A body is rolling without slipping on a horizontal plane. The rotational energy of the body is `40% `of the total kinetic energy. Identify the body.

Options

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

Explanation

To solve the problem, we need to analyze the relationship between translational and rotational kinetic energy for a body rolling without slipping. Let's break down the solution step by step.

### Step 1: Understand the Kinetic Energy Components
The total kinetic energy (K) of a rolling body is the sum of its translational kinetic energy (K_trans) and rotational kinetic energy (K_rot):
\[ K = K_{\text{trans}} + K_{\text{rot}} \]
Where:
- \( K_{\text{trans}} = \frac{1}{2} m v^2 \) (translational kinetic energy)
- \( K_{\text{rot}} = \frac{1}{2} I \omega^2 \) (rotational kinetic energy)

### Step 2: Relate Linear and Angular Velocity
For a body rolling without slipping, the relationship between linear velocity (v) and angular velocity (ω) is given by:
\[ v = r \omega \]
Where r is the radius of the body.

### Step 3: Substitute ω in the Rotational Kinetic Energy
Using the relationship \( \omega = \frac{v}{r} \), we can express the rotational kinetic energy in terms of v:
\[ K_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \frac{I}{r^2} v^2 \]

### Step 4: Express Total Kinetic Energy in Terms of v
Now, substituting K_trans and K_rot into the total kinetic energy equation:
\[ K = \frac{1}{2} m v^2 + \frac{1}{2} \frac{I}{r^2} v^2 \]
Factoring out \( \frac{1}{2} v^2 \):
\[ K = \frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right) \]

### Step 5: Set Up the Given Condition
According to the problem, the rotational kinetic energy is 40% of the total kinetic energy:
\[ K_{\text{rot}} = 0.4 K \]
Substituting for K:
\[ \frac{1}{2} \frac{I}{r^2} v^2 = 0.4 \left(\frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right)\right) \]

### Step 6: Simplify the Equation
Cancelling \( \frac{1}{2} v^2 \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{I}{r^2} = 0.4 \left(m + \frac{I}{r^2}\right) \]
Expanding and rearranging gives:
\[ \frac{I}{r^2} - 0.4 \frac{I}{r^2} = 0.4 m \]
\[ 0.6 \frac{I}{r^2} = 0.4 m \]
\[ \frac{I}{mr^2} = \frac{0.4}{0.6} = \frac{2}{3} \]

### Step 7: Identify the Body
The ratio \( \frac{I}{mr^2} = \frac{2}{3} \) corresponds to a hollow sphere. Therefore, the body rolling without slipping on a horizontal plane, with rotational energy being 40% of the total kinetic energy, is a **hollow sphere**.

### Final Answer
The body is a **hollow sphere**.