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Rigid body rotation

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data

    imagine a thin length of wood supported by two supports, one at either end. One support dissapears and the other turns into a pivot (as gravity acts on the wood). Show that the load supported by the pivot is Mg/4.


    3. The attempt at a solution

    i dont know where to start due to my confusion on forces with rotation. i know the plank is acted on by gravity at its CM, and its moment of inertia about its CM = I=1/12*M*l^2.
    I know there is a reacton force at the Pivot, and this is the force i need to calculate, but i dont know what to consider first, a nudge in the right direction would be helpful!
     
  2. jcsd
  3. Nov 25, 2008 #2

    Doc Al

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    Start by figuring out the acceleration of the center of mass. Hint: Apply Newton's 2nd law to the rotation.
     
  4. Nov 25, 2008 #3
    F=ma -> TAU = I*ALPHA

    TAU = (1/12*M*l*l)*ALPHA

    so ALPHA = TAU/(1/12*M*l*l)

    so a = (TAU/(1/12*M*l*l)) * (0.5*l) ? where a=ALPHA*(0.5l)
     
  5. Nov 25, 2008 #4

    Doc Al

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    (1) It pivots about one end, not the center of mass, so correct the rotational inertia.
    (2) What's the torque about the pivot?
     
  6. Nov 25, 2008 #5
    TAU = (1/3*M*L*L)*ALPHA

    the torque about the pivot is mgl/2 ?
     
  7. Nov 25, 2008 #6

    Doc Al

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    Good. Keep going.
     
  8. Nov 25, 2008 #7
    so the acceleration due to gravity is a = 3*TAU/2*ML at the CM
    and the torque about the pivot is mgl/2

    so m*a = TAU*(1/2*l) ?
     
  9. Nov 25, 2008 #8

    Doc Al

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    Good. So what's the acceleration of the center of mass? (Don't call it "acceleration due to gravity".)

    Once you have the acceleration of the center of mass, apply Newton's 2nd law for translation. What forces act on the wood?
     
  10. Nov 25, 2008 #9
    is the acceleration ofthe CM just 3TAU/2mL again
    the reacton force at the pivot and mg at the CM are the forces on the wood

    R = 0 because mgl/2 (where l = 0) = 0

    therfore ma=mg ?
     
  11. Nov 25, 2008 #10
    acceleration at CM = 3/4*g
     
  12. Nov 25, 2008 #11

    Doc Al

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    But you know the torque, so eliminate Tau from this expression.
    Good.
    No. Apply Newton's 2nd law. What's net force? The acceleration?
     
  13. Nov 25, 2008 #12

    Doc Al

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    Good!
     
  14. Nov 25, 2008 #13
    the force at the CM=3/4*mg

    Net force at pivot = force at CM ?
     
  15. Nov 25, 2008 #14

    Doc Al

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    That's the "ma" part of "Fnet = ma". But what's Fnet?
    No. Two forces act on the wood (you named them earlier). Combine them to get the net force.
     
  16. Nov 25, 2008 #15
    Fnet = mg + mgl/2
    ?
     
  17. Nov 25, 2008 #16
    FNet = N - mg = 3/4mg

    therfore N = 1/4mg ?
     
  18. Nov 25, 2008 #17

    Doc Al

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    Careful. As written, this implies that N = 7/4mg. But the acceleration is downward, thus negative: N - mg = -3/4mg.

    Yep.
     
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