A uniform solid sphere of mass m and radius R rests on a rough horizontal floor; its centre is therefore at he

NCERT Exemplar · Q18

m

(a)

h = R/2

;

(b)

h = R

;

(c)

h = 3R/2

;

(d)

h = 7R/5

.
Column II:

(i) Sphere rolls without slipping at constant velocity, with no loss of energy.

(ii) Sphere spins in the forward (clockwise) sense and loses energy by friction.

(iii) Sphere spins in the backward (anti-clockwise) sense and loses energy by friction.

(iv) Sphere has only translational motion (no spin) and loses energy by friction.

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Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

Imagine you're trying to open a heavy door. You push near the hinge — it barely moves. Push near the handle — it swings open easily. Same force, different result. That's the first clue: rotation isn't just about how much you push, but where and in what direction.

Now think about a spinning bicycle wheel. Why is it so hard to tilt it sideways when it's spinning fast? And why does a figure skater spin faster when she pulls her arms in? These are the questions rotational dynamics answers.

The Core Idea

Rotational dynamics is the study of why things rotate and how their rotation changes. It's the spinning-world equivalent of Newton's laws for straight-line motion.

In linear motion, you have:

Force ( $F$ ) causes acceleration ( $a$ )
Mass ( $m$ ) resists acceleration

In rotational motion, you have:

Torque ( $\tau$ ) causes angular acceleration ( $\alpha$ )
Moment of inertia ( $I$ ) resists angular acceleration

The master equation is:

\tau_{\text{net}} = I \alpha

This is the rotational version of $F = ma$ . Every term has a direct parallel.

Breaking It Down

Torque — The Rotational "Push"

Torque isn't just force — it's force multiplied by the distance from the pivot point (the lever arm). That's why the door handle works better than the hinge.

\tau = r F \sin\theta

Where $r$ is the distance from the axis, $F$ is the force, and $\theta$ is the angle between them. Maximum torque happens when you push perpendicular to the lever arm ( $\theta = 90^\circ$ ).

Tip

Think of torque as "twisting effectiveness." A wrench works because the handle gives you a long lever arm. A short wrench needs more force to do the same job.

Moment of Inertia — The Rotational "Mass"

Mass resists linear acceleration. Moment of inertia resists angular acceleration. But unlike mass, moment of inertia depends on how the mass is distributed relative to the axis of rotation.

For a point mass $m$ at distance $r$ from the axis:

I = mr^2

For extended objects, you sum (or integrate) over all mass elements:

I = \sum m_i r_i^2

Note

| Object | Axis | Moment of Inertia |

|--------|------|-------------------|

| Thin hoop | Through center, perpendicular to plane | $MR^2$ |

| Solid disk | Through center, perpendicular to plane | $\frac{1}{2}MR^2$ |

| Solid sphere | Through center | $\frac{2}{5}MR^2$ |

| Thin rod | Through center, perpendicular to rod | $\frac{1}{12}ML^2$ |

Notice: a hoop has more moment of inertia than a disk of the same mass and radius because its mass is farther from the axis. That's why a hoop is harder to start spinning.

Angular Acceleration — How Fast Rotation Changes

Just as acceleration is the rate of change of velocity, angular acceleration $\alpha$ is the rate of change of angular velocity $\omega$ :

\alpha = \frac{d\omega}{dt}

And angular velocity is the rate of change of angular displacement $\theta$ :

\omega = \frac{d\theta}{dt}

The Complete Picture: Rotational Analogues

Linear Quantity	Rotational Analogue
Displacement $x$	Angular displacement $\theta$
Velocity $v$	Angular velocity $\omega$
Acceleration $a$	Angular acceleration $\alpha$
Mass $m$	Moment of inertia $I$
Force $F$	Torque $\tau$
Newton's 2nd law: $F = ma$	$\tau = I\alpha$
Kinetic energy: $\frac{1}{2}mv^2$	$\frac{1}{2}I\omega^2$
Momentum: $p = mv$	Angular momentum: $L = I\omega$

The Key Insight: Conservation of Angular Momentum

This is where rotational dynamics gets beautiful. Just as linear momentum is conserved when no external force acts, angular momentum is conserved when no external torque acts:

$L = I\omega = \text{constant}$ …

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The blow $J$ gives translation $v=J/m$ and, because it acts a height $(h-R)$ above the centre, a spin $\omega = J(h-R)/I$ with $I=\tfrac25 mR^2$ . Requiring $v=\omega R$ (pure rolling, no friction needed) gives the special height $h=7R/5$ . Other heights leave a mismatch between $v$ and $\omega R$ , so friction acts and energy is lost; the spin is backward for $h<R$ , zero for $h=R$ , forward for $h>R$ .

