1.

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

Answer»

Solution :(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`


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