A sphere of mass `M` rolls without slipping on rough surface with centre of mass has constant speed `v_0`. If its radius is R`, then the angular momentum of the sphere about the point of contact is.

Options

A. `(5)/(7) Mv_(0)R`
B. `(7)/(5)Mv_(0)R`
C. `(3)/(5)Mv_(0)R`
D. `(1)/(2)Mv_(0)R`

Explanation

To find the angular momentum of a sphere of mass \( M \) rolling without slipping on a rough surface about the point of contact, we can follow these steps:

### Step 1: Understand the System
The sphere is rolling without slipping, which means that the point of contact with the surface has zero velocity at that instant. The center of mass of the sphere is moving with a constant speed \( v_0 \).

### Step 2: Identify the Instantaneous Center of Rotation
Since the sphere is rolling without slipping, the point of contact (let's call it point P) acts as the instantaneous center of rotation. This means we can consider the sphere as rotating about point P.

### Step 3: Use the Formula for Angular Momentum
The angular momentum \( L \) about point P can be calculated using the formula:
\[ L = I_P \cdot \omega \]
where \( I_P \) is the moment of inertia about point P and \( \omega \) is the angular velocity of the sphere.

### Step 4: Calculate the Moment of Inertia
The moment of inertia of a solid sphere about its center of mass (point O) is given by:
\[ I_O = \frac{2}{5} M R^2 \]
To find the moment of inertia about point P, we can use the parallel axis theorem:
\[ I_P = I_O + M d^2 \]
where \( d \) is the distance from the center of mass to point P, which is equal to the radius \( R \) of the sphere. Thus,
\[ I_P = \frac{2}{5} M R^2 + M R^2 = \frac{2}{5} M R^2 + \frac{5}{5} M R^2 = \frac{7}{5} M R^2 \]

### Step 5: Relate Angular Velocity to Linear Velocity
For rolling without slipping, the relationship between the linear velocity \( v_0 \) and the angular velocity \( \omega \) is given by:
\[ v_0 = \omega R \]
From this, we can express \( \omega \) as:
\[ \omega = \frac{v_0}{R} \]

### Step 6: Substitute into the Angular Momentum Formula
Now we substitute \( I_P \) and \( \omega \) back into the angular momentum formula:
\[ L = I_P \cdot \omega = \left(\frac{7}{5} M R^2\right) \cdot \left(\frac{v_0}{R}\right) \]
This simplifies to:
\[ L = \frac{7}{5} M R v_0 \]

### Conclusion
Thus, the angular momentum of the sphere about the point of contact is:
\[ L = \frac{7}{5} M v_0 R \]

### Answer
The correct option is \( \frac{7}{5} M v_0 R \).

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