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

Options

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

Answer

C

Explanation

`omega_(f)=omega_(i)-alphatrArr0=omega_(i)-alphat`
`thereforealpha=omega_(i)//t,` where `alpha` is retardation.
The torque on the wheel is given by
`tau=Ialpha=(Iomega_(i))/t=(I.2piv)/t=(2xx2xxpixx60)/(60xx60)`
`rArr=pi/15` Nm
This is the torque required to stop the wheel in 1 min. (or 60 sec.).