Question

A high-speed flywheel in a motor is spinning at 500 rpm (revolutions per minute) when a power failure suddenly occurs. The flywheel has mass 39.0 kg and diameter 77.0 cm . The power is off for 25.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 160 complete revolutions.

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

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?

How many revolutions would the wheel have made during this time?

Answer 1

Part A.

for constant angular acceleration we know that:

wi = Initial angular speed = 500 rpm = 500*2*pi/60 = 52.36 rad/sec

wf = final angular speed = ?

t = time interval = 25.0 sec

= 160 rev = 160*2*pi rad, So

wf = [2*()/t] - wi

wf = [2*160*2*pi/25.0] - 52.36

wf = 28.06 rad/sec

Part B.

Using 1st rotaional kinematic equation:

wf = wi + *t

= (wf - wi)/t = (28.06 - 52.36)/25.0

= -0.972 rad/sec^2

Now when flywheel completely stops, then using 1st rotational kinematic equation again:

w1f = wi + *t1

w1f = 0 rad/sec

t1 = (0 - 52.36)/(-0.972)

t1 = 53.87 sec = total time taken by flywheel to stop

Part C.

Using 2nd rotational kinematic equation:

= wi*t1 + (1/2)**t1^2

= 52.36*53.87 + (1/2)*(-0.972)*53.87^2

= 1410.3 rad = 1410.3/(2*pi)

= 224.45 rev

Let me know if you've any query.