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Rotational Kinetic Energy of sphere

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A solid sphere of mass 4 kg rolls w/o slipping UP an incline with an angle of 30 degrees. the radius of the sphere is 0.5 m and its moment of Inertia is I = 2/5(m)R^2. At the bottom of the incline the center of mass of the sphere has a translational speed of 5 m/s.
    a) What is the total kinetic energy of the sphere at the bottom of the incline?
    b) How far does the sphere travel up the incline before coming to rest and starting to roll back down?
    c) Does the answer to b) depend on the mass?

    3. The attempt at a solution

    a) K = Krotational + Ktranslational
    = 1/2Iw^2 + 1/2mv^2
    w = v/R = 5/0.5
    = 1/2(2/5)(4)(o.5^2)(10^2) + 1/2(4)(5^2)
    = 70 J

    b) Kf + Uf = Ki + Ui
    0? (Not sure about this, because not sure if there is still some rotational kinetic energy?) + mgh = 70 J + 0J
    (5)(9.8)(h) = 70
    h = 1.43 m

    h = xsin30
    1.43 = xsin30
    x = 2.86 m

    c) So this is where I screw up... because I know from my theory classes that the answer SHOULDNT depend on the mass because all spheres roll down the same regardless of mass... but mine does so I think it's wrong. Thanks everyone!
  2. jcsd
  3. Feb 6, 2008 #2
    b.) When the sphere starts to roll back down, it means that it has lost all its kinetic energy to potential energy

    c.) no, because when you do b, you will notice that m cancels out from the equation. Do part b and you will udnerstand.
  4. Feb 6, 2008 #3

    Shooting Star

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    Homework Helper

    Since the sphere is rolling without slipping, if it moves, it has to spin. That's why both the translational movement and rotation stops at the same time.

    How do you know that your answer does depend on the mass? This is what happens if you plug in numerical values early in the problems. As mentioned by Oerg, do it taking m as the mass and see whether it cancels out.
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