A flywheel rotates with a uniform angular acceleration. Its angular speed increases from `2pirad//s` to `10pirad//s` in `4 s`. Find the number of revolutions in this period.

Explanation

`omega_(0)=2pi rad//s, omega=10 pi rad//s`
`omega=omega_(0)+alpha t`
` 10 pi = 2 pi+alphaxx4implies alpha=2pi rad//s^(2)`
`theta = omega_(0) t + (1)/(2)alpha t^(2)`
`= 2pixx4+(1)/(2)xx2pixx(4^(2))`
`= 8pi+16pi=24 pi rad`
Number of revolutions
`N=(theta)/(2pi)=(24pi)/(2pi)=12`