A loop and a disc have same mass and roll without slipping with the same linear velocity `v`. If the total kinetic energy of the loop is `8 J`, the kinetic energy of the disc must be.

Options

A. 6 J
B. 8 J
C. 12 J
D. 16 J

Explanation

To solve the problem, we need to find the kinetic energy of the disc given that the total kinetic energy of the loop is 8 J and both the loop and the disc have the same mass and roll without slipping with the same linear velocity \( v \).

### Step-by-Step Solution:

1. **Understanding Kinetic Energy Components**:
The total kinetic energy \( KE \) of a rolling object is the sum of its translational kinetic energy and rotational kinetic energy:
\[ KE = KE_{\text{trans}} + KE_{\text{rot}} \]
Where:
\[ KE_{\text{trans}} = \frac{1}{2} mv^2 \]
\[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 \]

2. **Moment of Inertia**:
- For the loop (ring), the moment of inertia \( I \) is given by:
\[ I_{\text{loop}} = mR^2 \]
- For the disc, the moment of inertia \( I \) is given by:
\[ I_{\text{disc}} = \frac{1}{2} mR^2 \]

3. **Relating Angular Velocity**:
Since both objects roll without slipping, their angular velocities \( \omega \) are related to their linear velocities \( v \) by:
\[ \omega = \frac{v}{R} \]

4. **Total Kinetic Energy of the Loop**:
Substituting \( I \) and \( \omega \) into the kinetic energy equation for the loop:
\[ KE_{\text{loop}} = \frac{1}{2} m v^2 + \frac{1}{2} (mR^2) \left(\frac{v}{R}\right)^2 \]
Simplifying this:
\[ KE_{\text{loop}} = \frac{1}{2} mv^2 + \frac{1}{2} m v^2 = mv^2 \]
Given \( KE_{\text{loop}} = 8 \, J \), we have:
\[ mv^2 = 8 \, J \quad (1) \]

5. **Total Kinetic Energy of the Disc**:
Now, substituting \( I \) and \( \omega \) into the kinetic energy equation for the disc:
\[ KE_{\text{disc}} = \frac{1}{2} mv^2 + \frac{1}{2} \left(\frac{1}{2} mR^2\right) \left(\frac{v}{R}\right)^2 \]
Simplifying this:
\[ KE_{\text{disc}} = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \]

6. **Substituting from Equation (1)**:
From equation (1), we know \( mv^2 = 8 \, J \):
\[ KE_{\text{disc}} = \frac{3}{4} \times 8 \, J = 6 \, J \]

### Final Answer:
The kinetic energy of the disc is \( \boxed{6 \, J} \).