Physics help?

A student on a piano stool rotates freely with an angular speed of 2.95 rev/s . The student holds a 1.35 kg mass in each outstretched arm, 0.769 m from the axis of rotation. The combined moment of inertia of the student and the stool, ignoring the two masses, is 5.03 kg⋅m2 , a value that remains constant.

As the student pulls his arms inward, his angular speed increases to 3.64 rev/s . How far are the masses from the axis of rotation at this time, considering the masses to be points?

Calculate the initial kinetic energy of the system.

Calculate the final kinetic energy of the system.

1 Answer

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  • oubaas
    Lv 7
    3 weeks ago

    ω1 = 2PI*n1 = 6.2832*2.95 = 18.54 rad/sec

    Js = 5.03 kg*m^2

    Jm = 2*1.35*0.769^2 = 1.596 kg*m^2

    L = ω1*(Js+Jm) = 6.628*18.54 = 122.8 kg*m^2/sec

    ω2 = ω1*n2/n1 = 18.54*3.64/2.95 = 22.87 rad/sec

    L is conserved :

    L = ω2*(Js+Jm')

    (Js+Jm') = 122.8/22.87 = 5.37 kg*m^2

    Jm' = 5.37 - 5.03 = 0.34 kg*m^2 = 1.35*2*d^2

    d = √0.34/1.35 = 0.501 m

    Ei = ω1^2*(Js+Jm)/2 = 1138 joule

    Ef = ω2^2*(Js+Jm')/2 = 1405 joule

    • Pinkgreen
      Lv 7
      3 weeks agoReport

      According to d=0.501, Ef=(22.87^2)[5.03+2.7*0.501^2]/2=1492.7 joules.

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