A solid sphere of mass M and radius R is in pure rolling as shown in figure. The angular momentum of the sphere about the origin is (linear velocity of centre of mass is `v_0`, in positive x-direction)

Options

A. ` 3MR V_0 /5 (-K) `
B. ` 3MR V_0 /5 (+K) `
C. ` 7MR V_0 /5 (-K) `
D. ` 7MR V_0 /5 (+K) `

Answer

To find the angular momentum of a solid sphere of mass \( M \) and radius \( R \) that is rolling without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Mass of the sphere: \( M \)
- Radius of the sphere: \( R \)
- Linear velocity of the center of mass: \( v_0 \) (in the positive x-direction)
- Angular velocity \( \omega \) can be related to the linear velocity as \( v_0 = R \omega \).

2. **Calculate the Moment of Inertia**:
The moment of inertia \( I \) of a solid sphere about its center is given by:
\[ I = \frac{2}{5} M R^2 \]

3. **Determine Angular Momentum about the Center of Mass**:
The angular momentum \( L_{CM} \) of the sphere about its center of mass is given by:
\[ L_{CM} = I \omega \]
Substituting the expression for \( I \):
\[ L_{CM} = \frac{2}{5} M R^2 \cdot \omega \]
Since \( \omega = \frac{v_0}{R} \):
\[ L_{CM} = \frac{2}{5} M R^2 \cdot \frac{v_0}{R} = \frac{2}{5} M R v_0 \]

4. **Calculate the Angular Momentum due to the Translation of the Center of Mass**:
The angular momentum due to the translational motion of the center of mass about the origin is given by:
\[ L_{trans} = M \cdot R \cdot v_0 \]
Here, \( R \) is the distance from the origin to the center of mass.

5. **Combine Both Contributions**:
The total angular momentum \( L \) about the origin is the sum of the angular momentum about the center of mass and the angular momentum due to the translation:
\[ L = L_{CM} + L_{trans} \]
Substituting the values we calculated:
\[ L = \frac{2}{5} M R v_0 + M R v_0 \]
\[ L = \left( \frac{2}{5} M R v_0 + \frac{5}{5} M R v_0 \right) = \frac{7}{5} M R v_0 \]

6. **Direction of Angular Momentum**:
Since the sphere is rolling in the positive x-direction, the angular momentum vector will point in the negative z-direction (using the right-hand rule). Thus, we can express it as:
\[ L = \frac{7}{5} M R v_0 \hat{k} \text{ (in the negative z-direction)} \]
or
\[ L = -\frac{7}{5} M R v_0 \hat{k} \]

### Final Answer:
The angular momentum of the sphere about the origin is:
\[ L = -\frac{7}{5} M R v_0 \hat{k} \]