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Sphere rolling down incline then up incline?

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A sphere of mass M and radius R is not necessarily solid or hollow. It has moment of inertia I = cMR^2 . The sphere starts from rest and rolls without slipping down a ramp from height H. It then moves back up the other side with height h, but now with no friction at all between the sphere and the ramp.
    What height h does the sphere reach?
    (Answer should only include H and c.)

    2. Relevant equations
    I = cMR^2
    KE = 1/2Iω^2
    PEg = mgH

    3. The attempt at a solution
    So I tried setting up a conservation of energy for both sides of the ramp. For the first half I got
    mgH = 1/2mv^2 + 1/2Iω^2. I then solved for v^2.
    For the second half of the ramp (one without friction), I got
    1/2mv^2 + 1/2Iω^2 = mgh + 1/2Iω^2.
    I solved for h then plugged in v^2 from the first equation and got g^2H/c but my answer can't have g so I know this is wrong. Any pointers? I've never done a problem like this so I'm pretty sure I'm way off.
     
  2. jcsd
  3. Nov 4, 2016 #2
    In this first part, what did you get when you solved for v2?
     
  4. Nov 4, 2016 #3
    ]

    I got (gH)/c but I'm not entirely sure if it's correct.
     
  5. Nov 4, 2016 #4
    Yeah, I got something different for v2.

    In your equation for the second half of the ramp, you have 1/2Iω2 on both sides of the equation. I assume those are the same and they cancel out, true? At least that's how I see the problem. When the sphere starts up the other side of the ramp, none of the rotational kinetic energy converts back to potential energy. So it looks like your method is good. I think you are just having algebra issues. I was able to come up with a solution in terms of only H and c.

    Edit: I feel obligated to put in my standard disclaimer. I am prone to making mistakes. And even though I found a solution in terms of only H and c, it could still be incorrect. :)
     
  6. Nov 4, 2016 #5
    Haha no worries! So I tried it again and got 2gH/c+1. Is this correct?
     
  7. Nov 4, 2016 #6
    Plugged it in and got the right answer! Thanks Tom!
     
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