Question

A bicycle wheel with a radius of 0.42 m accelerates uniformly for 6.8 s from an initial angular velocity of 5.5 rad/s to a final angular velocity of 6.7 rad/s. What was its angular acceleration

Answer 1

The angular acceleration required is 0.1765 rad/
`s^2`

Step-by-step explanation:

The radius of the bicycle wheel has a radius of 0.42 m.

The acceleration is for time, t = 6.8 seconds.

Initial angular velocity is given as
`\omega_(0)` = 5.5 rad/s

Final angular velocity is given as
`\omega_(f)` = 6.7 rad/s

Therefore from the formula for angular speed we get

`\omega_(f)` =
`\omega_(0)` + (
`(d\omega)/(dt)`
`*` t), where t is the time in seconds.

Therefore we get

6.7 = 5.5 + (6.8 ×
`(d\omega)/(dt)` )

Therefore we get the angular acceleration,
`(d\omega)/(dt)` =
`((6.7 - 5.5 )/(6.8) = 0.1765 rad/ s^2`

The angular acceleration required is 0.1765 rad/
`s^2`