Q.(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2/5$ times the initial value? Assume that the turntable rotates without friction.

Concept: Rotational Dynamics — conservation of angular momentum in the absence of external torque.

(a) No external torque acts on the child–turntable system, so angular momentum $L = I \omega$ is conserved.

Let initial moment of inertia be $I$ and initial angular speed $\omega_1 = 40$ rev/min.

Final moment of inertia $I_2 = \frac{2}{5} I$.

From $I \omega_1 = I_2 \omega_2$: …

Why conservation of angular momentum works here

The child and turntable together form a system that rotates without friction. When the child folds his arms, no external torque acts on the system — the only forces are internal (the child's muscles). For any system with zero net external torque, angular momentum is conserved:

This is the rotational analogue of conservation of linear momentum. The moment of inertia $I$ changes because the child redistributes mass closer to the axis; angular speed $\omega$ must adjust to keep $L$ unchanged.

(a) Finding the new angular speed

1. Let the initial moment of inertia be $I_1$ and the initial angular speed be $\omega_1 = 40$ rev/min.

2. When the child folds his arms, the moment of inertia becomes:

3. Conservation of angular momentum gives:

Substitute $I_2$:

4. Cancel $I_1$ (non-zero) and solve:

(b) Comparing kinetic energies

1. Rotational kinetic energy is:

Initial kinetic energy:

Final kinetic energy:

2. Express $K_2$ in terms of $K_1$:

But $K_1 = \frac{1}{2} I_1 \cdot 40^2 = \frac{1}{2} I_1 \cdot 1600$. So:

Thus $K_2 = 2.5\, K_1$ — the kinetic energy increases by a factor of $2.5$. …

Concept: Conservation of Angular Momentum ($L=I\omega=$ constant)

No external torque acts on the child–turntable system (frictionless), so $L$ is conserved even though the child's own moment of inertia changes.

Step 1: Set up part (a)

Let initial $I_1$, $\omega_1=40$ rev/min; final $I_2=\frac25I_1$.

Step 2: Apply $I_1\omega_1=I_2\omega_2$

Step 3: Compare kinetic energies for part (b)

Step 4: Physical explanation …