Concept

Impulse $J$ at height $h$ from the floor is at height $(h-R)$ relative to the centre. It produces:

linear speed of the centre: $v = J/m$ ;
angular speed about the centre: $\omega = \dfrac{J(h-R)}{I}$ , with $I = \tfrac{2}{5}mR^2$ .

The rolling height

Pure rolling with no friction requires the contact point to be instantaneously at rest: $v = \omega R$ .

\frac{J}{m} = \frac{J(h-R)}{\tfrac25 mR^2}\,R \ \Rightarrow\ 1 = \frac{h-R}{\tfrac25 R} \ \Rightarrow\ h - R = \frac{2R}{5} \ \Rightarrow\ h = \frac{7R}{5}.

So (d) $h=7R/5$ gives rolling without slipping at constant velocity, no energy loss → (i).

The other cases

The sign of $(h-R)$ fixes the spin sense (taking the blow toward the right, forward rolling is clockwise): …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{N}-\mathrm{m}$
›Reveal solutionSolution
The torque required is the product of the moment of inertia of the four corner masses plus the central mass about the given axis and the angular acceleration. The result is $\frac{2\pi}{25}\,\mathrm{N\cdot m}$ , which corresponds to option (D).

Concept and intuition

Torque $\tau$ is the rotational analogue of force: $\tau = I \alpha$ , where $I$ is the moment of inertia about the rotation axis and $\alpha$ is the angular acceleration. Here the axis passes through the centre O and is perpendicular to the square frame. The frame itself is massless, so only the five point masses contribute to $I$ . For a point mass $m$ at distance $r$ from the axis, $I = m r^2$ . The four corner masses are all at the same distance from O (half the diagonal of the square), and the central mass is exactly on the axis, so its distance is zero. Summing these contributions gives the total $I$ ; multiplying by $\alpha$ yields the required torque.

Step-by-step solution

Find the distance from centre O to a corner. The square has side $a = 0.2\ \mathrm{m}$ . The distance from the centre to any corner is half the diagonal:

$r = \frac{a\sqrt{2}}{2} = \frac{0.2\sqrt{2}}{2} = 0.1\sqrt{2}\ \mathrm{m}.$

Moment of inertia of the four corner masses. Each corner mass is $m = 2\ \mathrm{kg}$ . For one corner:

$I_{\text{corner}} = m r^2 = 2 \times (0.1\sqrt{2})^2 = 2 \times (0.01 \times 2) = 2 \times 0.02 = 0.04\ \mathrm{kg\cdot m^2}.$
Four such masses:
$I_{\text{corners}} = 4 \times 0.04 = 0.16\ \mathrm{kg\cdot m^2}.$

Moment of inertia of the central mass. The central mass (40 kg) lies exactly on the axis, so its distance from the axis is $0$ . Hence:
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KCET 2025Set D-41 markMCQ
$I_1 = 4\, \mathrm{kg\,m^2}$
(a) Angular acceleration of smaller wheel
(b) Torque on the larger wheel
(c) Angular acceleration of larger wheel II Their numerical Values (in SI units)
(i) $\dfrac{5}{3}$
(ii) $100/3$
(iii) $5/2$ (A) a - ii, b - iii, c - i (B) a - iii, b - i, c - ii (C) a - ii, b - i, c - iii (D) a - iii, b - ii, c - i
›Reveal solutionSolution

Get $\alpha_1$ from $\tau = I\alpha$ , transfer it through the belt using the equal-rim-acceleration condition $\alpha_1 r_1 = \alpha_2 r_2$ , then get the large wheel's torque from $\tau_2 = I_2\alpha_2$ .

Data. $I_1 = 4\ \mathrm{kg\,m^2},\ r_1 = 0.20\ \mathrm{m},\ I_2 = 20\ \mathrm{kg\,m^2},\ r_2 = 0.30\ \mathrm{m},\ \tau_1 = 10\ \mathrm{N\,m}$ .

Step 1 — (a) Angular acceleration of the smaller wheel.

Rotational form of Newton's second law:
$\alpha_1 = \frac{\tau_1}{I_1} = \frac{10}{4} = \frac{5}{2}\ \mathrm{rad\,s^{-2}}$
So (a) → (iii).

Step 2 — Belt constraint.

