Answer to Question #270460 in Physics for Jane

Question

1. A wheel with a diameter of 80cm speeds up from 30 rad/s to 80 rad/s. The linear acceleration at the edge of the wheel is 15 m/s2. How many revolutions will the wheel go through during this period?

Answer 1

Let's first find the angular acceleration of the wheel:

$$a=r\alpha,$$ $$\alpha=\dfrac{a}{r}=\dfrac{15\ \dfrac{m}{s^2}}{0.4\ m}=37.5\ \dfrac{rad}{s^2}.$$

Also, we need to find the time that the wheel takes to accelerate:

$$t=\dfrac{\omega_f-\omega_i}{\alpha}=\dfrac{80\ \dfrac{rad}{s}-30\ \dfrac{rad}{s}}{37.5\ \dfrac{rad}{s^2}}=1.33\ s.$$

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

$$\theta=\omega_it+\dfrac{1}{2}\alpha t^2,$$ $$\theta=30\ \dfrac{rad}{s}\times1.33\ s+\dfrac{1}{2}\times37.5\ \dfrac{rad}{s^2}\times(1.33\ s)^2=73\ rad.$$

Finally, we can find the number of revolutions that the wheell will go through during  this period:

$$n=\dfrac{\theta}{2\pi\ \dfrac{rad}{rev}}=\dfrac{73\ rad}{2\pi\ \dfrac{rad}{rev}}=12\ rev.$$