A wheel having moment of inertia `2kgm^(2)` about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel.s rotation in one minute would be……………

Options

A. `(pi)/(18)N/m`
B. `(2pi)/(15)N.m`
C. `(pi)/(12)N.m`
D. `(pi)/(15)N.m`

Answer

D

Explanation

`Deltat=60sec`
speed of wheel = 60 rpm
`therefore v=("60 revolution")/("minute")`
`=(60xx2pir)/(60 sec)`
`v=2pir`
but `omega=(v)/(r )`
`omega=2pi`
For stopping wheel,
(Necessary moment of torque) = (angular momentum of wheel)
`therefore tau.Deltat=Iomega`
`therefore tau(60)=(2 kg m^(2))(2pi)`
`therefore tau=(4pi)/(60)N.m`
`=(pi)/(15)N.m`