A ballet dancer spins about a vertical axis at 120 rpm with arms out stretched. With her arms fold the moment of inertia about the axis of rotation decreases by 40% . What is new rate of revolution?

Options

A. 100 rpm
B. 150 rpm
C. 200 rpm
D. 250 rpm

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 torques are acting on it.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial angular velocity (ω₁) of the dancer is given as 120 revolutions per minute (rpm).
- Convert this to radians per second:
\[ \omega_1 = 120 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} = 12.5664 \, \text{rad/s} \]

2. **Determine the Moment of Inertia**:
- Let the initial moment of inertia be \( I_1 \).
- When the dancer folds her arms, the moment of inertia decreases by 40%. Therefore, the new moment of inertia \( I_2 \) is:
\[ I_2 = I_1 - 0.4 I_1 = 0.6 I_1 \]

3. **Apply Conservation of Angular Momentum**:
- According to the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
- Substituting \( I_2 \):
\[ I_1 \omega_1 = (0.6 I_1) \omega_2 \]

4. **Simplify the Equation**:
- Cancel \( I_1 \) from both sides (assuming \( I_1 \neq 0 \)):
\[ \omega_1 = 0.6 \omega_2 \]

5. **Solve for the New Angular Velocity \( \omega_2 \)**:
- Rearranging the equation gives:
\[ \omega_2 = \frac{\omega_1}{0.6} \]
- Substitute \( \omega_1 = 12.5664 \, \text{rad/s} \):
\[ \omega_2 = \frac{12.5664 \, \text{rad/s}}{0.6} = 20.944 \, \text{rad/s} \]

6. **Convert Back to Revolutions Per Minute**:
- To convert \( \omega_2 \) back to rpm:
\[ \omega_2 = 20.944 \, \text{rad/s} \times \frac{1 \, \text{revolution}}{2\pi \, \text{radians}} \times 60 \, \text{seconds} = 200 \, \text{rpm} \]

### Final Answer:
The new rate of revolution when the dancer folds her arms is **200 rpm**.