A flywheel rotating at the rate of 120 rpm slows down at a constant rate of `2 rad s^(-2)`. What time is required to stop the flywheel and how many rations does it make in the process ?

Explanation

Here, `n_(1) = 120 "rpm" = 2rps`.
`alpha =- 2 rad s^(-2), n_(2) = 0 , t = ?, theta = ?`
From, `omega_(2) = omega_(1) + alpha t`
`t = (omega_(2) - omega_(1))/(alpha) = (2pi(n_(2) - n_(1)))/(alpha) = (2pi(0- 2))/(-2) = 2pi sec`
From, `theta = omega_(1) t + (1)/(2) alpha t^(2) = 2pi (2).2pi + (1)/(2)`
`(-2) (2pi)^(2) = 4pi^(2)`
This is the angle traced by the flywheel before stopping. As angle traced in one revolution is `2 pi`
radian, therefore, number of revolutions complated
`= (theta)/(2pi) = (4pi^(2))/(2pi) = 2pi`