Find out the number of revolution:
A flywheel of mass 20 kg and radius 100 mm is made to rotate at 600 RPM. Determine the KE of the flywheel. If the frictional couple at its bearing is 10 N.m., find out the number of revolution it shall make before coming to rest.
Solution
ω = 600 rpm = 10 r.p s.
= (10 × 2 π) rad / sec. = 62.8 rad / sec.
I _{m} = M a^{ 2} / 2
= 20 × ((0.1)^{2} /2) = 0.1 kg m ^{2} = 197 N m.
If frictional couple at the bearing is CF, the W. D. by friction is (C_{F } × θ) to bring it to rest while
C _{F} × θ = KE of the flywheel
10 × θ = 197
∴ θ= 19.7 radians.
No. of cycles of rotations = θ /2 π = 3.14 cycles.