Two cylindrical hollow drums of radii $R$ and $2R$, and of a common height $h$, are rotating with angular velocities $\omega$ (anti-clockwise) and $\omega$ (clockwise), respectively. Their axes, fixed are parallel and in a horizontal plane separated by $(3R + \delta)$. They are now brought in contact $(\delta \to 0)$. (a) Show the frictional forces just after contact. (b) Identify forces and torques external to the system just after contact. (c) What would be the ratio of final angular velocities when friction ceases?

Two cylindrical hollow drums of radii R and 2R, and of a common height h, are rotating with angular velocities

NCERT Exemplar · Q26

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(a) Show the frictional forces just after contact.

(b) Identify forces and torques external to the system just after contact.

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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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Friction at the contact is tangential (vertical) and equal-and-opposite on the two drums. The fixed-axle (bearing) reactions are external, so total angular momentum is not conserved. Slipping stops when the contact points have equal linear speed: $R\omega_1' = 2R\omega_2'$ , giving $\omega_1':\omega_2' = 2:1$ .

Setup

Drum 1 has radius $R$ , drum 2 has radius $2R$ ; their parallel axes are $3R$ apart, so they just touch ( $\delta\to0$ ) at a point on the line joining the centres. Since the axes are horizontal and fixed, the common tangent to both surfaces at the contact is vertical.

(a) Frictional forces just after contact

Just before friction acts, the two surfaces slide past each other along the (vertical) tangent. Their contact-point speeds are

v_1=R\omega \quad(\text{drum 1}),\qquad v_2=2R\omega\quad(\text{drum 2}),

and for the given senses of rotation they point the same way along the tangent. Since $v_2>v_1$ , drum 2's surface slides relative to drum 1's.

Kinetic friction opposes this relative sliding:

on drum 2 it acts so as to retard its surface,
on drum 1 it acts in the opposite direction (Newton's third law), dragging its surface along.

So the two frictional forces are equal in magnitude ( $f$ ), opposite in direction, and tangential (vertical) at the contact.

(b) External forces and torques

Take the two drums together as the system.

The mutual normal push and the friction pair at the contact are internal action-reaction pairs.
External to the system are the weights of the two drums and the reaction forces at the two fixed axles/bearings that hold the axes in place.

These axle reactions can exert a net external torque on the system, so the total angular momentum of the system is not conserved - it cannot be used to find the final state. (About each drum's own axis the axle reaction passes through the axis and gives no torque; only friction torques each drum: $fR$ on drum 1 and $f(2R)$ on drum 2, each opposing the relative slipping.) …

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Concept understanding — Rotational Dynamics

Rotational Dynamics: The Physics of Spinning Things

The Core Idea

Breaking It Down

Torque — The Rotational "Push"

Moment of Inertia — The Rotational "Mass"

Angular Acceleration — How Fast Rotation Changes

The Complete Picture: Rotational Analogues

The Key Insight: Conservation of Angular Momentum

Concept: Rotational Inertia Comparison and Angular Momentum Conservation

(a) Frictional forces just after contact

(b) External forces and torques

Setup

(a) Frictional forces just after contact

(b) External forces and torques

Concept: Contact Friction Is Internal; Fixed-Axle Reactions Are External — So Angular Momentum Is NOT Conserved Here

Setup

(a) Frictional forces just after contact

(b) External forces and torques

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