Question

A circular loop of rope of length l rotates with uniform angular velocity about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with respect to rope) produced due to slight radial displacement is given by

Answer 1

Answer:

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

r

vt

.

Explanation:

Given that,

Length = L

Angular velocity = \omegaω

We know that,

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

v= r\omegav=rω

\omega = \dfrac{v}{r}ω=

r

v

The radial displacement is defined as:

\theta = \omega tθ=ωt ....(I)

Here, \thetaθ = radial displacement

\omegaω = angular velocity

t= time

Put the value of \omegaω in equation (I)

\theta = \dfrac{vt}{r}θ=

r

vt

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

r

vt

.

Answer 2

Answer: wL/(2π)

Explanation:

Pls check the attachment. Hope it helps. Mark me brainliest.