ : the axis OO′ lies in the plane of the disc, parallel to a diameter through its centre and at a perpendicular distance R from it. Since the moment of inertia of a disc about a diameter is $\frac{1}{4}\text{MR}^2$, the theorem of parallel axes gives

$\text I_2 = \text I_3 = \frac{1}{4}\text{MR}^2 + \text{MR}^2 = \frac{5}{4}\text{MR}^2$

That is, each of them is turning about a tangent lying in its own plane.

By the additive theorem of moment of inertia, the total moment of inertia of the system about OO′ is

$\text I = \text I_1 + \text I_2 + \text I_3 = \frac{1}{2}\text{MR}^2 + \frac{5}{4}\text{MR}^2 + \frac{5}{4}\text{MR}^2$

$= \frac{2\text{MR}^2 + 5\text{MR}^2 + 5\text{MR}^2}{4} = \frac{12}{4}\text{MR}^2$

$\text I = 3\text{MR}^2$

6 kg-m²/s

Reason —

Given,

• Mass of the particle, m = 20 g = 0.02 kg
• Initial velocity at A, u = 5 m/s
• Height of A above B, h = 10 m
• Distance a = 10 m
• g = 10 m/s²

The surface is frictionless, so the mechanical energy is conserved between A and B,

$\frac{1}{2}\text{mu}^2 + \text{mgh} = \frac{1}{2}\text{mv}^2$

Cancelling m and substituting the values,

$\text v^2 = \text u^2 + 2\text{gh} = (5)^2 + 2(10)(10) = 25 + 200 = 225$

$\text v = 15\ \text{m/s}$

At B the particle moves horizontally, and the perpendicular distance of its line of motion from O is a = 10 m. Hence the angular momentum about O is

$\text L = \text{mv}(\text a + \text h) = 0.02 \times 15 \times 20 = 6\ \text{kg-m}^2/\text s$

2.0 × 10^-5 Nm

Reason —

Given,

• Mass of the coin, m = 5 g = 5 × 10^-3 kg
• Radius of the coin, r = 1 cm = 10^-2 m
• Final rate of rotation, n = 25 rotations per second
• Time, t = 5 s

From the figure the stick AB is tangential to the coin, so the axis of rotation lies along a tangent to the disc in its own plane. The moment of inertia about such an axis is

$\text I = \frac{5}{4}\text{mr}^2 = \frac{5}{4} \times (5 \times 10^{-3}) \times (10^{-2})^2$

$= \frac{5}{4} \times 5 \times 10^{-7} = 6.25 \times 10^{-7}\ \text{kg m}^2$

The final angular velocity is

$\omega = 2\pi \text n = 2\pi \times 25 = 50\pi\ \text{rad s}^{-1}$

The system starts from rest, so the angular acceleration is

$\alpha = \frac{\omega - \omega_0}{\text t} = \frac{50\pi}{5} = 10\pi\ \text{rad s}^{-2}$

Therefore the required torque is

$$\begin{aligned}\tau = \text I \alpha = (6.25 \times 10^{-7})(10\pi) \\[1em] = 6.25 \times 10 \times 3.14 \times 10^{-7} = 1.96 \times 10^{-5}\end{aligned}$$

$\tau \approx 2.0 \times 10^{-5}\ \text{Nm}$

$\frac{14}{15}$

Reason —

Given, a solid sphere and a solid cylinder of identical radii approach the incline with the same linear velocity v and roll without slipping.

The total kinetic energy of a rolling body is converted into potential energy at the maximum height,

$\frac{1}{2}\text{mv}^2\left(1 + \frac{\text k^2}{\text R^2}\right) = \text{mgh}$

$\text h = \frac{\text v^2}{2\text g}\left(1 + \frac{\text k^2}{\text R^2}\right)$

For the solid sphere, $\frac{\text k^2}{\text R^2} = \frac{2}{5}$,

$\text h_{sph} = \frac{\text v^2}{2\text g}\left(1 + \frac{2}{5}\right) = \frac{7\text v^2}{10\text g}$

For the solid cylinder, $\frac{\text k^2}{\text R^2} = \frac{1}{2}$,

$\text h_{cyl} = \frac{\text v^2}{2\text g}\left(1 + \frac{1}{2}\right) = \frac{3\text v^2}{4\text g}$

Therefore,

$\frac{\text h_{sph}}{\text h_{cyl}} = \frac{7/10}{3/4} = \frac{7}{10} \times \frac{4}{3} = \frac{14}{15}$

1 : 1

Reason — Given, two particles A and B move in concentric circles of radii r_A and r_B with the same time period of rotation T.

