An ice skater spins at `3pi" rad s"^(-1)` with hers arms extended. If her moment of inertia with arms folded is `75%` of that with arms extended, her angular velocity when she fold her arms is

Options

A. `pi rad s^(-1)`
B. `2pi rad s^(-1)`
C. `3pi rad s^(-1)`
D. `4pi rad s^(-1)`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of the ice skater must remain constant when she folds her arms, as there is no external torque acting on her.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Let the moment of inertia with arms extended be \( I_1 \).
- The initial angular velocity is given as \( \omega_1 = 3\pi \, \text{rad/s} \).

2. **Determine the Moment of Inertia with Arms Folded:**
- The moment of inertia when the arms are folded is given as \( I_2 = 0.75 I_1 \).

3. **Apply Conservation of Angular Momentum:**
- According to the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
- Substituting the values we have:
\[ I_1 (3\pi) = (0.75 I_1) \omega_2 \]

4. **Simplify the Equation:**
- We can cancel \( I_1 \) from both sides (assuming \( I_1 \neq 0 \)):
\[ 3\pi = 0.75 \omega_2 \]

5. **Solve for \( \omega_2 \):**
- Rearranging the equation to solve for \( \omega_2 \):
\[ \omega_2 = \frac{3\pi}{0.75} \]
- This simplifies to:
\[ \omega_2 = \frac{3\pi}{\frac{3}{4}} = 3\pi \times \frac{4}{3} = 4\pi \, \text{rad/s} \]

### Final Answer:
The angular velocity when she folds her arms is \( \omega_2 = 4\pi \, \text{rad/s} \).

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