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)

Question

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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RuslanHeatt 4 hours 2021-07-22T04:58:21+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-22T04:59:42+00:00

    Answer:

    (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

    0
    2021-07-22T04:59:54+00:00

    Answer:

    Explanation:

    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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