If rotational kinetic energy is `50%` of total kinetic energy then the body will be

Options

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

Answer

To solve the given problem, we need to analyze the relationship between rotational kinetic energy and total kinetic energy. Here’s a step-by-step breakdown of the solution:

### Step 1: Define Rotational and Translational Kinetic Energy
- The rotational kinetic energy (KE_rot) of a body is given by the formula:
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity.

- The translational kinetic energy (KE_trans) of a body is given by:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]
where \(m\) is the mass and \(v\) is the linear velocity.

### Step 2: Express Angular Velocity in Terms of Linear Velocity
- For a body rolling without slipping, the relationship between linear velocity \(v\) and angular velocity \(\omega\) is:
\[ v = R \omega \quad \Rightarrow \quad \omega = \frac{v}{R} \]

### Step 3: Substitute Angular Velocity into Rotational Kinetic Energy
- Substituting \(\omega\) in the expression for \(KE_{\text{rot}}\):
\[ KE_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{R}\right)^2 = \frac{I v^2}{2 R^2} \]

### Step 4: Calculate Total Kinetic Energy
- The total kinetic energy (KE_total) is the sum of rotational and translational kinetic energies:
\[ KE_{\text{total}} = KE_{\text{rot}} + KE_{\text{trans}} = \frac{I v^2}{2 R^2} + \frac{1}{2} mv^2 \]

### Step 5: Set Up the Given Condition
- According to the problem, the rotational kinetic energy is 50% of the total kinetic energy:
\[ KE_{\text{rot}} = 0.5 \times KE_{\text{total}} \]
Substituting the expressions we derived:
\[ \frac{I v^2}{2 R^2} = 0.5 \left(\frac{I v^2}{2 R^2} + \frac{1}{2} mv^2\right) \]

### Step 6: Simplify the Equation
- Multiply both sides by 2 to eliminate the fraction:
\[ \frac{I v^2}{R^2} = \frac{I v^2}{2 R^2} + \frac{1}{2} mv^2 \]
- Rearranging gives:
\[ \frac{I v^2}{R^2} - \frac{I v^2}{2 R^2} = \frac{1}{2} mv^2 \]
- Factor out \(v^2\):
\[ v^2 \left(\frac{I}{R^2} - \frac{I}{2 R^2}\right) = \frac{1}{2} mv^2 \]
- Cancel \(v^2\) (assuming \(v \neq 0\)):
\[ \frac{I}{R^2} - \frac{I}{2 R^2} = \frac{1}{2} m \]
- This simplifies to:
\[ \frac{I}{2 R^2} = \frac{1}{2} m \]

### Step 7: Solve for Moment of Inertia
- Rearranging gives:
\[ I = m R^2 \]
- This equation indicates that the moment of inertia \(I\) corresponds to that of a ring or hollow cylinder.

### Conclusion
- Since we derived that \(I = m R^2\), the body must be a **ring** or **hollow cylinder**. The correct answer is **ring**.