A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by `(omegaL)/(2pi)` , if the motion of the pulse and rotation of the loop, both are in same direction then the velocity of the pulse w.r.t. to ground will be:

Options

A. `omegaL`
B. `(omegaL)/(2pi)`
C. `(omegaL)/(pi)`
D. `(omegaL)//(4pi^(2))`

Answer

To solve the problem, we need to find the velocity of a pulse in a rotating circular loop of rope with respect to the ground. Let's break it down step by step.

### Step 1: Understand the given information
We have a circular loop of rope of length \( L \) that rotates with a uniform angular velocity \( \omega \). The velocity of the pulse with respect to the rope due to a slight radial displacement is given as:
\[ v_{pr} = \frac{\omega L}{2\pi} \]

### Step 2: Determine the relationship between the pulse velocity and the rope's velocity
The velocity of the pulse \( v_p \) with respect to the ground can be expressed as:
\[ v_p = v_{pr} + v_r \]
where \( v_r \) is the velocity of the rope at the radius \( r \).

### Step 3: Find the radius of the circular loop
The length of the rope \( L \) is related to the radius \( r \) of the circular loop by the formula:
\[ L = 2\pi r \]
From this, we can solve for \( r \):
\[ r = \frac{L}{2\pi} \]

### Step 4: Calculate the velocity of the rope
The velocity of the rope \( v_r \) at radius \( r \) can be calculated using the formula:
\[ v_r = \omega \cdot r \]
Substituting the value of \( r \) we found:
\[ v_r = \omega \cdot \left(\frac{L}{2\pi}\right) = \frac{\omega L}{2\pi} \]

### Step 5: Substitute values into the pulse velocity equation
Now we can substitute \( v_{pr} \) and \( v_r \) back into the equation for \( v_p \):
\[ v_p = v_{pr} + v_r = \frac{\omega L}{2\pi} + \frac{\omega L}{2\pi} \]
This simplifies to:
\[ v_p = \frac{\omega L}{2\pi} + \frac{\omega L}{2\pi} = \frac{2\omega L}{2\pi} = \frac{\omega L}{\pi} \]

### Step 6: Final result
Thus, the velocity of the pulse with respect to the ground is:
\[ v_p = \frac{\omega L}{\pi} \]

### Conclusion
The correct answer is \( \frac{\omega L}{\pi} \).

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