A non-slipping belt has the same linear (tangential) speed, hence the same tangential acceleration, at both rims:
$a = \alpha_1 r_1 = \alpha_2 r_2$ $a = \frac{5}{2}\times 0.20 = 0.5\ \mathrm{m\,s^{-2}}$
Step 3 — (c) Angular acceleration of the larger wheel. …
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COMEDK 2025Set 2025-M1 markMCQ
$-\mathrm{F} \hat{\mathbf{i}}$
›Reveal solutionSolution
With $\vec{r}$ drawn from $P=(0,1,-1)$ to the origin, $\vec{r} = (0,-1,1)$ and $\vec{\tau} = \vec{r}\times\vec{F} = (0,-1,1)\times(-F,0,0) = -F(\hat{\jmath}+\hat{k})$ — option (B).

Concept

Torque about a chosen point is $\vec{\tau} = \vec{r} \times \vec{F}$ , where $\vec{r}$ points from that point to the point of application of the force. Here the force acts at the origin, so $\vec{r}$ runs from the reference point $P$ to the origin.

Solution

Position vector:

$\vec{r} = (0,0,0) - (0,1,-1) = (0,\,-1,\,1).$

Cross product with $\vec{F} = (-F,0,0)$ :

$\vec{\tau} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 0 & -1 & 1\\ -F & 0 & 0\end{vmatrix}.$

Components:

$\tau_x = (-1)(0)-(1)(0) = 0,\quad \tau_y = (1)(-F)-(0)(0) = -F,\quad \tau_z = (0)(0)-(-1)(-F) = -F.$

Assemble:
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COMEDK 2024Set 2024-A1 markMCQ
$1 \mathrm{~kg}$
›Reveal solutionSolution
The block's fall gives $a = 2\text{ m/s}^2$ . Combining Newton's law for the block with $\tau = I\alpha$ for the wheel yields $R = 0.5$ m — option (D).

The string wound on the rim links the block's linear acceleration to the wheel's angular acceleration by $a = \alpha R$ ; the tension supplies the only torque.

Linear acceleration from the fall ( $s = \tfrac12 a t^2$ ):

$a = \frac{2s}{t^2} = \frac{2\times 4}{2^2} = 2\text{ m/s}^2.$

Newton's law for the hanging block (taking $g = 10\text{ m/s}^2$ ):

$mg - T = ma \;\Rightarrow\; T = m(g-a) = 1\times(10-2) = 8\text{ N}.$

Rotational law for the wheel, with $\alpha = a/R$ and torque $TR$ : …
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COMEDK 2024Set 2024-M1 markMCQ
$5 \mathrm{~kg}$
›Reveal solutionSolution
The work done equals the change in rotational kinetic energy. Using torque, time, and moment of inertia, we find the final angular velocity and then the work: 6250 J, which corresponds to option (C).

The key idea is that work done by a constant torque equals the change in rotational kinetic energy. Since the body starts from rest, the work done is simply the final rotational kinetic energy:
$W = \frac{1}{2} I \omega^2$
We know the moment of inertia $I = 5\ \mathrm{kg\,m^2}$ and the torque $\tau = 10\ \mathrm{N\,m}$ applied for $t = 25\ \mathrm{s}$ . To find $\omega$ , we use the rotational analogue of Newton’s second law:
$\tau = I \alpha$
where $\alpha$ is the angular acceleration. Once we have $\alpha$ , we can find $\omega$ from $\omega = \alpha t$ (starting from rest). Then plug into the work formula.

Find angular acceleration From $\tau = I \alpha$ :

$\alpha = \frac{\tau}{I} = \frac{10}{5} = 2\ \mathrm{rad/s^2}$

Find final angular velocity Since the body starts from rest ( $\omega_0 = 0$ ) and accelerates uniformly:

$\omega = \alpha t = 2 \times 25 = 50\ \mathrm{rad/s}$

Compute work done Work = change in rotational kinetic energy:
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KCET 2023Set A-31 markMCQ
$'m'$
›Reveal solutionSolution

The full 7-disc system's moment of inertia adds the central disc's own spin to the six outer discs' parallel-axis contributions, giving $\frac{55mR^2}{2}$ .

Step 1 — Central disc's own moment of inertia.

The central disc's axis passes through its own center, so it simply spins about itself: $I_{central} = \dfrac{1}{2}mR^2$ .

Step 2 — Each outer disc, by the parallel-axis theorem.

Every outer disc touches the central disc, so its center sits $2R$ from the axis:

$I_{outer} = \dfrac{1}{2}mR^2 + m(2R)^2 = \dfrac{1}{2}mR^2 + 4mR^2 = \dfrac{9}{2}mR^2$

Step 3 — Sum all six outer discs plus the central disc.

$I_{total} = \dfrac{1}{2}mR^2 + 6 \times \dfrac{9}{2}mR^2 = \dfrac{1}{2}mR^2 + 27mR^2 = \dfrac{55}{2}mR^2$ …
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COMEDK 2023Set 2023-E1 markMCQ
$200 \mathrm{~N}$
›Reveal solutionSolution
The angle between the position vector and force is $30^\circ$ , giving $\tau=rF\sin30^\circ=2\ \text{N m}$ .

