A uniform ring of radius `R` is given a back spin of angular velocity `V_(0)//2R` and thrown on a horizontal rough surface with velocity of centre to be `V_(0)`. The velocity of the centre of the ring when it starts pure rolling will be

Options

A. `(V_(0))/2`
B. `(V_(0))/4`
C. `(3V_(0))/4`
D. `0`

Explanation

To solve the problem, we need to analyze the motion of the uniform ring as it transitions from sliding to pure rolling. Here are the steps to find the velocity of the center of the ring when it starts pure rolling:

### Step 1: Understand the Initial Conditions
- The ring has a radius \( R \).
- It is given a backspin with an angular velocity \( \omega_0 = \frac{V_0}{2R} \).
- The initial velocity of the center of the ring is \( V_0 \).

### Step 2: Define the Final Conditions
- Let the final velocity of the center of the ring be \( V \).
- Let the final angular velocity when the ring starts pure rolling be \( \omega \).

### Step 3: Apply the Condition for Pure Rolling
For pure rolling, the relationship between the linear velocity \( V \) and the angular velocity \( \omega \) is given by:
\[ V = R \omega \]

### Step 4: Conservation of Angular Momentum
We will consider the angular momentum about the point of contact with the ground. The initial angular momentum \( L_i \) is given by:
\[ L_i = I_{cm} \omega_0 + m V_0 R \]
where \( I_{cm} \) for a ring is \( mR^2 \). Thus,
\[ L_i = mR^2 \left(-\frac{V_0}{2R}\right) + m V_0 R = -\frac{m V_0 R}{2} + m V_0 R = \frac{m V_0 R}{2} \]

### Step 5: Final Angular Momentum
The final angular momentum \( L_f \) when the ring is rolling without slipping is:
\[ L_f = I_{cm} \omega + m V R \]
Substituting \( I_{cm} = mR^2 \) and \( V = R \omega \):
\[ L_f = mR^2 \omega + m V R = mR^2 \omega + m R^2 \omega = 2mR^2 \omega \]

### Step 6: Set Initial and Final Angular Momentum Equal
Since no external torques act on the system, we have:
\[ L_i = L_f \]
Thus,
\[ \frac{m V_0 R}{2} = 2mR^2 \omega \]
Canceling \( m \) and \( R \) (assuming \( R \neq 0 \)):
\[ \frac{V_0}{2} = 2R \omega \implies \omega = \frac{V_0}{4R} \]

### Step 7: Substitute Back to Find \( V \)
Using the relationship \( V = R \omega \):
\[ V = R \left(\frac{V_0}{4R}\right) = \frac{V_0}{4} \]

### Conclusion
The velocity of the center of the ring when it starts pure rolling is:
\[ \boxed{\frac{V_0}{4}} \]