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A constant force of 20N is applied tangentially to a string wound on the rim of a 40 cm diameter wheel, initially at rest.
(i) How much work does this force do as it turns the wheel through 45„a?
(3 %)
(ii) If the wheel has mass 8.0 kg and radius of gyration 15 cm, how long does it take the wheel to rotate through this angle?

  • physics -

    (i) 20N * 0.2 m * (angle in radians).
    I cannot interpret your angle symbol, which shows up here as ,,a.

    (ii) Angular acceleration rate = (Torque)/(Moment of inertia)
    = (Torque)/[(Mass)*Rg^2]
    The torque is 4 N*m
    Angular acceleration =
    4 N*m/(8*.15^2 kg m^2) = 22.2 rad/s^2

    Time required to rotate angle theta =
    2*(theta)/(angular acceleration rate)

  • physics -

    so for the first part 20N * 0.2m * 2pi is that correct?

    also how did you get your torque?

    (iii) What is the angular momentum of the wheel after this time?

  • physics -

    You never said the angle turned was 2 pi. You typed 45,,a

    Torque is force times lever arm (in this case, wheel radius)

    (iii) Angular momentum after time T is
    (Torque)*(Time) =
    (Moment of inertia)*(angular velocity)

  • physics -

    no i mean to convert 45degrees into rads you multiply by 2pi
    thanks for your help

  • physics -

    45 degrees is pi/4 radians. You multiply by pi/180 radians per degree, NOT 2 pi

  • physics -

    how did u find the angular velocity?
    im tryin to use v = wr but i don't know where to find the normal velocity of the rim

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