Question

A flywheel in a motor is spinning at 500rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.0cm. The power is off for 40.0s, and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions.

A. At what rate is the flywheel spinning when the power comes back on?

B. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer 1

500 rpm = 8.33rps (as in, revs per second).

Considering that the slowdown is linear, we can just treat the revolutions in the same way that we normally would metres, or whatever. So, 210 = 8.33*40 - 0.5*a*40^2

==> a = 0.17rps^2 (that is, every second the number of revs per second drops by 0.41)

So, if v = 8.33- 0.17 *40, we have that v = 1.53, in other words our rpm is down to 91.3

For the second question, we just have 0 = 8.33 - 0.17t, which implies that t is 49 seconds.

Total revolutions =8.33*49-0.5*0.17*40^2=272.17