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= 2kg m^(2), f = (60)/(60)= 1r ps`
`omega= 2pi` rad/sec
`alpha= (2pi)/(60)` rad/sec
`tau= I alpha = 2 xx (2pi)/(60) = (pi)/(15) N-m`