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.

Answer 2

Answer:

• The rope is $0.837meters$ long

Explanation:

$w = 0.20 rev/sec\\\\w = 0.20 * 271\\\\ w = 0.4\pi rad/sec$

using conservation of angular momentum

$L = mr^2w\\\\L = m * 2^2 * (0.4\pi)\\\\L = 5.026m$

and,

$L = mvr$

therefore,

$5.026m = 5*m*r\\\\5.026 = 5 * r\\\\r = \frac{5.026}{6}\\\\r = 0.837meters$

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