Answer to Question #197174 in Physics for gibson

Question

A motor turning at 1800rev/min rotates uniformly to a stop in 20s.

(a) Find its angular deceleration in rev/s2 and rad/s2.

(b) The motor has a wheel of radius 10cm attached to its shaft. What length does the wheel

wind in the 20s?

Answer 1

(a) The angular deceleration can be found from the kinematic equation:

$$\omega=\omega_0+\alpha t,$$ $$\alpha=\dfrac{\omega-\omega_0}{t},$$ $$\alpha=\dfrac{0-1800\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}}{20\ s}=-1.5\ \dfrac{rev}{s^2},$$ $$\alpha=\dfrac{0-1800\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}}{20\ s}=-9.42\ \dfrac{rad}{s^2}.$$

The sign minus means that the motor decelerates.

(b) Let's find the initial linear velocity of the wheel:

$$v=\omega r=1800\ \dfrac{rev}{min}\cdot\dfrac{1\ min}{60\ s}\cdot\dfrac{2\pi\ rad}{1\ rev}\cdot0.1\ m=18.85\ \dfrac{m}{s}.$$

Let's also find the deceleration of the wheel in m/s²:

$$a=\dfrac{v-v_0}{t}=\dfrac{0-18.85\ \dfrac{m}{s}}{20\ s}=-0.94\ \dfrac{m}{s^2}.$$

Then, we can find the length that the wheel winds in 20 seconds from the kinematic equation:

$$L=d=v_0t+\dfrac{1}{2}at^2,$$ $$L=18.85\ \dfrac{m}{s}\cdot20\ s+\dfrac{1}{2}\cdot(-0.94\ \dfrac{m}{s^2})\cdot(20\ s)^2=189\ m.$$