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

### Step 1: Identify the Moment of Inertia
The moment of inertia (I) of a solid sphere about its center is given by the formula:
\[ I_{\text{center}} = \frac{2}{5} m r^2 \]
where \(m\) is the mass of the sphere and \(r\) is its radius.

### Step 2: Use the Parallel Axis Theorem
Since the axis of rotation is not through the center of the sphere but rather touching its surface, we need to use the parallel axis theorem to find the moment of inertia about this new axis. The parallel axis theorem states:
\[ I = I_{\text{center}} + md^2 \]
where \(d\) is the distance from the center of mass to the new axis. For a solid sphere, this distance is equal to the radius \(r\).

Thus, we can calculate:
\[ I_{\text{new}} = \frac{2}{5} m r^2 + m r^2 = \frac{2}{5} m r^2 + \frac{5}{5} m r^2 = \frac{7}{5} m r^2 \]

### Step 3: Calculate Angular Momentum
The angular momentum (L) is given by the formula:
\[ L = I \omega \]
where \(\omega\) is the angular velocity. Substituting the moment of inertia we found:
\[ L = \left(\frac{7}{5} m r^2\right) \omega \]

### Final Result
Thus, the angular momentum of the solid sphere rotating about an axis touching its surface is:
\[ L = \frac{7}{5} m r^2 \omega \]