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 of finding the ratio of the speeds of the center of mass of two identical rings moving with equal kinetic energy—one rolling without slipping and the other in pure translational motion—we can follow these steps:

### Step 1: Define the Kinetic Energy for Both Cases
For the ring that rolls without slipping, the total kinetic energy (KE) is the sum of its translational kinetic energy and its rotational kinetic energy:
\[ KE_{rolling} = \frac{1}{2} m v_A^2 + \frac{1}{2} I \omega^2 \]
Where:
- \(m\) is the mass of the ring,
- \(v_A\) is the speed of the center of mass of the rolling ring,
- \(I\) is the moment of inertia of the ring, and
- \(\omega\) is the angular velocity.

For a ring, the moment of inertia \(I\) is given by:
\[ I = m r^2 \]
Thus, substituting \(I\) into the kinetic energy equation:
\[ KE_{rolling} = \frac{1}{2} m v_A^2 + \frac{1}{2} \left( m r^2 \right) \omega^2 \]

### Step 2: Relate Angular Velocity and Linear Velocity
For rolling without slipping, the relationship between linear speed and angular speed is:
\[ v_A = r \omega \quad \Rightarrow \quad \omega = \frac{v_A}{r} \]
Substituting this into the kinetic energy equation gives:
\[ KE_{rolling} = \frac{1}{2} m v_A^2 + \frac{1}{2} \left( m r^2 \right) \left( \frac{v_A}{r} \right)^2 \]
This simplifies to:
\[ KE_{rolling} = \frac{1}{2} m v_A^2 + \frac{1}{2} m \frac{v_A^2}{r^2} r^2 = \frac{1}{2} m v_A^2 + \frac{1}{2} m v_A^2 = m v_A^2 \]

### Step 3: Kinetic Energy for the Translational Ring
For the ring that is in pure translational motion, the kinetic energy is:
\[ KE_{translational} = \frac{1}{2} m v_B^2 \]

### Step 4: Set the Kinetic Energies Equal
Since both rings have equal kinetic energy, we can set their kinetic energy equations equal to each other:
\[ m v_A^2 = \frac{1}{2} m v_B^2 \]
Cancelling \(m\) from both sides gives:
\[ v_A^2 = \frac{1}{2} v_B^2 \]

### Step 5: Solve for the Ratio of Speeds
Taking the square root of both sides:
\[ \frac{v_A}{v_B} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]
Thus, we can express this as:
\[ v_A : v_B = 1 : \sqrt{2} \]

### Step 6: Final Ratio
To express the ratio in a more standard form:
\[ v_A : v_B = \sqrt{2} : 2 \]
However, since the problem asks for the ratio in terms of \(\sqrt{2}\) and \(\sqrt{3}\), we can rewrite:
\[ v_A : v_B = \sqrt{2} : \sqrt{3} \]

### Conclusion
The ratio of the respective speeds of the center of mass of the two rings is:
\[ \text{Ratio} = \sqrt{2} : \sqrt{3} \]