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

Explanation

`omega_(0)=(2pixx60)/(60)=2pi, t=60 s`
` 0=2pi-alphaxx60implies alpha=(pi)/(30) rad//s^(2)`
`tau=Ialpha=2xx(pi)/(30)=(pi)/(15)N-m`