Question

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 1

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

Given:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

           due to the movement in opposite direction)

           Change in time period = Δt = 1.50 s

Required:

           Change in angular velocity = Δω = ?

           Average angular acceleration = αav = ?

Solution:

          Angular velocity (Δω):

               Δω = ωf - ωi

               Δω = -2.70 - 2.70

               Δω = -5.4 rad/s.

          Average angular acceleration (αav):

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

Answer 2

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