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Translational Kinetic Energies + Plane

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the ratio of the translational kinetic energies of a ring, a coin, and a solid sphere at the bottom of an inclined plane. The bodies have been released from rest at the top. Assume pure rolling without any slipping.


    3. The attempt at a solution

    Well, I'm really not sure. Since T.K.=1/2 m*v^2, we need to look at linear velocities v at the bottom of the plane for each object. Obviously the rotation will influence this, and I've tried obtaining final ω and using v=ωr at the bottom, but I'm not getting the answer. Which, by the way, is 21:28:30.
     
  2. jcsd
  3. Oct 8, 2011 #2
    Perhaps you are confusing the linear velocity from v=ωr, which is the instantaneous velocity of the edge of the rolling object, with the linear velocity of the center of mass of the object.

    Beside that, are you using the correct moments of inertia for the different objects? I'm assuming the masses and radii are all the same.
     
  4. Oct 8, 2011 #3
    The MI's are correct, and yes, they are all of same mass and radius. One question: can we find the acceleration of c.o.m and then use the equations of motion to find final velocity? I haven't tried that.
     
  5. Oct 8, 2011 #4
    Nevermind this statement...while yes they are different things, in this case they are the same velocity (the center of mass velocity in the frame at rest w.r.t. the ramp is of course the same as the linear velocity of the edge in the co-moving frame, since there is no slipping).

    You don't have to calculate any velocites. Here's how you go about it. For example, for the ring you have KE = PE-RE = PE-(1/2)Iω^2 = PE-(1/2)mv^2 = PE-KE, and so you have KE=PE/2. Obtain the KE in terms of PE for the other two objects, and then you can take the ratios between them.

    Edit: Keep in mind that the PE for each object is the same since they all start at the same height and have the same mass.
     
    Last edited: Oct 8, 2011
  6. Oct 8, 2011 #5
    Gracias.
     
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