A flywheel in the form of a disc is initially at rest. When a constant torque acts on it for one minute, it attains a speed of 600 rpm. Calculate the torque, if mass and radius of the disc are 10 kg and 0.4m respectively.

Explanation

M = 10kg, R=0.4m
`I=(MR^(2))/(2)=10((0.4)^(2))/(2)0.8kgm^(2)`
`omega_(i)=0,omega_(f)=(2piN)/(t)=(2pi(600))/(60)=20pirads^(-1)`
time taken for the change of speed, t=60 s.
`alpha=(omega_(f)-omega_(i))/(t)=(20pi-0)/(60)=(pi)/(3)rads^(-1)`
The torque, `tau=Ialpha=0.8((pi)/(3))=0.8379Nm`.