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Torque question

  1. May 20, 2007 #1
    When a door is pushed at the edge with a force F, a torque is exerted on the door of magnitude Fr (r=level arm).
    The torque compel the door to rotate at an angular acceleration alpha.
    So the torque=(moment of inertia)*(angular acceleration).

    Now the question is : is Fr=(moment of inertia)*(angular acceleration)?

    I think it is incorrect for "torque=I*alpha" is derived by summing Fr of every single point mass over the whole body.

    Your response will broaden my view toward physics.
     
  2. jcsd
  3. May 20, 2007 #2

    Doc Al

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    Assuming that the applied force F is the only force creating a torque on the door (frictionless hinges), then yes it is correct.

    I don't understand what you mean. There's only one force (F) and it is applied at a particular distance from the axis (r).

    Perhaps you are thinking of the derivation of I, which is obtained by integrating R^2dm for every mass element in the body.
     
  4. May 20, 2007 #3
    How only one force is attributed on every mass element in the body? Does a force acting on a particular point effect every other point in a same way? I mean does every point mass experience the same force F?
    If they do then why torques vary when we exert the same force on different point along the length (r)?
     
    Last edited: May 20, 2007
  5. May 20, 2007 #4

    malawi_glenn

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    have you carefully studied the derivation of moment of inertia and torque?
     
  6. May 20, 2007 #5

    Doc Al

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    Applying a force F to a point r does not mean you have applied that same force to all points--the torque is transmitted through the body, not the force. The force applies a single torque to the entire body, since it is applied at a single point. You can think of it as applying a torque to each mass point individually, and then adding up all the contributions--but that's been done for you by using the moment of inertia for the body.

    Realize that the "point masses" that compose the body are connected by internal forces. (Just like if you push a block, the whole block accelerates even though you just pushed one side.)
    For rotational purposes, it's the torque that counts, not the force. If you apply the same force at different points it produces a different torque.
     
  7. May 20, 2007 #6
    You raised an important question. Probably I didn't. Thnx, now I should indulge myself on that .......
     
  8. May 20, 2007 #7

    malawi_glenn

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    also Doc_Al's answer is very good, you must take account for that the "points" that the solid body conssist of is connected with forces (electromagnetic if we are talking of atoms)..
     
  9. May 24, 2007 #8
    :uhh:
    I have made some careful study and found out the followings.
    When we push a block with a force of F. We exert the force at one point but the whole block accelerates. Each tiny mass of the block doesn't experience the force F but a small portion of it. If the block were composed of three mass element m1, m2 and m3, then F=m1a+m2a+m3a=a*(m1+m2+m3)=a*M

    So when we push the door with a force F, the door moves. It means that each mass element have experienced a force but these forces are not equal. Because the linear accelerations (ai) vary for different points (F=ma, a is different so F must be different).
    When the door begins to rotate with an angular acceleration of @ different points feel different forces according to their position. Let these forces on different point with masses m1,m2...........mi situated at different perpendicular distance from the axis r1, r2................ri be F1, F2, .................Fi
    So
    F1=m1a1=m1@r1 (here @=angular acceleration)
    F2=m2a2=m2@r2
    ..................
    ..................
    Fi=miai=mi@ri


    So SUMMATION of Firi= SUMMATION of @(miri^2)=@*SUMMATION of (miri^2)=@*(moment of inertia)

    When a block accelerate linearly the exerted force F=(summation of forces experienced by every mass element).
    So when the door rotates at @ angular acceleration: Is the exerted force
    F=(summation of mi@ri)?
    If the answer is yes then the point of exertion of force should have no value in determining the torque.
    I am actually trying to understand how exertion of the same force on different point along the length alter the turning effect (here @). Torque=@*(moment of inertia) doesn't tell us anything about this. Torque=Fr tell us that torque is greater when the level arm is longer. But it is nothing more than a mathematical expression to me (so far).
    I expect your answer to be less mathematical but more intuitional.
    Thnx in advance.
     
    Last edited: May 24, 2007
  10. May 25, 2007 #9
    Work done by torque=torque*(angle of rotation)
    Work done by Force=Force*(displacment toward the direction of force)

    So when we push a door at r with a force of magnitude F and an angular displacement of "theta" is observed. Here the force is always changing direction as to be parrallel to the displacement. Now the work done by the force is F*displacement=F*(theta)*r =work done by the torque

    As we are exerting the force to a more far away point from the axis of rotation the displacement is higher for a given angle, so more work is done.
    Since torque is something related with work, more torque is applied when more work is done.
    Now I am wandering why there is more angular acceleration when we exert the same force to a more terminal point.
    Suppose we have exerted two same force on two different point r1 and r2, where r2>r1.
    So for a given angular displacement more work is done for the second case.
    More work is done for the same displacement, when applied force is greater. In a simmilar way a greater torque and angular acceleration is involved when there is more work done for the same angular displacement.
    I am still unable to convince myself that I understand why there is more angular acceleration when the applied force move to a more terminal point.
    I think your response will help me (it always does).
     
    Last edited: May 25, 2007
  11. May 25, 2007 #10
    You've convinced yourself that more work is done by applying a force at the larger radius (through some angular displacement). So, how does doing work on a rotating body relate to changing the body's rotational kinetic energy? How does a change in rotational kinetic energy relate to angular acceleration?
     
  12. May 26, 2007 #11
    Actually you are right.
    For a given angular displacement "theta" from rest more work is done when the rotating object is given more kinetic energy. Rotational kinetic energy = 1/2*(moment of Inertia)*(angular velocity)^2

    So to recieve greater angular velocity from stationary position for fixed angle of theta , greater angular acceleration is required. That's why greater angular acceleration is involved with greater torque.
    Now it is enough for me and I fully undrestand the concept of torque.
    Thank you very much for all of your help. :-)
     
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