Two identical rings are moving with equal kinetic energy One ring rolls without slipping and other ring is in pure translational motion. The ratio of their respective speeds of centre of mass is

Options

A. ` 1 : sqrt(2)`
B. `sqrt (2) : sqrt (3) `
C. ` 1 : sqrt(3)`
D. ` 1 : 1`

Answer

To solve the problem, we need to analyze the kinetic energies of the two identical rings, one rolling without slipping and the other moving with pure translational motion.

### Step-by-Step Solution:

1. **Identify the Kinetic Energy of Each Ring**:
- For the ring that rolls without slipping, the total kinetic energy (KE) can be expressed as:
\[ KE_{\text{rolling}} = \frac{1}{2} mv_1^2 + \frac{1}{2} I \omega^2 \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a ring, \( I = m r^2 \) and \( \omega = \frac{v_1}{r} \).

- Therefore, substituting \( I \) and \( \omega \):
\[ KE_{\text{rolling}} = \frac{1}{2} mv_1^2 + \frac{1}{2} (m r^2) \left(\frac{v_1}{r}\right)^2 \]
Simplifying this gives:
\[ KE_{\text{rolling}} = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_1^2 = mv_1^2 \]

2. **Kinetic Energy of the Translating Ring**:
- For the ring that is in pure translational motion, the kinetic energy is:
\[ KE_{\text{translational}} = \frac{1}{2} mv_2^2 \]

3. **Setting the Kinetic Energies Equal**:
- Since both rings have equal kinetic energies, we set:
\[ mv_1^2 = \frac{1}{2} mv_2^2 \]
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ v_1^2 = \frac{1}{2} v_2^2 \]

4. **Solving for the Ratio of Speeds**:
- Taking the square root of both sides gives:
\[ v_1 = \frac{v_2}{\sqrt{2}} \]
- Therefore, the ratio of their respective speeds of the center of mass is:
\[ \frac{v_1}{v_2} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The ratio of their respective speeds of center of mass is \( \frac{1}{\sqrt{2}} \).