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25 January, 09:14

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.)

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  1. 25 January, 09:41
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    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
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