A solid cylinder of mass M and radius R rotates about its axis with angular speed `omega`. Its rotational kinetic energy is

Options

A. `1/2 MR^(2)omega^(2)`
B. `MR^(2)omega^(2)`
C. `1/4 MR^(2)omega^(2)`
D. `1/8 MR^(2)omega^(2)`

Explanation

To find the rotational kinetic energy of a solid cylinder rotating about its axis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Rotational Kinetic Energy**:
The formula for rotational kinetic energy (K.E.) is given by:
\[ K.E. = \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular speed.

2. **Determine the Moment of Inertia for a Solid Cylinder**:
The moment of inertia \( I \) for a solid cylinder rotating about its axis is given by:
\[ I = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the cylinder and \( R \) is its radius.

3. **Substitute the Moment of Inertia into the K.E. Formula**:
Now, substitute the expression for \( I \) into the kinetic energy formula:
\[ K.E. = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \omega^2 \]

4. **Simplify the Expression**:
Simplifying the expression gives:
\[ K.E. = \frac{1}{2} \cdot \frac{1}{2} M R^2 \cdot \omega^2 = \frac{1}{4} M R^2 \omega^2 \]

5. **Final Result**:
Therefore, the rotational kinetic energy of the solid cylinder is:
\[ K.E. = \frac{1}{4} M R^2 \omega^2 \]