Torque about O is $\tau = rF\sin\theta$ , where $\theta$ is the angle between $\vec{OP}$ and $\vec F$ .

$\vec{OP}$ is at $55^\circ$ to the x-axis.

$\vec F$ is at $25^\circ$ to the horizontal (x-axis).

Angle between them: $\theta = 55^\circ - 25^\circ = 30^\circ$ . …
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COMEDK 2021Set 20211 markMCQ
${{dp} \over {dt}} = {\tau _{ext}}$
›Reveal solutionSolution

(dp/dt = F_ext is the translational law, not the rotational one.)

Concept: rotational analogue of Newton's second law for a system of particles.

For a system, the internal torques cancel in pairs (Newton's third law, forces along the line joining the particles), so only the EXTERNAL torque changes the total angular momentum:

dL/dt = tau_ext. …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. A ring starts from rest and acquires an angular speed of 20 rad/s in 4 seconds. The mass of the ring is 250 g and its radius is 20 cm, the torque of the ring is (A) 0.5 Nm (B) 0.025 Nm (C) 0.25 Nm (D) 0.05 Nm$
›Reveal solutionSolution

With $\alpha = 5\ \text{rad/s}^2$ and ring moment of inertia $I = mr^2 = 0.01\ \text{kg·m}^2$ , the torque is $\tau = I\alpha = 0.05$ Nm.

Angular acceleration:
$\alpha = \frac{\omega - \omega_0}{t} = \frac{20 - 0}{4} = 5\ \text{rad/s}^2.$
Moment of inertia of a ring about its axis: …
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KCET 2020Set A-11 markMCQ
$25 \pi$
›Reveal solutionSolution
The wheel undergoes uniform angular acceleration from rest. Using the kinematic equation for angular displacement under constant acceleration, the total angle turned is 25 rad.

The problem gives you a wheel starting from rest, reaching 10 rad/s in 5 seconds under uniform angular acceleration. The key is to recognise that this is the rotational analogue of linear motion with constant acceleration. The same kinematic equations apply, just with angular variables: $\theta$ for displacement, $\omega$ for velocity, $\alpha$ for acceleration, and $t$ for time.

Since the acceleration is uniform, the average angular velocity is simply the arithmetic mean of the initial and final velocities. For motion from rest, that average is half the final velocity. The total angle turned is then average angular velocity multiplied by time — a clean, intuitive shortcut.

Let’s work through it step by step.

Identify the known quantities.

Initial angular velocity: $\omega_0 = 0$ (starts from rest).

Final angular velocity: $\omega = 10\ \text{rad/s}$ .

Time interval: $t = 5\ \text{s}$ .

Acceleration is uniform (constant $\alpha$ ).

Find the angular acceleration.

Using the definition $\alpha = \frac{\omega - \omega_0}{t}$ :

$\alpha = \frac{10 - 0}{5} = 2\ \text{rad/s}^2.$

Use the angular displacement equation for constant acceleration. The standard kinematic equation is:

$\theta = \omega_0 t + \frac{1}{2} \alpha t^2.$
Substitute $\omega_0 = 0$ , $\alpha = 2$ , $t = 5$ :

$\theta = 0 + \frac{1}{2} \times 2 \times (5)^2 = 1 \times 25 = 25\ \text{rad}.$ …
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KCET 2018Set A-11 markMCQ
$20\, \mathrm{kg\, m}^2$
›Reveal solutionSolution

Apply the perpendicular axis theorem, $I_z = I_x + I_y$ , valid for a planar lamina.

Step 1 — Identify the applicable theorem.

The body is a lamina (a plane body), the two given axes $X$ and $Y$ lie in its plane and are mutually perpendicular, and the required axis is perpendicular to the plane through their point of intersection. These are exactly the conditions of the perpendicular axis theorem:
$\boxed{I_z = I_x + I_y}$
Step 2 — Why the theorem is true.

For a particle of mass $m_i$ at $(x_i, y_i)$ in the plane, its distance from the $Z$ -axis is $r_i$ with $r_i^2 = x_i^2 + y_i^2$ . Hence
$I_z = \sum m_i r_i^2 = \sum m_i (x_i^2 + y_i^2) = \sum m_i y_i^2 + \sum m_i x_i^2 = I_x + I_y$
(The distance of a point from the $X$ -axis is $|y_i|$ and from the $Y$ -axis is $|x_i|$ .) Note it works only because $z_i = 0$ for every particle — i.e. only for a lamina.

Step 3 — Substitute. …
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Concept

The rolling height

The other cases

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