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Force at the end of a pivoted object

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    (a) Find the acceleration at the tip of the length L rod
    (b) Find the normal force at the hinge

    2. Relevant equations
    Moment of Inertia of rod: I = mL2/3

    3. The attempt at a solution

    Since the rod is rotating, there is centripetal acceleration throughout the rod. This points toward the hinge.
    acen = - L ω2 er ... (1)

    The fact that the rod is attached to the hinge would cancel all translational forces.

    The only remaining force comes from the net torque. The net torque at the hinge is:

    τnet = - τ + (L/2) mg sinθ ... (2)

    The net torque is related to the angular acceleration α by the moment of inertia. And α is related to the tangential a by the rod length.

    τnet = I α
    atan = L α
    atan = L τnet/I = (3/mL)τnet eθ ... (3)

    This is counter-clockwise.

    The total a is acen + atan ... (4)

    The only force is mg, so the normal force at the hinge is mg (pointing up).

    This seems right to me, what do you think?

    Now there is a weird problem:

    Suppose I also calculate the net force at the tip. Since the translational forces are cancelled, the net force at the tip comes purely from the net torque:

    Ftip = τnet/L ... (5)

    This force is entirely tangential. Since I also had the acceleration at the tip, I could compute the "mass" at the tip:

    Ftip = mtip atip ... (6)

    Now, this is completely bizzare, since F and a do not point in the same direction, and m is supposed to be 0 at the tip. What is the logical reason that (6) is not valid? (Or perhaps (5) is not valid (?))

    - Thanks
  2. jcsd
  3. Dec 8, 2008 #2


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    Homework Helper

    Wow! You are really tearin' this one apart. Excellent! BTW, I must admit that I would have gotten this one wrong b/c I would have forgotten to account for the centripetal acceleration. Whoops! Great job on your part! Also, I didn't actually check to see if you did the problem correctly; instead I want to address the following excellent question:

    Simply put: b/c (5) is invalid.

    You are correct in thinking that, in practice, an applied torque is actually an applied linear force at a given distance and in a given direction wrt a center of rotation. However, just as in Newton's second law, where on one side of the equation you have applied forces, and on the other side of the equation you have ma, here you should treat tau_net as I x alpha, the result of the applied torque. This takes into account the object as a whole, and is not specific to any point on the rod, tip or otherwise.

    A different way to look at it is that, if you are going to start talking about forces and masses at all of the specific points on the rod, then you must also account for all of the intermolecular forces that hold the rod together that make it a rod in the first place, which you have not done.
    Last edited: Dec 8, 2008
  4. Dec 8, 2008 #3
    Thank you, I think I understand it now. When I use F = T/L in (5), the force I get is not a force that is acting on the tip, but a force that the tip has to act on other object. So when I do

    F = ma ... (6)

    m and a are not the mass and acceleration of the tip, but of the hypothetical object that the tip would hit if there was such an object near the tip:

    The ball and the rod are initially at rest. Then a torque is applied to the rod at B. What is the acceleration of the ball when the rod hits the ball?

    F = T/L = ma

    a = T/mL
  5. Dec 9, 2008 #4


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    Homework Helper

    Basically, yes. Of course, if the rod is massive, then you have to account for that as well ...
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