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    Rotational Work and Kinetic Energy

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    A 0.050kg phonograph record with a radius of 0.15m drops onto a turntable and is soon rotating at 33 1/3 rpm. How much work must be supplied to get the record to rotate at this speed, and what supplies the work?

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf
    https://brainmass.com/physics/rotation/rotational-work-and-kinetic-energy-38303

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    A 0.050 kg phonograph record with a radius of 0.15m drops onto a turntable and is soon rotating at 33 1/3 rpm. How much work must be supplied to get the record to rotate at this speed, and what supplies the work?

    KE of rotation= 1/2 (I w ^2)
    where I= rotational inertia
    w = angular speed

    I of a disc= 1/2 (M r^2)
    r= 0.15 m
    M= 0.05 Kg
    Therefore I = 1/2 M r^2= 0.0005625 kg ? m^2 =1/2 x 0.05x 0.15^2

    w =2 pi f
    f= 33 1/3 rpm
    = 5/9 revolutions per second =100/3 /60

    Therefore w = 3.4907 rad/s =2 pi x 5/9

    And KE of rotation = 0.003427 Joules =1/2 x 0.0005625x 3.4907^2

    Work supplied = Final Kinetic energy - Initial Kinetic energy= Final KE since Initial Kinetic energy=0
    as the record starts from rest

    Therefore work done= 0.003427 Joules
    This work is supplied by the turntable

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf>
    https://brainmass.com/physics/rotation/rotational-work-and-kinetic-energy-38303

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