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No-Slip Rolling & Conservation of Energy

  1. Apr 8, 2014 #1
    I wanted to just make sure I was doing this right. The problem:

    A marble of mass M and radius R rolls without slipping down a track from height h1. The marble then goes up a frictionless track to a height h2. Find h2.

    I figure this has to be in terms of other variables, here is my work.

    initial mechanical energy = final mechanical energy

    Mgh1 = Mgh2 + (1/2)Mv^2 + (1/2)Iω^2

    I for sphere = (2/5)MR^2

    Plug in for I and cancel M's. Plug in v = ωR

    gh1 = gh2 + (1/2)v^2 + (1/5)v^2

    g(h1 - h2) = (7/10)v^2

    h2 = h1 - (7/10)(v^2 / g)

    Can the answer be expressed simpler?
  2. jcsd
  3. Apr 8, 2014 #2


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    yes. What is the value of v?
  4. Apr 9, 2014 #3
    angular velocity times radius. That's pretty much all I can think of in terms of substituting in something.
  5. Apr 9, 2014 #4
    Hello Kavorka

    This is incorrect.

    I would suggest you to break the problem in two stages .In the first part the marble rolls from the top at height h1 to the bottom .In the second part it moves up to a height h2.

    Just write the energy conservation equation for the first part .What do you get ?
  6. Apr 9, 2014 #5
    The same thing, minus the Mgh2.
  7. Apr 10, 2014 #6


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    Read carefully the problem: the second track is frictionless. Is it possible rolling without slipping on a frictionless track? Think of a car moving on wet ice...

  8. Apr 10, 2014 #7
    How am I suppose to find the 2nd height then?
  9. Apr 10, 2014 #8
    This is a two step problem. 1st find the speed at the bottom of the track h2 = 0. Than use energy conservation again on the way up the second track where due to the lack of friction there is no torque and the angular kinetic energy doesn't change.
  10. Apr 11, 2014 #9


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    When the marble reaches height h2, its translational motion comes to a temporary stop. So at that point, the velocity of its center of mass is__?___ and ω is (___see hints given by others____).
  11. Apr 11, 2014 #10
    Why does it stop? The problem doesn't say that, it just lists an arbitrary height up a frictionless slope.
  12. Apr 11, 2014 #11


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    I asumed it implied it stopped otherwise the solution must be left in terms of the unknown variable 'v' but also in which case your solution is not correct because you have not identified ω properly as noted by others' hints (no friction on incline).
  13. Apr 11, 2014 #12
    Yes it stops. The problem doesn't state it explicitly but it is in between the lines. If the mass hadn't stopped, it would still be moving upwards reaching even higher elevations. It is implicit the h2 is the highest point of the trajectory up the second ramp, so it must stop there.
  14. Apr 13, 2014 #13
    Do you end up with height 1 = height 2 then? If not I have no idea what I'm doing.
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