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Rolling Motion of a Rigid Object

  1. Apr 5, 2005 #1
    I have no clue where to start on this problem. I appreciate any help. Thanks.

    A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from rest and roll without slipping, which object reaches the bottom first? Verify you answer by calculating their speeds when they reach the bottom in terms of h.
     
  2. jcsd
  3. Apr 5, 2005 #2
    We assume no friction ? No ? Ok then use conservation of energy :

    [tex]MgLsin( \theta) = \frac{1}{2}I \omega^2 + \frac{1}{2}Mv^2[/tex]

    v is the speed, omega the angular speed and L is the length of the incline which makes an angle theta with the horizontal.

    The clue really is to see the motion of a rigid object as the sum of a translation and a rotation : hence the two terms on the RHS, which express the kinetic energy

    Keep in mind that [tex]v = \omega R[/tex] where R is the radius of the object

    Just fill in the I (rotational inertia about the center of mass of the rotating object) for both objects and solve for v so you can compare

    oh i just realized, the Lsin(theta) is equal to the height h...so you can replace them...

    marlon
     
    Last edited: Apr 5, 2005
  4. Apr 5, 2005 #3
    Ok, I can see that now. h=Lsin(theta)

    How do I fill in the I for both objects when I'm given no values at all?
     
  5. Apr 5, 2005 #4
    They should have been given. You can calculate them yourselves by using 3-D-integration, but let's keep it simple

    For the hoop I = MR²
    For the disk I = (1/2)MR²
    For a sphere I =(2/5)MR²

    So assume all objects have same radius R and same mass M, which one will come down first and which one last ???

    Be sure you write v as a function of h, I and R and then substitute the I-values for each object...


    marlon
     
    Last edited: Apr 5, 2005
  6. Apr 5, 2005 #5

    dextercioby

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    *For a 2-sphere of mass M & radius R,the moment of inertia wrt a rotation axis (chosen as Oz) passing through the center of the 2-sphere is

    [tex] I_{Oz}^{2-sphere} = \frac{2MR^{2}}{3} [/tex]

    *For a ball in [itex]\mathbb{R}^{3} [/itex] of mass M and radius R,the moment of inertia wrt a rotation axis (chosen as Oz) passing through the center of the ball is

    [tex] I_{Oz}^{ball \subset \mathbb{R}^{3}} =\frac{2MR^{2}}{5} [/tex]


    Daniel.
     
  7. Apr 5, 2005 #6
    :rofl: :rofl: :rofl: :rofl:

    What the hell is a two dimensonal sphere ???
    You are talking about a spherical shell...that is not the same...aahhhhh

    TO the OP : let's also study the spherical shell... :rofl:

    marlon
     
  8. Apr 5, 2005 #7
    A 2-sphere...yes yes yes :) :) :)

    marlon
     
  9. Apr 5, 2005 #8

    dextercioby

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    2-Sphere is the surface of (implicit) equation (in cartesian coordinates)

    [tex] x^{2}+y^{2}+z^{2}=R^{2} [/tex]

    and it's not funny not to know what a 2-sphere is...:rolleyes:

    Daniel.
     
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