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

Answer

C

Explanation

`I=2 "kgm"^(2), f=(60)/(60)=1 rps`
`omega=2pi` rad/sec
`alpha=(2pi)/(60)` rad/sec
`tau=I alpha=2xx(2pi)/(60)=(pi)/(15) N-m`