Question

A circular loop of rope of length L rotates with uniform angular velocity w about an axis through its center on a horizontal smooth platform. Velocity of pulse produced due to slight radial displacement is given by.....
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Answer 1

L=2πr>>r=L/2π

v=rw>>v=w×L/2π

Answer 2

Answer:

The radial displacement is $\theta = \dfrac{vt}{r}$.

Explanation:

Given that,

Length = L

Angular velocity = $\omega$

We know that,

The liner velocity is the product of the angular velocity and radius of the circle.

$v= r\omega$

$\omega = \dfrac{v}{r}$

The radial displacement is defined as:

$\theta = \omega t$....(I)

Here, $\theta$= radial displacement

$\omega$= angular velocity

t= time

Put the value of $\omega$ in equation (I)

$\theta = \dfrac{vt}{r}$

Hence, The radial displacement is $\theta = \dfrac{vt}{r}$.