A rigid body of mass m rotates with angular velocity `omega` about an axis at a distance a from the centre of mass G. The radius of gyration about a parallel axis through G is k. The kinetic energy of rotation of the body is

Options

A. `1/2 mk^2 omega^2`
B. `1/2 ma^2 omega^2`
C. `1/2 m(a^2 + k^2)omega^2`
D. `1/2 m (a+k)^2 omega^2`

Answer

To solve the problem, we need to find the kinetic energy of a rigid body rotating about an axis that is a distance \( a \) from its center of mass \( G \). The radius of gyration about a parallel axis through \( G \) is given as \( k \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a rigid body with mass \( m \) rotating with an angular velocity \( \omega \).
- The axis of rotation is at a distance \( a \) from the center of mass \( G \).
- The radius of gyration about the axis through \( G \) is \( k \).

2. **Using the Parallel Axis Theorem**:
- The parallel axis theorem states that the moment of inertia \( I \) about an axis parallel to one through the center of mass is given by:
\[ I = I_G + m \cdot d^2 \]
where \( I_G \) is the moment of inertia about the center of mass axis, \( m \) is the mass of the body, and \( d \) is the distance between the two axes (in this case, \( a \)).
- Here, we can express \( I_G \) in terms of the radius of gyration \( k \):
\[ I_G = m k^2 \]
- Therefore, the moment of inertia about the new axis is:
\[ I = mk^2 + m a^2 \]

3. **Calculating the Kinetic Energy**:
- The rotational kinetic energy \( K \) of the body is given by:
\[ K = \frac{1}{2} I \omega^2 \]
- Substituting the expression for \( I \):
\[ K = \frac{1}{2} (mk^2 + ma^2) \omega^2 \]
- This simplifies to:
\[ K = \frac{1}{2} m (k^2 + a^2) \omega^2 \]

4. **Final Expression**:
- The final expression for the kinetic energy of rotation of the body is:
\[ K = \frac{1}{2} m (k^2 + a^2) \omega^2 \]