A particle travels in a circle of radius 20 cm at a uniformly increasing speed. If the speed changes from `8 ms^(-1)` to `9 ms^(-1)` in 2s, what would be the angular acceleration in `"rad s"^(-2)` ?

Options

A. `1.5 "rad s"^(-2)`
B. `2.5 "rad s"^(-2)`
C. `3.5 "rad s"^(-2)`
D. `4.5 "rad s"^(-2)`

Answer

B

Explanation

Given, radius of the circle `= 20 cm = 20 xx 10^(-2) m`
When speed is `8 ms^(-1) and 9 ms^(-1)`, then angular speeds are respectively
`omega_(1)=(8)/(20xx10^(-2))=40 "rads"^(-1)`
and `omega_(2)=(9)/(20xx10^(-2))=45 "rads"^(-1)`
`:.` Angular acceleration `= (omega_(2)-omega_(1))/(t)=(45-40)/(2)=2.5 "rad s"^(-2)`.