Question

A flywheel in a motor is spinning at 550 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , 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? Answer in rpm
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 in seconds
C) N = ? rev. Solve for N.

Answer 1

550rpm = 550/60 rps = 9.167 rps

s = ut + 0.5at^2

210 revs = 9.167rps * 40s + 0.5*a*40^2

a = - 0.195 revs/sec^2 ( -ve sign indicates slowing down)

a.)

v = u + at

v = 9.167 + (-0.195)*40 = 1.333 rps

So when the power comes on the flywheel is doing 1.333 revs/sec = 79.98 rpm

b.)

Time to stop = u / a = 9.167rps / 0.195 rpsps = 46.80 seconds

Number of revs in 46.80 seconds

s = 9.167*46.80 + 0.5*(-0.195)*(46.80)^2

s = 429.07 - 214.479 revs = 214.59 revolutions