The angular speed of a moton wheel is increased from `1200` rpm to 3120 rpm in `16` seconds. (i) What is the angular acceleration, assuming the acceleration to be uniform ? (ii) How many revolutions does the engine make during this time ?

Explanation

(i) We shall use `ω= ω_(0) + α t`
`ω_(0)` = initial angular speed in rad/s
= `2 π ×` angular speed in rev/s
= `(2pixx" angular speed in rev/min")/(60s//min)`
`=(2pixx1200)/(60)"rad/s"`
`=40pi" rad/s"`
Similarly `omega` = final angular speed in rad/s
`=(2pixx3120)/(60)"rad/s"`
`=2pixx52" rad/s"`
`=104pi" rad/s"`
`therefore` Angular acceleration
`alpha=(omega-omega_(0))/(t)=4pi" rad/s"^(2)`
The angular acceleration of the engine = `4π "rad/s"^(2)`
(ii) The angular displacement in time t is given by
`theta=omega_(0)t+(1)/(2)alphat^(2)`
`=(40pixx16+(1)/(2)xx4pixx16^(2))rad`
`=(640pi+512pi)rad`
`=1152pirad`
Number of revolutions = `(1152pi)/(2pi)=576`