Question

A flywheel of mass 35.0 kg and diameter 60.0 cm spins at 400 rpm when it experiences a sudden power loss. The flywheel slows due to friction in its bearings during the 20.0 s the power is off. If the flywheel makes 200 complete revolutions during the power failure, (a) at what rate is the flywheel spinning when the power comes back on? (b) How long would it have taken for the flywheel to come to a complete stop?

Answer 1

To solve this problem it is necessary to apply the concepts of displacement, speed and acceleration given by classical mechanics.

Our previously given values correspond to,

$\omega =400rpm = 400*\frac{2*pi}{60}= 41.9 rad/s$

$\theta=200 revs = 200*2*pi= 1257.14 rad$

By definition we know that angular displacement is defined by,

$\theta = \theta_0 +\omega_0*t+\frac{1}{2}\alpha t^2$

Here we know that there is no initial deployment so replacing the values we have to,

$1257.14 = 0+41.9*20+0.5*\alpha*20^2$

Solving for $\alpha$

$\alpha = 2.0957 rad/s^2$

From the acceleration description equations we know that,

$\omega_f = \omega_i+\alpha t$

$\omega_f = 41.9+2.0957*20 = 83.814 rad/s$

The moment the body reaches the final velocity 0, it will take a period of time and a couple of turns the point at which it stops, then

$\omega_f = \omega_i+\alpha t$

$0=41.9-2.0957*t$

$t = \frac{41.9}{2.0957} = 19.99 s$

The angular displacement will then be given by

$\theta = \omega_0*t-\frac{1}{2}\alpha t$

$\theta=41.9*19.99-0.5*2.0957*19.992$

$\theta= 418.86 rad$