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6 September, 20:13

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s.

A) How much work must be done to stop it?

B) What is the required average power?

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Answers (2)
  1. 6 September, 20:40
    0
    (a) Workdone = - 27601.9J

    (b) Average required power = 1314.4W

    Explanation:

    Mass of hoop, m = 40kg

    Radius of hoop, r=0.810m

    Initial angular velocity Winitial=438rev/min

    Wfinal=0

    t = 21.0s

    Rotation inertia of the hoop around its central axis I = mr²

    I = 40 * 0.810²

    I=26.24kg. m²

    The change in kinetic energy = K. E final - K. E initail

    Change in K. E = 1/2I (Wfinal² - Winitial²)

    Change in K. E = 1/2 * 26.24[0 - (438*2π/60) ²]

    Change in K. E = - 27601.9J

    (a) Change in Kinetic energy = Workdone

    W = 27601.9J (since work is done on hook)

    (b) average required power = W/t

    =27601.9/21 = 1314.4W
  2. 6 September, 21:09
    0
    Given that,

    Mass of wheel M = 40kg

    Radius of thin hoop R = 0.81m

    Angular speed ωi = 438 rev/mins

    1rev = 2πrad

    ωi = 438 * 2π/60 rad/sec

    ωi = 45.87 rad/s

    Time take to stop t = 21s

    Moment of inertial of a hoop around it central axis can be calculated using

    I = MR²

    I = 40 * 0.81²

    I = 26.24 kgm²

    Then,

    Change in kinetic energy is given as

    ∆K. E = Kf - Ki

    ∆K. E = ½Iωf² - ½Iωi²

    Then final angular velocity is zero, since the wheel comes to rest

    ∆K. E = ½Iωf² - ½Iωi²

    ∆K. E = ½I * 0² - ½ * 26.24 * 45.87²

    ∆K. E = 0 - 27,609.43

    ∆K. E = - 27,609.43 J

    B. Power

    Power = Workdone/Time take

    The change in kinetic energy of the loop is equal to the net work done on it

    ∆K•E = W = - 27,609.43

    Power = |W| / t

    P = 27,609.43/21

    P = 1314.73 W

    The average power to stop the loop is 1314.73 watts
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