The angular speed of a particle in uniform circular motion is

$\omega = \frac{2\pi}{\text T}$

This expression depends only on the time period and not on the radius of the circle or the linear speed. Since both particles have the same time period T,

$\frac{\omega_A}{\omega_B} = \frac{2\pi/\text T}{2\pi/\text T} = 1$

$\omega_A : \omega_B = 1 : 1$

Their linear speeds v = rω do differ, being proportional to their radii, but their angular speeds are equal.

3 J

Reason — Given,

• Radius of the disc, R = 2 m
• Mass of the disc, M = 100 kg
• Speed of the centre of mass, v = 20 cm/s = 0.2 m/s

The disc rolls on a horizontal floor, so it possesses both translational and rotational kinetic energy. For a disc $\frac{\text k^2}{\text R^2} = \frac{1}{2}$, so the total kinetic energy is

$\text E = \frac{1}{2}\text{Mv}^2\left(1 + \frac{\text k^2}{\text R^2}\right) = \frac{1}{2}\text{Mv}^2\left(1 + \frac{1}{2}\right) = \frac{3}{4}\text{Mv}^2$

Substituting the values,

$\text E = \frac{3}{4} \times 100 \times (0.2)^2 = \frac{3}{4} \times 100 \times 0.04$

$\text E = 3\ \text J$

The work needed to stop the disc is equal to its total kinetic energy, that is, 3 J.

2 s

Reason — Given,

• Moment of inertia, I = 1.5 kg m²
• Rotational kinetic energy to be produced, K = 1200 J
• Angular acceleration, α = 20 rad/s²
• Initial angular velocity, ω₀ = 0 (body initially at rest)

From the kinetic energy of rotation,

$\text K = \frac{1}{2}\text I \omega^2$

$\omega^2 = \frac{2\text K}{\text I} = \frac{2 \times 1200}{1.5} = 1600$

$\omega = 40\ \text{rad/s}$

Using the first equation of rotational motion,

$\omega = \omega_0 + \alpha \text t$

$40 = 0 + 20\text t \quad \Rightarrow \quad \text t = 2\ \text s$

$\sqrt{2}$ : 1

Reason — For a thin uniform disc of mass M and radius R :

About an axis through the centre and normal to its plane,

$\text I_1 = \frac{1}{2}\text{MR}^2 = \text{MK}_1^2 \quad \Rightarrow \quad \text K_1 = \frac{\text R}{\sqrt{2}}$

About a diameter, by the theorem of perpendicular axes,

$\text I_2 = \frac{1}{4}\text{MR}^2 = \text{MK}_2^2 \quad \Rightarrow \quad \text K_2 = \frac{\text R}{2}$

Therefore the required ratio is

$\frac{\text K_1}{\text K_2} = \frac{\text R/\sqrt{2}}{\text R/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$\text K_1 : \text K_2 = \sqrt{2} : 1$

4 π

Reason — Given,

• Initial angular speed, ω₀ = 1200 rpm
• Final angular speed, ω = 3120 rpm
• Time, t = 16 s

Converting the angular speeds into rad/s,

$\omega_0 = \frac{2\pi \times 1200}{60} = 40\pi\ \text{rad/s}$

$\omega = \frac{2\pi \times 3120}{60} = 104\pi\ \text{rad/s}$

Using the first equation of rotational motion,

$\omega = \omega_0 + \alpha \text t$

$104\pi = 40\pi + \alpha(16)$

$16\alpha = 64\pi \quad \Rightarrow \quad \alpha = 4\pi\ \text{rad/s}^2$

8.5 cm

Reason —

Given,

• Moment of inertia, I = 2400 g cm²
• Mass of the rod, M = 400 g
• The axis passes through the mid-point and is perpendicular to the rod

