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 step-by-step, we will analyze the given information about a body rolling without slipping on a horizontal plane and determine which type of body it is based on the provided conditions.

### Step 1: Understand 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):
\[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} \]
Where:
- \( KE_{\text{trans}} = \frac{1}{2} mv^2 \)
- \( KE_{\text{rot}} = \frac{1}{2} I \omega^2 \)

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

### Step 3: Express Rotational Kinetic Energy in Terms of Linear Velocity
Substituting \( \omega = \frac{v}{r} \) into the rotational kinetic energy expression:
\[ KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \frac{I}{r^2} v^2 \]

### Step 4: Substitute into Total Kinetic Energy
Now, we can express the total kinetic energy in terms of \( v \):
\[ KE_{\text{total}} = \frac{1}{2} mv^2 + \frac{1}{2} \frac{I}{r^2} v^2 \]
Factoring out \( \frac{1}{2} v^2 \):
\[ KE_{\text{total}} = \frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right) \]

### Step 5: Use the Given Condition
According to the problem, the rotational kinetic energy is 40% of the total kinetic energy:
\[ KE_{\text{rot}} = 0.4 \times KE_{\text{total}} \]
Substituting the expressions we derived:
\[ \frac{1}{2} \frac{I}{r^2} v^2 = 0.4 \times \left(\frac{1}{2} v^2 \left(m + \frac{I}{r^2}\right)\right) \]

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

### Step 7: Identify the Body
The moment of inertia \( I \) for a hollow sphere is given by:
\[ I = \frac{2}{3}mr^2 \]
Since we derived \( \frac{I}{r^2} = \frac{2}{3}m \), this matches the moment of inertia of a hollow sphere.

### Conclusion
Thus, the body that is rolling without slipping on a horizontal plane, with rotational energy being 40% of the total kinetic energy, is a **hollow sphere**.

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