Question

A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming to be uniform?
A. `60 pi rad s^(-2)`
B. `90 pi rad s^(-2)`
C. `2 pi rad s^(-2)`
D. `40 pi rad s^(-2)`

Answer 1

Correct Answer - C
We know that
`omega=2pi n rArromega_(1)=2pin_(1)`
where, `n_(1) = 1800` rpm
`n_(2) = 3000` rpm
`omega_(1)=2pixx(1800)/(60)=2pixx30=60pi`
Similarly, `omega_(1)=2pin_(2)=2pixx(3000)/(60)`
`rArr = 2pi xx 50 =100pi `
If the angular velocity of rotating wheel about an axis changes by change in angular velocity in a time intervval `Delta t`, then the angular acceleration of rotating wheel about that axis is
`alpha=("Change in angular velocity")/("Time interval")`
`alpha=(omega_(2)-omega_(1))/(Deltat)rArr alpha=(100pi-60pi)/(20)`
`alpha=(40pi)/(20)=2pi "rad s"^(-2)`.