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Force on rigid body: translation or rotational acceleration?

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A perpendicular force F acts on the tip of a thin rod of length L and mass M in the ij-plane which is not fixed. What is the translational and rotational acceleration of the rod about the center of the rod?


    2. Relevant equations
    At first it seems like an easy problem.

    a-linear = F/m (NSL, system : the rod)

    t == r X F == rFsin 90 k == rF k
    so a-rot == t/I == rf/I k

    3. The attempt at a solution

    The equations are easy to solve but that's not what I'm having trouble with. I don't know if NSL for linear acceleration is valid if the force doesn't act on the Center of Mass.

    If it is valid, that means that if F acted on the center, there would be only linear acceleration, but if F acted on the tip, there would be the same linear acceleration in addition to rotational acceleration.
    Thus the force, if applied for a short distance, does more work when it is on the tip then when it is on the center. This is unintuitive.
     
  2. jcsd
  3. Nov 27, 2012 #2

    Dick

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    Newtons second law doesn't apply easily to extended objects. If you are exerting a force on the end of a rod, the rest of the rod is also exerting a force on the end of the rod you are pulling to accelerate along with the end you are pulling. That's Newton's third law. You would have to account for all of these forces. Does that make it seem a little more intuitive why NSL doesn't directly apply?
     
  4. Nov 27, 2012 #3
    Thanks for the reply.

    1. Can I still use NSL for the rotation about the center of mass? (i.e. t = Ia)
    2. Can I pick a random reference point, find the torque of the force about that point, get angular acceleration about that point, and use [a-trans = a-rot * r] to get the linear acceleration?
     
  5. Nov 28, 2012 #4

    SammyS

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    Newton's Second Law holds.

    Fexternal = M aCM.
     
  6. Nov 28, 2012 #5

    Dick

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    SammyS is right, of course. I managed to confuse myself. The force applied to two different points for the same distance does the same work. Better to think of the force applied to the two different point for the same short time. The final CM velocity will be the same but the force applied at the end will need to be applied over a larger distance. That's where the extra work for the rotational motion comes from.
     
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