A ring is rolling without slipping. Its energy of translation is E. Its total kinetic energy will be :-

Options

A. E
B. 2E
C. 3E
D. 4E

Answer

To determine the total kinetic energy of a ring rolling without slipping, we will analyze both its translational and rotational kinetic energies.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A ring rolling without slipping has both translational and rotational motion. The translational motion is due to the center of mass moving, while the rotational motion is due to the ring spinning around its center.

2. **Translational Kinetic Energy**:
- The translational kinetic energy (TKE) of the ring is given by the formula:
\[ TKE = \frac{1}{2} m v^2 \]
- According to the problem, this translational energy is given as \( E \):
\[ E = \frac{1}{2} m v^2 \]

3. **Rotational Kinetic Energy**:
- The rotational kinetic energy (RKE) of the ring is given by the formula:
\[ RKE = \frac{1}{2} I \omega^2 \]
- For a ring, the moment of inertia \( I \) about its center is:
\[ I = m r^2 \]
- The angular velocity \( \omega \) can be related to the linear velocity \( v \) by the equation:
\[ \omega = \frac{v}{r} \]

4. **Substituting Values**:
- Substitute \( I \) and \( \omega \) into the RKE formula:
\[ RKE = \frac{1}{2} (m r^2) \left(\frac{v}{r}\right)^2 \]
- Simplifying this gives:
\[ RKE = \frac{1}{2} m r^2 \cdot \frac{v^2}{r^2} = \frac{1}{2} m v^2 \]

5. **Total Kinetic Energy**:
- The total kinetic energy (TKE) is the sum of the translational and rotational kinetic energies:
\[ \text{Total KE} = TKE + RKE \]
- Substituting the values we have:
\[ \text{Total KE} = E + \frac{1}{2} m v^2 \]
- Since \( \frac{1}{2} m v^2 = E \):
\[ \text{Total KE} = E + E = 2E \]

6. **Final Answer**:
- Therefore, the total kinetic energy of the ring is:
\[ \text{Total KE} = 2E \]