An electric motor of power 75 W rotates a flywheel of moment of inertia 0.36 kg - `m^(2)` at a constant rate of 100 rad `s^(-1)`. If the power is switched off, the time taken for the wheel to come to rest is

Options

A. 12 s
B. 24 s
C. 36 s
D. 48 s

Answer

D

Explanation

Given, `I=0.36 kg-m^(2), omega_(i)=100 rad//sec " and " P=75 W`
`because" Angular acceleration, "alpha=(omega_(f)-omega_(i))/t`
or `" "alpha=-omega_(i)/t " "(because omega_(f)=0)`
or `" "t=omega_(i)/alpha " ...(i)"`
`because` Power of an electric motor, `P=tauomega_(i)`
Torque, `" "tau=Ialpha`
Where, I is moment of inertia,
`therefore " "P=Ialphaomega_(i)`
`rArr " "alpha=P/(omega_(i)) " ...(ii)"`
From Eqs. (i) and (ii), we get
`" "t=omega_(i)/((P//Iomega_(i)))=(Iomega_(i)^(2))/P`
or `" "t=(0.36 times (100)^(2))/75=48 s`