Question

A tetherball is attached to a pole with a 2.0-m rope. It is circling at 0.20 rev/s. As the rope wraps around the pole it shortens. How long is the rope when the ball is moving at 5.0 m/s?

Answer 1

Answer:

The length of the rope  is 1.005 m.

Explanation:

Given that,

Length of rope = 2.0 m

Angular speed = 0.20 rev/s

Linear speed = 5.0 m/s

Since there are no external torques about the rotation  axis

We need to calculate the length of the rope

Using conservation of angular momentum

$I_{1}\omrga_{1}=I_{2}\omega_{2}$

$mr_{1}^2\times\omega_{1}=mr_{2}^2\times\omega_{2}$

$r_{1}^2\times\omega_{1}=r_{2}^2\times\dfrac{v_{2}}{r_{2}}$

$r_{2}=\dfrac{r_{1}^2\times\omega_{1}}{v_{2}}$

Put the value into the formula

$r_{2}=\dfrac{2.0^2\times0.20\times2\pi}{5.0}$

$r_{2}=1.005\ m$

Hence, The length of the rope  is 1.005 m.