1.

A flywheel rotating at `420 rpm` slows down at a constant rate of `2 rad s^-2` The time required to stop the flywheel is.A. 22 sB. 11 sC. 44 sD. 12 s

Answer» Correct Answer - A
(a) Here, `v_0 = 420 rpm = 7 rps`
:. `omega_0 = 2 pi v_0 = 2 xx (22)/(7) xx 7 = 44 rad s^-1`
`omega = 0, prop = -2 rad s^-2`
`t = (omega - omega_0)/(prop) = (-44)/(-2) = 22 s`.


Discussion

No Comment Found

Related InterviewSolutions