A uniform smooth rod (mass `m` and length `l`) placed on a smooth horizontal floor is it by a particle (mass `m`) moving on the floor, at a distance `l/4` from one end elastically `(e = 1)`. The distance travelled by the centre of the rod after the collision when it has completed three revolutions will be

Options

A. `2pil`
B. cannot be determined
C. `pil`
D. none of the above

Explanation

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the Setup
We have a uniform smooth rod of mass `m` and length `l` placed on a smooth horizontal floor. A particle of mass `m` hits the rod at a distance of `l/4` from one end. The collision is elastic, meaning both momentum and kinetic energy are conserved.

### Step 2: Determine the Impulse and Angular Impulse
When the particle collides with the rod, it exerts an impulse on the rod. The impulse can be expressed as:
\[ I = m \cdot v_0 \]
where \( v_0 \) is the initial velocity of the particle before the collision.

The angular impulse about the center of mass of the rod is given by:
\[ \text{Angular Impulse} = I \cdot \text{distance from center} \]
The distance from the center of the rod to the point of impact is \( l/4 \) (since the center is at \( l/2 \) and the impact point is at \( l/4 \)):
\[ \text{Angular Impulse} = I \cdot \left(\frac{l}{4}\right) \]

### Step 3: Relate Linear and Angular Quantities
The angular impulse can also be expressed in terms of the final angular velocity \( \omega_f \) of the rod:
\[ I \cdot \frac{l}{4} = I_{cm} \cdot \omega_f \]
where \( I_{cm} \) is the moment of inertia of the rod about its center of mass, given by:
\[ I_{cm} = \frac{1}{12} m l^2 \]

### Step 4: Calculate the Final Angular Velocity
From the angular impulse equation, we can solve for \( \omega_f \):
\[ I \cdot \frac{l}{4} = \frac{1}{12} m l^2 \cdot \omega_f \]
Substituting \( I = m v_0 \):
\[ m v_0 \cdot \frac{l}{4} = \frac{1}{12} m l^2 \cdot \omega_f \]
Cancelling \( m \) from both sides:
\[ v_0 \cdot \frac{l}{4} = \frac{1}{12} l^2 \cdot \omega_f \]
Rearranging gives:
\[ \omega_f = \frac{3 v_0}{l} \]

### Step 5: Calculate the Time to Complete Three Revolutions
The angle rotated in three revolutions is:
\[ \theta = 3 \cdot 2\pi = 6\pi \]
Using the relationship between angular displacement, angular velocity, and time:
\[ \theta = \omega_f \cdot t \]
Substituting for \( \omega_f \):
\[ 6\pi = \left(\frac{3 v_0}{l}\right) t \]
Solving for \( t \):
\[ t = \frac{6\pi l}{3 v_0} = \frac{2\pi l}{v_0} \]

### Step 6: Calculate the Distance Traveled by the Center of the Rod
The distance traveled by the center of the rod while it completes three revolutions is given by:
\[ \text{Distance} = v_{cm} \cdot t \]
Where \( v_{cm} \) is the velocity of the center of mass of the rod after the collision. The velocity of the center of mass can be found using:
\[ v_{cm} = \frac{v_0}{2} \]
Thus, the distance traveled is:
\[ \text{Distance} = \frac{v_0}{2} \cdot \frac{2\pi l}{v_0} = \pi l \]

### Final Answer
The distance traveled by the center of the rod after it has completed three revolutions is:
\[ \text{Distance} = \pi l \]

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