Answer to Question #270458 in Physics for Jane

Question

A disk with diameter of 60 cm speeds up from 20 rad/s to 40 rad/s in 5 seconds. (a) How many revolutions will the disk go through during that time period? (b) What is the average angular acceleration of the disk? (c) What is the average linear acceleration of the disk?

Answer 1

(a)-(b) Let's first find the average angular acceleration of the disk:

$\alpha=\dfrac{\omega_f-\omega_i}{t}=\dfrac{40\ \dfrac{rad}{s}-20\ \dfrac{rad}{s}}{5\ s}=4\ \dfrac{rad}{s^2}.$

Then, we can find the angular displacement of the disk:

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2,$ $\theta=20\ \dfrac{rad}{s}\times5\ s+\dfrac{1}{2}\times4\ \dfrac{rad}{s^2}\times(5\ s)^2=150\ rad.$

Finally, we can find the number of revolutions that the disk will go through during 5 seconds:

$n=\dfrac{\theta}{2\pi\ \dfrac{rad}{rev}}=\dfrac{150\ rad}{2\pi\ \dfrac{rad}{rev}}=24\ rev.$

(c) We can find the average linear acceleration of the disk as follows:

$a=r\alpha=0.3\ m\times4\ \dfrac{rad}{s^2}=1.2\ \dfrac{m}{s^2}.$