For a thin uniform rod about an axis through its centre and perpendicular to its length,

$\text I = \frac{1}{12}\text{ML}^2$

Rearranging for L,

$\text L^2 = \frac{12\text I}{\text M}$

Substituting the values,

$\text L^2 = \frac{12 \times 2400}{400} = \frac{28800}{400} = 72$

$\text L = \sqrt{72} = 8.49 \approx 8.5\ \text{cm}$

Point P moves faster than Q

Reason —

In rolling motion the velocity of any point on the wheel is the vector sum of the velocity of the centre of mass and the tangential velocity due to rotation.

At the highest point P : the tangential velocity and the velocity of the centre of mass are in the same direction,

$\text v_P = \text v_{cm} + \omega \text R = \text v + \text v = 2\text v$

At the lowest point Q : the tangential velocity is opposite to the velocity of the centre of mass,

$\text v_Q = \text v_{cm} - \omega \text R = \text v - \text v = 0$

The point of contact Q is therefore momentarily at rest relative to the ground, while the topmost point P moves with twice the speed of the centre of mass. Hence point P moves faster than Q.

20 rad/s

Reason —

Given,

• Mass of the wheel, M = 10 kg
• Radius of the wheel, R = 10 cm = 0.1 m
• Steady pull on the cord, F = 20 N
• Length of cord unwound, s = 1 m

The wheel is supported by spokes of negligible mass, so its entire mass lies on the rim and its moment of inertia is

$\text I = \text{MR}^2 = 10 \times (0.1)^2 = 0.1\ \text{kg m}^2$

The work done by the steady pull in unwinding 1 m of cord is

$\text W = \text{Fs} = 20 \times 1 = 20\ \text J$

The wheel rotates without friction and is initially at rest, so all this work appears as its kinetic energy of rotation,

$\text W = \frac{1}{2}\text I \omega^2$

$20 = \frac{1}{2} \times 0.1 \times \omega^2 = 0.05\omega^2$

$\omega^2 = \frac{20}{0.05} = 400 \quad \Rightarrow \quad \omega = 20\ \text{rad/s}$

10 N

Reason —

Given, the square lamina OABC has length 10 cm and is pivoted at O. Forces of 10 N act on the lamina as shown, together with the unknown force F.

For the lamina to remain stationary it must be in rotational equilibrium, so the net torque about the pivot O must be zero,

$\sum \tau_O = 0$

From the figure, the 10 N forces acting at the sides of the square exert moments about O. Taking the side of the square as the moment arm wherever the force acts perpendicular to a side through the far edge, and equating the sum of the anticlockwise moments to the sum of the clockwise moments,

$\text F \times 0.10 = 10 \times 0.10$

$\text F = 10\ \text N$

The forces whose lines of action pass through the pivot O contribute no moment, since for them the perpendicular distance from the axis of rotation is zero.

α/4

Reason — Given, the moment of inertia of a rod of mass M and length L about an axis through its centre and normal to its length is

$\alpha = \frac{1}{12}\text{ML}^2$

The rod is cut into two equal parts, so each part has

• mass $\frac{\text M}{2}$
• length $\frac{\text L}{2}$

These two parts are joined symmetrically at their centres to form a cross. The axis passes through the centre of the cross and is normal to the plane containing it. For each part this axis passes through its own centre and is perpendicular to its length, so

$\text I_1 = \frac{1}{12}\left(\frac{\text M}{2}\right)\left(\frac{\text L}{2}\right)^2 = \frac{1}{12} \times \frac{\text M}{2} \times \frac{\text L^2}{4} = \frac{\text{ML}^2}{96}$

The moment of inertia of the cross is the sum of the moments of inertia of the two parts,

$\text I = 2 \times \frac{\text{ML}^2}{96} = \frac{\text{ML}^2}{48}$

Expressing this in terms of α,

$\text I = \frac{\text{ML}^2}{48} = \frac{1}{4}\left(\frac{\text{ML}^2}{12}\right) = \frac{\alpha}{4}$