A wheel of mass 40 kg and radius of gyrating 0.5 m comes to rest from a speed of 180 revolutions per minute in 3 s. Assuming that the retardation is uniform, then the value of retarding torque `tau` in Nm is

Options

A. `10 pi`
B. `20 pi`
C. `30 pi`
D. `40 pi`

Answer

B

Explanation

`I=m K^(2)=40xx0.25=10 kg m^(2)`
`alpha = (omega_(2)-omega_(1))/(t)=(0-2pi xx 3)/(3)= - 2pi`
`tau = I alpha =-2pi xx 10=-20 pi`