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

Answer

C

Explanation

( c) Given : `I = 2 kg-m^2, omega_0 = (60)/(60) xx 2 pi rad//s`,
`omega = 0, t = 60 s`
The torque required to stop the wheel's rotation is
`tau = I prop = I ((omega_0 - omega)/(t))`
:. `tau = (2 xx 2 pi xx 60)/(60 xx 60) = (pi)/(15) N-m`.