A uniform disc of mass $m$ and radius $R$ stands vertically on its rim on a horizontal table; the coefficient of friction between the disc and the table is $\mu$. A horizontal force $\vec{F}$ is applied at the centre (the axle) of the disc, in the plane of the disc. Find the maximum value of $F$ for which the disc rolls without slipping.

A uniform disc of mass m and radius R stands vertically on its rim on a horizontal table; the coefficient of f

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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 applied force at the centre exerts no torque, so friction alone must spin the disc up to roll. Newton's second law for translation and rotation, combined with the rolling condition and $I=\tfrac12 mR^2$ , gives the needed friction $f=F/3$ . Since static friction cannot exceed $\mu N=\mu mg$ , rolling without slipping survives only up to $F=3\mu mg$ .

Set-up

Let the disc have mass $m$ , radius $R$ , and moment of inertia about its centre $I = \tfrac{1}{2}mR^2$ . The horizontal force $F$ acts at the centre; friction $f$ acts at the contact point at the bottom. The normal reaction is $N$ .

Equations of motion

Vertical balance: $N = mg$ (the force $F$ is horizontal).
Translation: the net horizontal force accelerates the centre,

F - f = m a.

Rotation about the centre: $F$ passes through the centre, so only friction gives a torque,

fR = I\alpha = \frac{1}{2}mR^2\,\alpha.

Rolling condition: $a = \alpha R$ .

Solve

From (3) and (4): $fR = \tfrac12 mR^2\cdot\dfrac{a}{R} = \tfrac12 mRa$ , so

$f = \frac{1}{2}m a.$ …

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

KEAM 2026Set pha-2026-0418F4 marksMCQ
$^{-1}$
›Reveal solutionSolution

Rolling ring: $E = \tfrac12 mv^2 + \tfrac12 I\omega^2 = mv^2 = 1.25\times10^{-2}\,J$ .

For a ring, moment of inertia $I = mr^2$ and rolling gives $\omega = v/r$ . Total energy $= \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 mv^2 + \tfrac12 (mr^2)(v/r)^2 = \tfrac12 mv^2 + \tfrac12 mv^2 = mv^2$ . With $m = 0.05\,kg$ and …
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KEAM 2026Set pha-2026-0419F4 marksMCQ
$Q. In a steam engine, the flywheel is used to resist (A) the slight increase or decrease of the speed of the vehicle (B) the sudden increase or decrease of the speed of the vehicle (C) the sudden stoppage of the vehicle (D) only the sudden increase of the speed of the vehicle (E) only the slight decrease of the speed of the vehicle$
›Reveal solutionSolution

A flywheel's large moment of inertia stores rotational KE and resists sudden changes in speed, evening out the fluctuating drive of the piston.

In a steam engine the driving torque from the piston is highly non-uniform over a cycle (it peaks during the power stroke and drops near the dead centres). A flywheel is a heavy rotating disc with a large moment of inertia $I$ . Its stored kinetic energy is $KE = \frac{1}{2} I \omega^2$ , so a change in speed requires a change in this stored energy; the flywheel therefore absorbs energy when the drive momentarily speeds up and releases it when …
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KEAM 2025Set eng-2025-04254 marksMCQ
$I\omega^{2}$
›Reveal solutionSolution
The correct standard relation for rigid-body rotation is angular momentum $L = I\omega$ ; the other options misstate KE, work, power and angular velocity.

Check each option against the standard rotational formulae:

(A) wrong — kinetic energy is $\tfrac{1}{2}I\omega^2$ , not $I\omega^2$ .

(B) correct — angular momentum $L = I\omega$ .

(C) wrong — work done is $W = \tau\,\theta$ (for constant torque), not $\tau^2\omega^2$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$Q. If a satellite of mass M is spinning about its own axis and revolves around the earth in a circular orbit, then it does not have (A) moment of inertia (B) potential energy (C) rotational kinetic energy (D) vibrational energy (E) angular momentum$
›Reveal solutionSolution
The satellite spins and orbits as a rigid body, so it possesses moment of inertia, potential energy, rotational kinetic energy and angular momentum; it does not have vibrational energy.

