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Disks and Hoops

  1. Oct 21, 2004 #1
    I'm having some trouble with a conceptual problem. Your help or input would be greatly appreciated. Here's the problem.

    A disk and a hoop, both having the same mass, are rolled down a hill. The disk rolls faster than the hoop. Why is this so?

    Thank you for your time!
     
  2. jcsd
  3. Oct 21, 2004 #2

    Doc Al

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    Which one has the greater rotational inertia? (And thus is harder to get rolling.)
     
  4. Oct 21, 2004 #3
    I guess the hoop would have a greater moment of inertia, but why is it greater? How is rotational inertia calculated?
     
  5. Oct 21, 2004 #4

    Pyrrhus

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    Moment of Inertia

    [tex] I = \sum_{i=1}^{n} m_{i} r_{i}^2 [/tex]

    For a a system composed of particles with defined mass the above can be calculated easily, but for a continous mass system we will take a small mass [itex] \Delta m [/tex] and applying the moment of inertia definition we should get.

    [tex] I = \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \Delta m_{i} r_{i}^2 [/tex]

    Which is the integral of an infinitessimal mass

    [tex] I = \int r^2 dm [/tex]

    Now the moment of inertia around the z axis for a Uniform hoop is

    [tex] I_{z} = \int r^2 dm [/tex]

    every particle with mass dm is at a constant distance R, so

    [tex] I_{z} = R^2 \int dm [/tex]

    [tex] I_{z} = MR^2 [/tex]

    For more detail we could calculate the moment of inertia for ring shaped cylinder will be:

    [tex] I_{z} = \int r^2 dm [/tex]

    Using

    [tex] \rho = \frac{dm}{dV} [/tex]

    [tex] I_{z} = \int r^2 \rho dV [/tex]

    Applying cylindric shells we get, where L is the height of our cylinder

    [tex] dV = 2 \pi r dr L [/tex]

    so

    [tex] I_{z} = 2 \pi L \rho \int^{r}_{r_{o}} r^3 dr [/tex]

    we get

    [tex] I_{z} = \frac{1}{2} M(r^2 + r_{o}^2) [/tex]

    A hoop can be considered a ring shaped cylinder of very thin walls so we could say

    [tex] I_{z} = MR^2 [/tex]

    Now an uniform disk could be considered a solid cylinder so

    [tex] I_{z} = \int r^2 dm [/tex]

    r being the radius of our cylinder

    Using
    [tex] \rho = \frac{dm}{dV} [/tex]


    [tex] I_{z} = \int r^2 \rho dV [/tex]

    Applying [itex] dV = dA L [/itex]

    [tex] I_{z} = \int r^2 \rho 2 r \pi L dr [/tex]

    [tex] I_{z} = \int r^3 \rho 2 \pi L dr [/tex]

    [tex] I_{z} = \rho 2 \pi L \int^{r}_{0} r^3 dr [/tex]

    [tex] I_{z} = \frac{1}{2} MR^2 [/tex]

    Note: All the z- axis go throught the center of mass

    Also:

    Moment of Inertia for the disk
    [tex] I_{z} = \frac{1}{2} MR^2 [/tex]

    Moment of Inertia for the hoop
    [tex] I_{z} = MR^2 [/tex]

    From this two moment of inertia you can see which will get down before the other.

    and i found this movie for this:
    http://solomon.physics.sc.edu/~tedeschi/demo/demo12.html
     
    Last edited: Oct 21, 2004
  6. Oct 21, 2004 #5

    Tide

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    Did the problem specify that the disk and hoop have the same radius?
     
  7. Oct 22, 2004 #6

    Doc Al

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    Cyclovenom gave the details, but the gist of it is that rotational inertia depends on how the mass is distributed: The more mass farther from the rotational axis, the greater the rotational inertia. It's harder to rotate something if its mass is far from the axis of rotation.
     
  8. Oct 22, 2004 #7

    Doc Al

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    It doesn't matter.
     
  9. Oct 22, 2004 #8

    Tide

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    I thought it would be something Geoff should think about! :smile:
     
  10. Oct 22, 2004 #9

    Doc Al

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    An excellent point! :biggrin:
     
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