A uniform ring of mass 'm' and radius 'R' is projected horizontally with velcoty `v_(0)` on a rough horizontal floor, so that it starts off with a purely sliding motion and it acquires a pure rolling motion after moving a distance d. If the coefficient of friction between the ground and ring is `mu`, then work done by the friction in the process is

Options

A. `- mu m g d`
B. `- ( 1)/( 4) mv_(0)^(2)`
C. `mu m g d `
D. `- ( 1)/( 8) mv_(0)^(2)`

Answer

To find the work done by friction on a uniform ring of mass 'm' and radius 'R' projected horizontally with an initial velocity \( v_0 \) on a rough horizontal floor, we can follow these steps:

### Step 1: Understand the Initial Conditions
Initially, the ring is sliding with a velocity \( v_0 \) and has no rotational motion. The initial kinetic energy (KE_initial) of the ring is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m v_0^2 \]

### Step 2: Final Conditions for Pure Rolling
After traveling a distance \( d \), the ring transitions to pure rolling motion. In pure rolling, the relationship between the translational velocity \( v \) and the angular velocity \( \omega \) is:
\[ v = R \omega \]
Let the final translational velocity be \( v \). The final kinetic energy (KE_final) when the ring is rolling is the sum of translational and rotational kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \]
For a ring, the moment of inertia \( I \) is:
\[ I = mR^2 \]
Substituting \( \omega = \frac{v}{R} \):
\[ KE_{\text{final}} = \frac{1}{2} m v^2 + \frac{1}{2} mR^2 \left(\frac{v}{R}\right)^2 = \frac{1}{2} m v^2 + \frac{1}{2} m \frac{v^2}{R^2} R^2 = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \]

### Step 3: Conservation of Angular Momentum
We can use the conservation of angular momentum about the point of contact. The initial angular momentum \( L_{\text{initial}} \) is:
\[ L_{\text{initial}} = mv_0 R \]
The final angular momentum \( L_{\text{final}} \) when rolling is:
\[ L_{\text{final}} = mvR + I \omega = mvR + mR^2 \left(\frac{v}{R}\right) = mvR + mvR = 2mvR \]
Setting them equal gives:
\[ mv_0 R = 2mvR \]
Cancelling \( mR \) from both sides (assuming \( R \neq 0 \)):
\[ v_0 = 2v \quad \Rightarrow \quad v = \frac{v_0}{2} \]

### Step 4: Calculate the Work Done by Friction
The work done by friction \( W \) is equal to the change in kinetic energy:
\[ W = KE_{\text{final}} - KE_{\text{initial}} = mv^2 - \frac{1}{2} m v_0^2 \]
Substituting \( v = \frac{v_0}{2} \):
\[ W = m\left(\frac{v_0}{2}\right)^2 - \frac{1}{2} m v_0^2 = m\frac{v_0^2}{4} - \frac{1}{2} m v_0^2 \]
\[ W = \frac{m v_0^2}{4} - \frac{2m v_0^2}{4} = -\frac{m v_0^2}{4} \]

### Final Answer
Thus, the work done by friction is:
\[ W = -\frac{1}{4} m v_0^2 \]