Consider what a spinning satellite in a circular orbit possesses:

(A) moment of inertia — yes, any extended body has one.

(B) potential energy — yes, gravitational PE in the earth's field.

(C) rotational kinetic energy — yes, it spins about its own axis. …
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KEAM 2025Set eng-2025-04264 marksMCQ
$4\ kg\,m^2$
›Reveal solutionSolution

Rotational kinetic energy is $KE = \tfrac{1}{2}I\omega^2$ . Solving for $\omega$ with $I=4\ \text{kg m}^2$ and $KE=50\ \text{J}$ gives $\omega = 5\ \text{rad s}^{-1}$ .

Formula:
$KE = \dfrac{1}{2}I\omega^2$
Rearrange: …
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KEAM 2024Set eng-2024-06054 marksMCQ
$Q. The analogy between linear motion and rotational motion are given. The FALSE one is (A) Force : Torque (B) Linear Displacement : Angular displacement (C) Mass : Moment of inertia (D) Linear momentum : Angular momentum (E) Translational energy : Vibrational energy$
›Reveal solutionSolution
Force $\leftrightarrow$ Torque, displacement $\leftrightarrow$ angular displacement, mass $\leftrightarrow$ moment of inertia, and momentum $\leftrightarrow$ angular momentum are all correct; translational energy pairs with rotational energy, not vibrational.

The valid linear $\to$ rotational analogies are:

Force $\leftrightarrow$ Torque

Linear displacement $\leftrightarrow$ Angular displacement

Mass $\leftrightarrow$ Moment of inertia

Linear momentum $\leftrightarrow$ Angular momentum …
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KEAM 2024Set eng-2024-06064 marksMCQ
$Q. The motion of a cylinder on an inclined plane is a (A) rotational but not translation (B) translation but not rotational (C) translational but not rolling (D) rotational, translational and rolling motion (E) rotational and rolling but not translational motion$
›Reveal solutionSolution

Rolling on an incline is simultaneous translation of the centre of mass and rotation about the axis, i.e. rolling motion.

As a cylinder moves on an inclined plane its centre of mass slides down the slope (translational motion) while it also spins about its own axis (rotational motion). When these occur together without …
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KEAM 2024Set eng-2024-06084 marksMCQ
$\dfrac{\pi}{4}$
›Reveal solutionSolution

Solid disc $I=\tfrac12 MR^2=0.3125$ ; final $\omega=4\pi$ rad/s in 5 s gives $\alpha=0.8\pi$ ; $\tau=I\alpha=\pi/4$ Nm.

Diameter $= 0.5$ m so radius $R = 0.25$ m. Moment of inertia of a uniform solid disc about its central axis:
$I = \tfrac12 MR^2 = \tfrac12(10)(0.25)^2 = \tfrac12(10)(0.0625) = 0.3125\ \text{kg·m}^2.$
Final angular speed: $120\ \text{rpm} = 120\times\dfrac{2\pi}{60} = 4\pi\ \text{rad/s}$ . …
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KEAM 2023Set eng-2023-P1-A14 marksMCQ
$2\sqrt{2}$
›Reveal solutionSolution
The centre speed is $v = 4/\sqrt2 = 2\sqrt2$ m/s.

Concept and Intuition

For a rolling wheel every rim point combines the translational velocity of the centre ( $v$ , horizontal) with a rotational velocity ( $\omega r = v$ ). At the point level with the centre (the side of the wheel) the rotational velocity is vertical, perpendicular to the translational one, so the two add in quadrature.

Step-by-Step Solution

At that point: horizontal component $= v$ , vertical component $= \omega r = v$ .

Speed $= \sqrt{v^2 + v^2} = v\sqrt2$ . …
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Set-up

Equations of motion

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