A Merry-go-round, made of a ring-like platform of radius $R$ and mass $M$, is revolving with angular speed $\omega$. A person of mass $M$ is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is (a) $2\omega$ (b) $\omega$ (c) $\dfrac{\omega}{2}$ (d) $0$

> [!TLDR] > The key is that the person jumps radially outward as seen from the rotating frame, which means they leave with zero tangential velocity relative to the ground. This removes no angular momentum from the system, so the merry-go-round’s angular speed remains unchanged at $\omega$. 1. **Understand the setup and the crucial phrase.** The problem says the person jumps “radially away from the centre … as seen from the round.” That means *from the perspective of someone sitting on the rotating platform*, the person moves straight outward along a radius. To a ground observer, this is **not** a straight radial path — because the platform is rotating, the person’s velocity relative to ground has two parts: the radial component they jump with, plus the tangential component they already had from riding the platform. 2. **What happens to angular momentum?** Angular momentum about the centre is conserved because no external torque acts on the system (the jump is an internal force). Before the jump, the person and the platform together have a certain angular momentum. After the jump, the person is moving away, and the platform spins alone. 3. **The critical insight: the person’s tangential velocity at the instant of leaving.** When the person jumps radially outward *as seen from the round*, they do **not** push off in a way that changes their tangential speed relative to the ground. They simply step off along a radius. At the exact moment they lose contact, their tangential velocity (the part that contributes to angular momentum) is still exactly what it was while standing on the platform: $v_\text{tan} = \omega R$. > [!WARNING] > A common mistake is to think the person “throws themselves off” and somehow takes extra angular momentum. But jumping radially, as seen from the rotating frame, means they leave with the same tangential speed they had — they don’t push tangentially. 4. **Write the angular momentum conservation.** Let the moment

A Merry-go-round, made of a ring-like platform of radius R and mass M, is revolving with angular speed ω. A pe

NCERT Exemplar · Q8

R

(a)

2\omega

(b)

\omega

(c)

\dfrac{\omega}{2}

(d)

0

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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 key is that the person jumps radially outward as seen from the rotating frame, which means they leave with zero tangential velocity relative to the ground. This removes no angular momentum from the system, so the merry-go-round’s angular speed remains unchanged at $\omega$ .

Understand the setup and the crucial phrase.

The problem says the person jumps “radially away from the centre … as seen from the round.” That means from the perspective of someone sitting on the rotating platform, the person moves straight outward along a radius. To a ground observer, this is not a straight radial path — because the platform is rotating, the person’s velocity relative to ground has two parts: the radial component they jump with, plus the tangential component they already had from riding the platform.
What happens to angular momentum?

Angular momentum about the centre is conserved because no external torque acts on the system (the jump is an internal force). Before the jump, the person and the platform together have a certain angular momentum. After the jump, the person is moving away, and the platform spins alone.
The critical insight: the person’s tangential velocity at the instant of leaving.

When the person jumps radially outward as seen from the round, they do not push off in a way that changes their tangential speed relative to the ground. They simply step off along a radius. At the exact moment they lose contact, their tangential velocity (the part that contributes to angular momentum) is still exactly what it was while standing on the platform: $v_\text{tan} = \omega R$ .

Watch out

A common mistake is to think the person “throws themselves off” and somehow takes extra angular momentum. But jumping radially, as seen from the rotating frame, means they leave with the same tangential speed they had — they don’t push tangentially.
Write the angular momentum conservation.

Let the moment of inertia of the ring-like platform be $I_\text{ring} = M R^2$ (since all mass is at radius $R$ ). The person, treated as a point mass at radius $R$ , also has moment of inertia $M R^2$ .

Initial angular momentum:

$L_i = (I_\text{ring} + I_\text{person})\,\omega = (M R^2 + M R^2)\,\omega = 2 M R^2 \omega.$ …

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