A flywheel of mass 100 kg and radius 1 m is rotating at the rate of 420 rev/min. Find the constant retarding torque to stop the wheel in 14 revolutions, the mass is concentrated at the rim. M.I. of the flywheel about its axis of rotation I = mr^(2).

A flywheel of mass 100 kg and radius 1 m is rotating at the rate of 42

1.	A flywheel of mass 100 kg and radius 1 m is rotating at the rate of 420 rev/min. Find the constant retarding torque to stop the wheel in 14 revolutions, the mass is concentrated at the rim. M.I. of the flywheel about its axis of rotation I = mr^(2).
Answer» Solution :Mass of flywheel = m = 100 kg Radius= R = 1m `I = mr^(2) = 100 xx 1^(2) = 100 kg m^(2)` Initial angular velocity `=omega_(0) = 2pir = 2 xx 3.14 xx 420/60 = 43.96 "rad"//s^(2)` Final angular velocity `=omega =0` Angular displacement in 14 REVOLUTION = `14 xx 2PI = 28 pi` radian. `alpha = (omega^(2) - omega_(0)^(2))/(20) = (0-43.96^(2))/(2 xx 28 pi) =-10.99 "rad"//s^(2)` Torque required to stop the flywheel `=tau = Ialpha = 100 xx 10.99 = 1099` Nm