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Conservation of Energy with Rotation

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A small circular object with mass m and radius r has a moment of inertia given by I = cmr^2. The
    object rolls without slipping along the track shown in the figure. The track ends with a ramp of height R = 2.5 m that launches the object vertically. The object starts from a height H = 6.0 m. To what maximum height will it rise after leaving the ramp if c = 0.40?


    3. The attempt at a solution

    My solution to this problem is 4.7 m.
    So I'm pretty sure my answer is right but the textbook indicates the answer is 5.0 m which confuses me.

    I applied conservation of energy from the starting point, mgH to the launch point (when it leave the ramp), mgR + 1/mv^2 + 1/2I[itex]\omega[/itex]^2 , and then solved for v obtaining 6.55m/s.
    Then, I used the equation v^2 = 2g(h-R), where h is the maximum height reached (note I'm
    pretty sure once the object leaves the ramp, it loses its rotational motion and maintains only
    linear motion after-though correct me if I'm wrong). Rearranging and substituting I obtain
    h=4.7m.
     
  2. jcsd
  3. Dec 4, 2013 #2
    The textbook is correct here. The magnitude of velocity at the end of ramp is not 6.55 m/s.

    That said, I suggest that you do not plug in the numbers until you solve everything symbolically. You will see that many things cancel each other out.
     
  4. Dec 4, 2013 #3
    Rather silly mistake on my part. I rearranged my equations and everything, but for some reason wrote c= 0.60 at the top of my page and kept referencing that value ....
     
  5. Dec 4, 2013 #4
    Which, I assume, means you got the correct result?
     
  6. Dec 4, 2013 #5
    oh yeah I did. Thanks.
     
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