A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed `omega_0`. Now, the child folds his hands back so that moment of inertia becomes `3` times the initial value. The new angular speed is.

Options

A. `3omega_o`
B. `omega_o`/3
C. `6omega_o`
D. `omega_o`/6

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of a system remains constant if no external torque acts on it.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Let the initial moment of inertia of the child with arms outstretched be \( I_0 \).
- The initial angular speed is given as \( \omega_0 \).

2. **Determine Final Conditions**:
- When the child folds his arms, the moment of inertia becomes \( I_f = 3 I_0 \).

3. **Apply Conservation of Angular Momentum**:
- According to the conservation of angular momentum:
\[ L_i = L_f \]
- Where \( L_i \) is the initial angular momentum and \( L_f \) is the final angular momentum.
- We can express angular momentum as:
\[ L = I \cdot \omega \]
- Therefore, we have:
\[ I_0 \cdot \omega_0 = I_f \cdot \omega_f \]

4. **Substitute Final Moment of Inertia**:
- Substitute \( I_f = 3 I_0 \) into the equation:
\[ I_0 \cdot \omega_0 = (3 I_0) \cdot \omega_f \]

5. **Simplify the Equation**:
- Cancel \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[ \omega_0 = 3 \cdot \omega_f \]

6. **Solve for Final Angular Speed**:
- Rearranging the equation gives:
\[ \omega_f = \frac{\omega_0}{3} \]

### Final Answer:
The new angular speed \( \omega_f \) is \( \frac{\omega_0}{3} \).

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