If a body is rolling on a surface without slipping such that its kinetic energy of translation is equal to kinetic energy of rotation then it is a

Options

A. disc
B. sphere
C. ring
D. cylinder

Answer

To solve the problem, we need to analyze the conditions given: a body is rolling on a surface without slipping, and its translational kinetic energy is equal to its rotational kinetic energy. We will derive the necessary relationships step by step.

### Step-by-Step Solution:

1. **Understand the Kinetic Energies**:
- The translational kinetic energy (TKE) of a body is given by the formula:
\[ \text{TKE} = \frac{1}{2} mv^2 \]
- The rotational kinetic energy (RKE) is given by:
\[ \text{RKE} = \frac{1}{2} I \omega^2 \]
where \( m \) is the mass of the body, \( v \) is the velocity of the center of mass, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.

2. **Condition of Rolling Without Slipping**:
- For a body rolling without slipping, the relationship between the linear velocity \( v \) and the angular velocity \( \omega \) is:
\[ v = \omega r \]
where \( r \) is the radius of the body.

3. **Set the Kinetic Energies Equal**:
- According to the problem, the translational kinetic energy is equal to the rotational kinetic energy:
\[ \frac{1}{2} mv^2 = \frac{1}{2} I \omega^2 \]
- We can simplify this by multiplying both sides by 2:
\[ mv^2 = I \omega^2 \]

4. **Substitute the Relationship Between \( v \) and \( \omega \)**:
- Substitute \( \omega = \frac{v}{r} \) into the equation:
\[ mv^2 = I \left(\frac{v}{r}\right)^2 \]
- This simplifies to:
\[ mv^2 = \frac{I v^2}{r^2} \]

5. **Cancel \( v^2 \) from Both Sides**:
- Assuming \( v \neq 0 \), we can divide both sides by \( v^2 \):
\[ m = \frac{I}{r^2} \]

6. **Rearrange to Find Moment of Inertia**:
- Rearranging gives us:
\[ I = m r^2 \]
- This indicates that the moment of inertia \( I \) is equal to \( m r^2 \).

7. **Identify the Type of Body**:
- The moment of inertia \( I = m r^2 \) corresponds to a ring (or hollow cylinder) where all the mass is concentrated at a distance \( r \) from the axis of rotation.

### Conclusion:
Thus, if a body is rolling on a surface without slipping such that its kinetic energy of translation is equal to its kinetic energy of rotation, it is a **ring**.