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Torque calculation

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end?


    2. Relevant equations

    [tex] \alpha L =a_t [/tex]

    [tex] \tau = I \alpha[/tex]


    3. The attempt at a solution

    The linear acceleration is not the same for all points in the rod but the angular acceleration is. So, I must find the angular acceleration and then calculate the linear acceleration on the right end.

    I find the angular acceleration through that second equation. I know that the moment of inertia is [tex] \frac{1}{3} M L^2[/tex] . But how can I calculate the net torque?
    My textbook says the net torque is equal to the torque on the center of mass considering it accumulates all the body´s mass? But why is this so?

    Can anybody give me a few tips on how to calculate net torques of rigid bodies?

    Thank you
     
  2. jcsd
  3. Feb 25, 2010 #2
    The net torque happens at the center of mass from a theorem known as parallel axis theorem. I know this isn't what you want to hear, but for now, just accept and learn that this is the case: the theorem says this is what you do, so this is what you do. So in that case, where would you say the center of mass is?

    In general, calculating the torque of a rigid body will be just like this. You find the moment of inertia (this will be given to you unless you want to do a double integral), and the center of mass. Then you use the [itex]\alpha I = \tau[/itex] formula.
     
  4. Feb 25, 2010 #3
    Are you sure? Doesn´t the parallel axis theorem simply say:

    [tex]I_z=I_CM + MD^2[/tex]

    where I_z is parallel to I_CM and D is distance from I_z to I_CM ? What does this have to do with calculating the torque?

    It´s in the center of the rod, at a distance L/2 to the pivot.
     
  5. Feb 26, 2010 #4
    Hmm, okay, sorry. Maybe I got my theorem's mixed up. You're right though, CM is at L/2. From here, it's plug and chug.
     
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