A body is rolling without slipping on a horizontal plane. If the rotational energy of the body is 40% of the total kinetic energy, then the body might be

Options

A. cylinder
B. hollow sphere
C. solid cylinder
D. ring

Answer

To solve the problem, we need to analyze the relationship between the translational kinetic energy and the rotational kinetic energy of a body rolling without slipping.

### Step-by-Step Solution:

1. **Understanding the Kinetic Energy Components**:
- The total kinetic energy (KE_total) of a rolling body is the sum of its translational kinetic energy (KE_trans) and rotational kinetic energy (KE_rot).
- Mathematically, this can be expressed as:
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} \]
- For a body rolling without slipping, the translational kinetic energy is given by:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]
- The rotational kinetic energy is given by:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
- Here, \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

2. **Relating Angular Velocity and Linear Velocity**:
- For a body rolling without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ v = r\omega \]
- Thus, we can express \(\omega\) in terms of \(v\):
\[ \omega = \frac{v}{r} \]

3. **Substituting into the Rotational Kinetic Energy**:
- Substituting \(\omega\) into the expression for \(KE_{\text{rot}}\):
\[ KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \frac{I}{r^2} v^2 \]

4. **Total Kinetic Energy Expression**:
- Now substituting \(KE_{\text{trans}}\) and \(KE_{\text{rot}}\) into the total kinetic energy equation:
\[ KE_{\text{total}} = \frac{1}{2} mv^2 + \frac{1}{2} \frac{I}{r^2} v^2 \]
- Factoring out \(\frac{1}{2} v^2\):
\[ KE_{\text{total}} = \frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right) \]

5. **Using the Given Information**:
- According to the problem, the rotational energy is 40% of the total kinetic energy:
\[ KE_{\text{rot}} = 0.4 KE_{\text{total}} \]
- Substituting the expressions we have:
\[ \frac{1}{2} \frac{I}{r^2} v^2 = 0.4 \left(\frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right)\right) \]

6. **Simplifying the Equation**:
- Canceling \(\frac{1}{2} v^2\) from both sides (assuming \(v \neq 0\)):
\[ \frac{I}{r^2} = 0.4 \left(m + \frac{I}{r^2}\right) \]
- Rearranging gives:
\[ \frac{I}{r^2} - 0.4 \frac{I}{r^2} = 0.4m \]
\[ 0.6 \frac{I}{r^2} = 0.4m \]
- Therefore:
\[ \frac{I}{r^2} = \frac{0.4m}{0.6} = \frac{2}{3}m \]

7. **Finding the Moment of Inertia**:
- Thus, we have:
\[ I = \frac{2}{3}mr^2 \]
- This indicates that the body is likely a hollow sphere, as this is the moment of inertia for a hollow sphere.

### Conclusion:
The body might be a hollow sphere, and the moment of inertia is given by:
\[ I = \frac{2}{3}mr^2 \]