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. `omega_0`
B. `(omega_0)/(3)`
C. `6 omega_0`
D. `(omega_0)/(6)`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. Here are the steps to find the new angular speed after the child folds his arms:

### Step-by-Step Solution:

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

2. **Calculate Initial Angular Momentum:**
- The initial angular momentum \( L_{\text{initial}} \) can be calculated using the formula:
\[ L_{\text{initial}} = I_0 \cdot \omega_0 \]

3. **Identify Final Conditions:**
- When the child folds his arms, the moment of inertia becomes \( 3 \) times the initial value:
\[ I_{\text{final}} = 3 I_0 \]
- Let the new angular speed be \( \omega_f \).

4. **Calculate Final Angular Momentum:**
- The final angular momentum \( L_{\text{final}} \) is given by:
\[ L_{\text{final}} = I_{\text{final}} \cdot \omega_f = 3 I_0 \cdot \omega_f \]

5. **Apply Conservation of Angular Momentum:**
- Since there is no external torque acting on the system, angular momentum is conserved:
\[ L_{\text{initial}} = L_{\text{final}} \]
- Substituting the expressions for angular momentum:
\[ I_0 \cdot \omega_0 = 3 I_0 \cdot \omega_f \]

6. **Solve for the Final Angular Speed:**
- We can simplify the equation by dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \omega_0 = 3 \cdot \omega_f \]
- Rearranging gives:
\[ \omega_f = \frac{\omega_0}{3} \]

### Final Answer:
The new angular speed \( \omega_f \) after the child folds his arms is:
\[ \omega_f = \frac{\omega_0}{3} \]