Answer to Question #197173 in Physics for gibson

Question

A wheel turning at 50rev/min has its speed increased to 65rev/min. If the increase occurred

in 20s, find the

(a) rate at which the angular speed changes.

(b) angular displacement during this time in

(i)

(ii)

(iii) radians

revolutions

degrees

Answer 1

(a) We can find the angular acceleration from the kinematic equation:

$$\omega=\omega_0+\alpha t,$$ $$\alpha=\dfrac{\omega-\omega_0}{t},$$ $$\alpha=\dfrac{(65\ \dfrac{rev}{min}-50\ \dfrac{rev}{min})\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{20\ s}=0.078\ \dfrac{rad}{s^2}.$$

(b) We can find the angular displacement during this time from the kinematic equation:

$$\theta=\omega_0t+\dfrac{1}{2}\alpha t^2.$$

(i)

$$\theta=50\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}\cdot20\ s+\dfrac{1}{2}\cdot0.078\ \dfrac{rad}{s^2}\cdot(20\ s)^2=120\ rad.$$

(ii)

$$\theta=120\ rad\cdot\dfrac{1\ rev}{2\pi\ rad}=19\ rev.$$

(iii)

$$\theta=120\ rad\cdot\dfrac{180^{\circ}}{\pi\ rad}=6875^{\circ}.$$