Question

The motor of an engine is rotating about its axis with angular velocity of `120 rpm`. It comes to rest in `10s`, after being switched off. Assuming constant deceleration, calculate the number of revolutions made by it before coming to rest.

Answer 1

Here `n=12 rpm=2rps`
`omega_(0)=2pin=4pirads^(-1), omega=0 ` and `t=10 s`
`omega=omega_(0)+alphat`
`0=4pi+alphaxx10` or `alpha=-0.4pirads^(-2)`
Also, the angle covered by the motor, `theta=omega_(0)t+1/2alphat^(2)`
`:. Theta=4pixx10+1/2xx(-0.4 pi)xx10^(2)=40 pi-20 pi=20 pi rad`
Hence, the number of revolutions completed.
`N=(theta)/(2pi)=(20 pi)/(2pi)=10`