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Help needed on a Mechanics problem

  1. Dec 8, 2005 #1
    A circular platform of radius b has a radial groove and is rotating with a speed w about central axis. A sphere of radius R and mass M rolls in the groove without slip. There is friction at the bottom. I need to calculate the time required for the ball to reach the edge. Any ideas on how to do this problem?
    Thanks.
     
  2. jcsd
  3. Dec 8, 2005 #2

    siddharth

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    You need to show your work.
    What have you done till now?
     
  4. Dec 8, 2005 #3

    Tide

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    This is "Introductory Physics?" What am I missing?
     
  5. Dec 9, 2005 #4

    siddharth

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    I thought so, but I might have misunderstood the question. A radial groove is simply a groove which runs straight(ie along the radius) from the center to the edge of the disc, right? What's the difficulty then?
     
  6. Dec 9, 2005 #5

    Tide

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    Sidd,

    My first thoughts on the problem were that you have to work out the dynamics of an object rolling without slipping to which a "force" is applied. This involves relating a torque to the position dependent angular acceleration of the sphere about a horizontal axis, i.e. a virtual rotation about an axis passing through the point of contact between the sphere and the surface of the groove). It seems somewhat advanced for "introductory physics" unless I am missing something. :)
     
  7. Dec 9, 2005 #6

    siddharth

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    Tide,
    The torque acting on the sphere will be due to friction. Will it not be easier to consider the angular acceleration about the center of the sphere?

    Since the sphere is rolling without slipping, it is easy to obtain the relation between the radial acceleration of the COM of the sphere and the angular acceleration of the sphere about it's center.

    Using that, one can represent the frictional force in terms of the acceleration and then using the second law, calculate the displacement from the center as a function of time.

    I don't know if it belongs in introductory physics, but unless I am making a mistake, it does not seem too advanced to me.
     
    Last edited: Dec 9, 2005
  8. Dec 9, 2005 #7

    Tide

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    Sidd,

    Nevermind! I just went ahead and solved it and it turns out to be not much of a challenge.
     
  9. Dec 9, 2005 #8
    My analysis is as follows,

    For a reference frame attached to the center of the sphere, a cenrtifulgal force mrw^2 acts radially outwards. Newton's law implies,

    mrw^2-mg*mu=ma

    Take torque about the center of the sphere,

    mumg*R=I*alpha

    alpha=a/R

    eliminating mumg, I get

    mrw^2=a(m+I/R^2)

    a=mrw^2/(m+I/R^2)=d^2r/dt^2

    I need to get r as a function of t from this equation. Am I correct so far?

    Thanks
     
  10. Dec 10, 2005 #9

    siddharth

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    Looks right to me, except that I wouldn't call the frictional force as [itex] \mu m g [/itex] but as [itex] f [/itex].Since static friction acts on the sphere, only the maximum static friction will be mumg(ie [itex] f_{max}= \mu mg [/itex]).
    Tide, is this what you got as well?
     
    Last edited: Dec 10, 2005
  11. Dec 10, 2005 #10

    Tide

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    I get the same thing:

    [tex]\frac {d^2 r}{dt^2} = \frac {\omega^2r }{1 + \frac {I}{mR^2}}[/tex]

    although I derived it without invoking the centrifugal force (the equivalent term comes out of handling the coordinates properly).

    Now, dilberg has to decide whether the sphere is hollow or solid and then decide what he will use for initial conditions. :)
     
  12. Dec 10, 2005 #11
    The radius of the platform is b. The sphere is released at r=ro at t=0. The problem says it rolls without slipping, but say if it were to slip after some time t then alpha=a/R is no longer valid. Any thoughts on how to approach this condition?

    I have a question about inertia tensor. How to superimpose two inertia tensors in the same co-ordiante system? I have to calculate the inertia tensor of a composite body made by welding a thin rod of mass m and length l to the center of a square plate of mass m and length s. The rod makes a angle beta with the plane of the plate.

    My Analysis
    The moment of inertia about the principal axis of the plate is 1/12m(a^2+b^2)
    in the plane of the plate. i.e Ixx=Iyy=1/12m(a^2+b^2). Izz=Ixx+Iyy for the plate. The procut inertia is zero. ie Ixy etc is zero about the principal axis. Similarly I can derive for the rod about its centre, Now how do I superimpose these two tensors? please see fig attached.
     

    Attached Files:

  13. Dec 10, 2005 #12

    Tide

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    You will use the parallel axis theorem to find the moment of inertia of a composite body.

    But I am waiting to see what you find for the original problem if the sphere starts at r = 0. :)
     
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