A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the opposite (clockwise) direction, also at 2.70 rad/s. Calculate the change in the tire's angular velocity Δω and the tire's average angular acceleration αav. (Indicate the direction with the signs of your answers.)

Answer: The change in angular velocity is - 5.40rad/s

And acceleration is - 3.6rad/s

Explanation:

Given that;

The initial angular velocity wi is 2.70rad/s and

The final angular velocity wf is - 2.70rad/s

The time taken ∆t is 1.50s.

(For angular velocity counterclockwise direction is positive while clockwise direction is negative)

The change in the angular velocity ∆w can be written as;

∆w = wf - wi

∆w = - 2.70 - 2.70

∆w = - 5.40rad/s

The angular acceleration Ar which is the change in angular velocity per unit time is;

Ar = ∆w/∆t

Ar = - 5.40/1.5

Ar = - 3.6rad/s^2

Therefore the change in angular velocity is - 5.40rad/s

And acceleration is - 3.6rad/s