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. `2 omega`
B. `omega`
C. `(omega)/(2)`
D. 0

Explanation

To solve the problem, we need to apply the principle of conservation of angular momentum. Here's a step-by-step solution:

### Step 1: Understand the System
We have a merry-go-round (ring-like platform) with mass \( M \) and radius \( R \) that is rotating with an angular speed \( \omega \). A person of mass \( m \) is standing on the edge of the merry-go-round.

### Step 2: Define Initial Angular Momentum
Before the person jumps off, the total angular momentum \( L_{\text{initial}} \) of the system (merry-go-round + person) can be calculated as follows:
- The angular momentum of the merry-go-round is given by:
\[ L_{\text{merry-go-round}} = I_{\text{merry-go-round}} \cdot \omega = M R^2 \cdot \omega \]
- The angular momentum of the person (considered as a point mass at a distance \( R \) from the center) is:
\[ L_{\text{person}} = m R^2 \cdot \omega \]
- Therefore, the total initial angular momentum is:
\[ L_{\text{initial}} = L_{\text{merry-go-round}} + L_{\text{person}} = (M + m) R^2 \omega \]

### Step 3: Define Final Angular Momentum
After the person jumps off, they move radially away from the center. The angular momentum of the merry-go-round after the jump (with the person no longer on it) is:
- The angular momentum of the merry-go-round becomes:
\[ L_{\text{final}} = M R^2 \cdot \omega' \]
where \( \omega' \) is the new angular speed of the merry-go-round after the person jumps off.

### Step 4: Apply Conservation of Angular Momentum
According to the conservation of angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
Substituting the expressions we derived:
\[ (M + m) R^2 \omega = M R^2 \cdot \omega' \]

### Step 5: Solve for the New Angular Speed
We can simplify this equation by canceling \( R^2 \) from both sides:
\[ (M + m) \omega = M \omega' \]
Now, solving for \( \omega' \):
\[ \omega' = \frac{(M + m) \omega}{M} \]

### Step 6: Substitute the Mass of the Person
In this case, since the mass of the person is also \( M \):
\[ \omega' = \frac{(M + M) \omega}{M} = \frac{2M \omega}{M} = 2\omega \]

### Conclusion
The speed of the merry-go-round after the person jumps off is:
\[ \omega' = 2\omega \]