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Kinetic and Rotational energy of a sphere

  1. Nov 5, 2011 #1
    Hi guys so this is my first time posting a question on this amazing forum, i hope i am doing this right since i read that no one will help me if i am missing something. Anyways, i am dealing with a physics problem that it driving me crazy (mainly because my teacher does not explain anything!) Here is the problem.

    1. The problem statement, all variables and given/known data

    An 7.50-cm-diameter, 320 g sphere is released from rest at the top of a 1.90-m-long, 20.0 incline. It rolls, without slipping, to the bottom.
    b. What fraction of its kinetic energy is rotational?
    So the known data is:
    mass= 0.0320 kg
    radius= 0.0375 m
    delta x= 1.90m
    wi = 0
    vf = 3.02 m/s
    wf = 80.5 rad/s
    Find ratio between rotational and kinectic energies.

    2. Relevant equations
    Kroll = 1/2 (I* w^2) + 1/2(M)(vf)^2 = Krot + Kcm

    3. The attempt at a solution
    ok so from part A of the problem i found the velocity at the bottom of the incline to be 3.02 m/s, then i found the final angular velocity. With that all i thought i had to do was plug the data into the previous equation and get an answer:
    Since moment of inertia of a cylinder is 2/5MR^2

    Kroll = 1/2 (2/5 ) (0.032kg)(0.0375m)^2 ( 80.5 rad/s)^2 + 1/2 ( 0.032 kg)(3.02 m/s)^2

    I am getting 0.204 for the Kroll and for the Krot i am getting 0.06... when i devide 0.06/0.204 i get 0.3 which is not the answer :( i have no idea what am i doing wrong! please help me, this homework was due 2 days ago, i am just doing it to understand the concept no for the points, so i would really appreciate your help, wonderful people, thanks!
  2. jcsd
  3. Nov 5, 2011 #2


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    The mass is 0.32kg, not 0.032kg, but that shouldn't affect the ratio.

    Maybe you have too much round-off error.

    .3 is about 5% different than the correct answer.
  4. Nov 5, 2011 #3
    ok but am i doing this correctly? i mean, am i supposed to divide Krot by Krolling to get a ratio? or is it the other way (Kroll/krot) ?

    I don't get it, no matter how i do it, i am still getting around 0.29... and when round i get 0.3 but it's wrong :/
  5. Nov 5, 2011 #4


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    The fraction of the kinetic energy that is rotational is what you are calling Krot / Kroll .

    Why are you rounding to 1 significant figure?
  6. Nov 6, 2011 #5
    i am rounding to two sig figures :( 0.30, cause i get 0.298
  7. Nov 6, 2011 #6
    please i really need to understand this, i am not getting the answer.. :/
  8. Nov 6, 2011 #7
    The question is What fraction of its kinetic energy is rotational?, your answer is the relationship between the rotational and translational kinetic energies.
  9. Nov 6, 2011 #8
    ok, i dont know what that means then, i thought you had to get the ratio between them,,, if so then i am getting 0.30, which is supposed to be wrong...
  10. Nov 6, 2011 #9


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    It's clear that what hyddro calls Kroll is the sum of the rotational KE, and the KE of the center of mass, the latter being the translational KE.

    So, Krot/Kroll is the fraction he needs.

    BTW: For a solid sphere rolling without slipping, it can be shown that this fraction is exactly 2/7 .
  11. Nov 6, 2011 #10
    hey bro, thanks for helping me, the answer was indeed 2/7 .. but can you explain me how you got that answer?
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