Question

A flywheel of moment of inertia 5.0 kg-m² 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.

Answer 1

A flywheel of moment of inertia 5 kg m is rotated at a speed of 60 rad/s. The flywheel comes to rest due to the friction at the axle after 5 minutes.
Therefore, the angular deceleration produced due to frictional force = ω = ω₀ + αt
ω₀ = –αt (ω = 0+)

α= –(60/5 × 60) = –1/5 rad/s².
a) Therefore total workdone in stopping the wheel by frictional force
W = 1/2 iω² = 1/2 × 5 × (60 × 60) = 9000 Joule = 9 KJ.
b) Therefore torque produced by the frictional force (R) is
I_R = I × α = 5 × (–1/5) = IN – m opposite to the rotation of wheel.
c) Angular velocity after 4 minutes
ω = ω₀ + αt = 60 – 240/5 = 12 rad/s
Therefore angular momentum about the centre = 1 × ω = 5 × 12 = 60 kg-m²/s.