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=2pinrArr omega_(1)=2pi n_(1)`
where, `n_(1)=1800 "rpm",n_(2)=3000 "rpm" rArr Delta t=20s`
`omega_(1)=2pixx(1800)/(60)=2pixx30=60pi`
Similarly, `omega_(2)=2pi n_(2)=2pixx(3000)/(60)=2pixx50=100pi`
If the angular velocity of a rotating wheel about an axis changes by change in angular velocity in a time interval `Delta t`, then the angular acceleration of rotating wheel about that axis is
`alpha=("Change in angular velocity")/("Timem interval")`
`alpha=(omega_(2)-omega_(1))/(Deltat)rArr alpha=(100pi-60pi)/(20)`
`alpha=(40pi)/(20)=2pi "rads"^(-2)`.