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Angular momentum with cylinders rolling

  1. Sep 6, 2008 #1

    lzh

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    1. The problem statement, all variables and given/known data
    A think cylindrical shell and a solid cylinder have the same mass and radius. The two are released side by side and roll down, w/o slipping, from the top of an inclined plane that is 4.9m above the ground.

    When the first object reaches the bottom, what is the height above the ground of the other object?


    2. Relevant equations
    (delta)x=.5(Vi+Vf)t


    3. The attempt at a solution
    The problem also asked for the final linear velocity of both cylinders. Which I found to be:
    shell: 6.92965m/s
    solid:8m/s

    I tried to use the delta x equation above, but time is not given so I had to use ratio to cancel out the time:
    z: how far down the shell is
    solid:
    .5(8)t=(delta)x
    =4t
    shell:
    .5(6.93)t=(delta)x
    =3.465t
    3.465t/4t = z/4.9
    z=4.2445

    answer: 4.9-4.2225=.655

    however, this is wrong. Did I miss something?
     
  2. jcsd
  3. Sep 6, 2008 #2

    Doc Al

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    Staff: Mentor

    OK.
    You found the time for the shell to reach the bottom (its final speed). What you need is its time and speed at the moment the solid hits the bottom.
     
  4. Sep 6, 2008 #3

    lzh

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    the solid's final velocity is 8m/s like I said, I know that my final velocity numbers are correct.
     
  5. Sep 6, 2008 #4

    Doc Al

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    Staff: Mentor

    Yes, your final velocity numbers are correct.

    Try this: Compare the acceleration of each cylinder.
     
  6. Sep 6, 2008 #5

    lzh

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    well, the problem is that I don't know what time is. I can do:
    8/t
    6.93/t
    for acceleration.
    If I knew how far it is to roll down to the bottom, I could figure out the time. But other than that I 'm not sure what else to do.
     
  7. Sep 7, 2008 #6

    Doc Al

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    Staff: Mentor

    That won't help since the times are different (of course). Instead, relate the final speed to the acceleration using distance, which you know is the same (call it x):
    [tex]v^2 = 2 a x[/tex]
    Just call the distance x, as you've been doing. How does distance relate to time for accelerated motion?
     
  8. Sep 7, 2008 #7

    lzh

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    I've tried using the equation you've given before. But the problem is that I don't have a number for the acceleration either. So I have two unknowns in one equation- if I could find another equation to use as a system I could find the answer. But eqns like:
    x=.5at^2
    Vf=at
    just can't work in the system of equation. All the equations that relates distance and time that I've tried does not work. However, I know for sure that:
    acceleration of solid=(4/3)acceleration of shell

    Thanks for all your help!
     
  9. Sep 7, 2008 #8

    Doc Al

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    Staff: Mentor

    Excellent. That's all you really need.

    Now make use of x = .5at^2. For the solid, call the acceleration a, the total distance x, and the time to reach the bottom t.

    Then find the distance that the shell reaches in that same time t in terms of x. (Same equation, different acceleration.)
     
  10. Sep 7, 2008 #9

    lzh

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    thanks I've figured it out! I just solve using proportions!
     
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