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Bowling balls rolling up a ramp (conservation of energy)

  1. Mar 25, 2013 #1
    This is from an old course I took. I'm not sure what I'm doing incorrectly.

    1. The problem statement, all variables and given/known data
    Two identical bowling balls are rolling on a horizontal floor without slipping. The initial speed of both balls is V = 9.9 m/s. Ball a encounters a frictionless ramp, reaching a maximum height H_a. Ball B rolls up a regular ramp (i.e. rolls without slipping), reaching a maximum vertical height H_b. Which ball goes higher, and by how much?


    2. Relevant equations
    [itex]GPE = mgh[/itex]
    [itex]KE_{trans} = \frac{mv^2}{2}[/itex]
    [itex]KE_{rot} = \frac{I\omega ^2}{2}[/itex]
    [itex]I_{solid-sphere} = \frac{2 M R^2}{5}[/itex]
    [itex] \omega = \frac{v}{R}[/itex]

    3. The attempt at a solution
    I just did conservation of energy. Ball a will still have rotational energy when it reaches the top, ball b will not. Both balls start with the same total kinetic energy. Ball a's final GPE will come entirely from its initial translation energy. Ball b's will come from both its rotational and translational kinetic energies. Solving:
    [itex] \frac{mv^2}{2} = mgH_a[/itex]
    [itex]\frac{mv^2}{2} (1 + \frac{1}{5}) = mgH_b[/itex]
    [itex]H_a = \frac{v^2}{2g}[/itex]
    [itex]H_b = \frac{3v^2}{5g}[/itex]

    However, the answer says that I should get that H_b is "2m" higher than H_a. Did I do something wrong here?
     
    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2
    I think you should be thinking about it like this: Ball A and B have some total energy, both equal to ##{\frac{1}{2}}mv^2##. The must have the same amount of energy at any point during the motion. Ball A's total energy will be converted entirely into potential energy. Which you have done. Ball B's total energy will be converted into rotational kinectic energy and gravitational potential energy so ##{\frac{1}{2}}mv^2=mgH_b+{\frac{1}{5}}mv^2##

    I'm not very familiar with the rotational kinectic energy, so I'm assuming you are right about it. I'm commenting on the setup of your equations.
     
  4. Mar 25, 2013 #3

    TSny

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    Check to make sure the fraction 1/5 is correct here.
     
  5. Mar 25, 2013 #4
    Except that both balls don't always have total energy equal to [itex]\frac{mv^2}{2}[/itex]. They're said to be rolling withoutt slipping initially, which means that they have both translattional and rotational kinetic energy, i.e. [itex]KE_{tot} = \frac{mv^2}{2} + \frac{I\omega^2}{2}[/itex].

    Ball a is still rolling with when it reaches its highest point, hence its full conservation of energy equation should look like this:
    [itex]KE_{trans} + KE_{rot} = GPE + KE_{rot}[/itex]
    [itex]\frac{mv^2}{2} + \frac{I\omega^2}{2} = mgH_a + \frac{I\omega^2}{2} [/itex]

    That's where my equations come from. I was fairly certain that part was okay but I could be wrong...
     
  6. Mar 25, 2013 #5

    rcgldr

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    If you follow your equations from above, you have:

    [itex]KE_{b} = \frac{mv^2}{2} + \frac{mv^2}{5}[/itex]

    but then you factor out [itex]\frac{mv^2}{2}[/itex] from [itex]\frac{mv^2}{5}[/itex] and get 1/5?
     
  7. Mar 25, 2013 #6
    @Tsny, I will try to be more explicit, let's see...

    For B:
    [itex]KE_{trans} + KE_{rot} = GPE[/itex]
    [itex]\frac{mv^2}{2} + \frac{1}{2} \frac{2mR^2\omega^2}{5} = mgH_b[/itex]
    [itex]v^2(\frac{1}{2} + \frac{1}{5}) = gH_b[/itex]
    [itex]H_b = \frac{7v^2}{10g}[/itex]
     
  8. Mar 25, 2013 #7
    Where does 1/5 come from?

    Why 3/5?

    What is 'm'? The unit meaning meters?
     
  9. Mar 25, 2013 #8

    TSny

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    Looks good.
     
  10. Mar 25, 2013 #9
    Thanks guys, I stink at algebra apparently haha. So is this stuff about the "2m" nonsense problaby just some sort of weird mistake?
     
  11. Mar 25, 2013 #10

    rcgldr

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    That fixes it, 7/10 instead of 3/5.
     
  12. Mar 25, 2013 #11
    If 'm' is 'meters', then it makes perfect sense. What number do you get?
     
  13. Mar 25, 2013 #12

    TSny

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    I think the "2m" means 2 meters and not 2 times the mass!
     
  14. Mar 25, 2013 #13
    No idea. It just says "2m higher". I didn't actually plug in the numbers though...

    H_a = 5 meters
    H_b = 7 meters.

    WELP. Mystery solved. It's listed in italics... so I didn't think it was a unit. Thanks again haha.
     
  15. Mar 25, 2013 #14
    Thanks for the tag team help everyone :). I feel slightly silly now.
     
  16. Mar 25, 2013 #15

    SammyS

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    Yup.

    (9.9 m/s )2 = 98.01 m2/s2,

    so v2/g is very close to 10 meters.
     
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