Question

A flywheel of mass `25 kg` has a radius of `0.2 m`. It is making 240 rpm. What is the torque necessary to bring it to rest in `20 s` ? If the torque is due to a force applies tangentially on the rim of the wheel, what is the magnitude of the force ? Assume that mass of flywheel is concentrated at its rim.

Answer 1

Here, `M = 25 kg, R = 0.2 m`,
`n_(1) = 240 "rpm" = (240)/(60) rps = 4 rps, n_(2) = 0`,
`t = 20 s`
Moment of inertia of flywheel,
`I = MR^(2) = 25 (0.2)^(2) = 1 kg m^(2)`
`tau = I alpha = (I (omega_(2) - omega_(1)))/(t) = I(2pi(n_(2) - n_(1)))/(t)`
`tau = (1 xx 2pi(0 - 4))/(20) = - 0.4 pi = - 1.257 N-m`
If `F` is the tangential force applied on the rim,
then as `F xx R = tau`
`:. F = (tau)/(R ) = (1.257)/(0.2) = 6.285 N`