A solid sphere is rotating with a constant angular velocity about an axis touching the surface. Its angular momentum will be:

Options

A. `(2)/(3)MR^(2)omega`
B. `(5)/(7)MR^(2)omega`
C. `(7)/(5)MR^(2)omega`
D. `(5)/(2)MR^(2)omega`

Answer

To solve the problem of finding the angular momentum of a solid sphere rotating about an axis touching its surface, we can follow these steps:

### Step 1: Understand the Problem
We have a solid sphere rotating with a constant angular velocity about an axis that touches its surface. We need to determine its angular momentum.

### Step 2: Identify the Moment of Inertia
The moment of inertia (I) of a solid sphere about an axis through its center of mass is given by the formula:
\[ I_{cm} = \frac{2}{5} m r^2 \]
where \( m \) is the mass of the sphere and \( r \) is its radius.

### Step 3: Apply the Parallel Axis Theorem
Since the axis of rotation is not through the center of mass but is touching the surface of the sphere, we will use the Parallel Axis Theorem to find the moment of inertia about the new axis. The theorem states:
\[ I' = I_{cm} + m d^2 \]
where \( d \) is the distance from the center of mass to the new axis. In this case, \( d = r \) (the radius of the sphere).

### Step 4: Calculate the Moment of Inertia about the New Axis
Substituting the values into the equation:
\[ I' = I_{cm} + m r^2 \]
\[ I' = \frac{2}{5} m r^2 + m r^2 \]
\[ I' = \frac{2}{5} m r^2 + \frac{5}{5} m r^2 \]
\[ I' = \frac{7}{5} m r^2 \]

### Step 5: Calculate Angular Momentum
The angular momentum (L) is given by the formula:
\[ L = I \cdot \omega \]
where \( \omega \) is the angular velocity. Substituting the moment of inertia we found:
\[ L = \left(\frac{7}{5} m r^2\right) \cdot \omega \]

### Step 6: Final Expression for Angular Momentum
Thus, the angular momentum of the solid sphere is:
\[ L = \frac{7}{5} m r^2 \omega \]

### Conclusion
The angular momentum of the solid sphere rotating about an axis touching its surface is given by:
\[ L = \frac{7}{5} m r^2 \omega \]