Answer to Question #172490 in Physics for Diocos

Question

A flywheel slows from 600 to 400 rev/min while rotating through 40 revolutions

1) Angular Acceleration of Flywheel

2) The Elapsed During 40 Revolutions

Answer 1

By definition, the angular acceleration is:

$$\varepsilon =\dfrac{\Delta \omega }{ t}$$

where $\Delta \omega = 400\space rev/min - 600\space rev/min = -200\space rev/min$ is the gain in angular velocity in time $t$. On the other hand, the number of revolutions under the constant acceleration are given as follows:

$$\varphi = \omega_0t+ \dfrac{\varepsilon t^2}{2}$$

where $\omega_0 = 600\space rev/min$ is the initial angular velocity. Substituting the expression for $\varepsilon$ and expressing $t$, obtain:

$$t= \dfrac{\varphi}{\omega_0 + \Delta \omega /2}$$

since $\varphi = 40$, obtain:

$$t= \dfrac{40}{600 -200 /2} = 0.08 min$$

The acceleration is then:

$$\varepsilon = -\dfrac{200\space rev/min}{0.08min} = -2500\space rev/min^2$$

Answer. 1) -2500 rev/min, 2) 0.08 min.