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