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21 November, 08:45

A Frisbee of radius 0.15 m is accelerating at a constant rate from 7.1 revolutions per second to 9.3 revolutions per second in 6.0 s. What is its angular acceleration?

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  1. 21 November, 10:38
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    Answer: 17.03 rad/s^2

    Explanation: One of the equation of motion that defines a circular motion with constant acceleration is given below as

    ω = ω0 + αt

    Where ω = final angular velocity = 9.3 rev/s

    ω0 = initial angular velocity = 7.1 rev/s

    α = angular acceleration = ?

    t = time taken = 6.0s

    By substituting the parameters, we have that

    9.3 = 7.1 + α (6)

    9.3 - 7.1 = 6α

    16.3 = 6α

    α = 16.3/6

    α = 2.71 rev/s^2

    We can also give the answer in rad/s^2 (another unit of angular acceleration) by multipying by 2π

    Hence, α = 2.71 * 2π, where π = 3.142

    α = 17.03 rad/s^2
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