Question

A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of the speck at t = 8.1 s ? Your answer should be between 0 and 2π rad. g

Answer 1

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

`\theta= \theta_0 +\omega t`

From our given data we know that,

`\omega = 76(rev)/(min)`

`\omega = 76(rev)/(min)((2\pi rad)/(1rev))((1 min)/(60s))`

`\omega = 7.958rad/s`

Moreover we know that

`\theta_0 = 0.47 rad`

Therefore for time t=8.1s we have,

`\theta= \theta_0+ \omega t`

`\theta= 0.47+(7.958)(8.1)`

`\theta = 64.9298rad`

That number in revolution is:

`\theta = 64.9298rad((1rev)/(2\pi))`

`\theta = 15.108 Revolutions`

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

`\theta_(net) = 0.108*2\pi=0.216\pi`

Therefore the angle of the speck at a time 8.1s is
`0.216\pi`