A Merry -go-round, made of a ring-like plarfrom of radius `R and mass M`, is revolving with angular speed `omega`. A person of mass `M` is standing on it. At one instant, the person jumps off the round, radially awaay from the centre of the round (as see from the round). The speed of the round after wards is

Options

A. `2omega`
B. `omega`
C. `omega`/2
D. 0

Answer

To solve the problem, we will apply the principle of conservation of angular momentum. Here’s the step-by-step solution:

### Step 1: Understand the System
We have a merry-go-round (ring-like platform) with:
- Mass of the platform (ring) = \( M \)
- Radius of the platform = \( R \)
- Initial angular speed = \( \omega \)

A person of mass \( M \) is standing on the platform and jumps off radially outward.

### Step 2: Initial Angular Momentum
The total initial angular momentum \( L_i \) of the system (merry-go-round + person) can be calculated as follows:
- Moment of inertia of the ring about its vertical axis:
\[ I_{\text{ring}} = M R^2 \]
- Moment of inertia of the person (considered as a point mass) at distance \( R \) from the center:
\[ I_{\text{person}} = M R^2 \]
- Therefore, the total initial moment of inertia \( I_i \) is:
\[ I_i = I_{\text{ring}} + I_{\text{person}} = M R^2 + M R^2 = 2 M R^2 \]
- The initial angular momentum \( L_i \) is:
\[ L_i = I_i \cdot \omega = (2 M R^2) \cdot \omega = 2 M R^2 \omega \]

### Step 3: Final Angular Momentum
After the person jumps off, only the ring is rotating. The moment of inertia of the ring remains the same, but the person no longer contributes to the moment of inertia:
- The final moment of inertia \( I_f \) is:
\[ I_f = M R^2 \]
- Let the final angular speed be \( \omega_f \). The final angular momentum \( L_f \) is:
\[ L_f = I_f \cdot \omega_f = (M R^2) \cdot \omega_f \]

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_i = L_f \]
Substituting the expressions we derived:
\[ 2 M R^2 \omega = M R^2 \omega_f \]

### Step 5: Solve for Final Angular Speed
We can simplify this equation:
\[ 2 \omega = \omega_f \]
Thus, we find:
\[ \omega_f = 2 \omega \]

### Conclusion
The speed of the merry-go-round after the person jumps off is \( 2\omega \).