A flywheel of moment of inertia 5.0 `kg-m^2` is rotated at a speed of 60 rad/s. Because of the friction at the axle, it comes to rest in 5.0 minutes. Find a. The average torque of the friction. B. the total work done by the friction and c. the angular momentum of the wheel 1 minute before it stops rotating.

Explanation

Here, `I = 5.0 kg m^(2), omega_(0) = 60 rad//s`
`omega = 0, t = 5 xx 60 = 300s, tau = ?`
From `omega = omega_(0) + alpha t`
`0 = 60 + alpha xx 300, alpha = (-1)/(5) rad//s^(2)`
(a) Average torque of friction `= I alpha = 5.0 (-(1)/(5))`
`=- N-m`
(b) Total work done by friction `= KE` of rotation
`W = (1)/(2)I omega_(0)^(2) = (1)/(2) xx 5.0 (60)^(2) = 9000 J = 9 kJ`
(c ) From `omega = omega_(0) + alpha t`
`= 60 - (1)/(5) xx (5 - 1) xx 60 = 12 rad//s`
`L = I omega = 5.0 xx 12 = 60 kg m^(2) s^(-1)`