A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occur

Question

A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 31.0 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 200 complete revolutions.

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?

N?

Explanation / Answer

2* rad/revolution
60 seconds/min

200 revs=1256 rad

530 rpm=55.50 rad/s

using

=0+0*t+0.5**t^2
0=0
0=55.50
t=31
=1256
1256=55.50*31+0.5**31^2
solve for
=-0.966 rad/s

now that we have , we can solve for using

=55.50-0.9667*31
25.53 rad/sec
or
243.79 rpm


If the power had stayed off

0=47.1-0.4806*t
solve for t
t=98 seconds

and

=47.1898*98-0.5*0.4806*98^2
2316.8 rad
or 369 revs

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