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Angular accerleration

  1. Jun 20, 2010 #1
    say i have two rods of the same length but different masses. each rod can rotate around a fixed point, when the same amount of force is applied on the rod at the same distance from the center, which would have the greatest angular acceleration? I think its the one with the greater mass, but I'm not positive
     
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  3. Jun 20, 2010 #2

    Doc Al

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    Staff: Mentor

    So you think the heavier one will have the greater acceleration? Take the linear analog: Push two masses, one more massive than the other, with the same force. Which has the greater linear acceleration?

    What determines the angular acceleration?
     
  4. Jun 20, 2010 #3
    You are applying the same force, and hence, the same torque.
    [tex]\alpha[/tex] depends on moment of inertia. [tex]\alpha = \frac{\tau}{I}[/tex].
    So the lighter one has greater angular acceleration.
     
  5. Jun 21, 2010 #4

    I don't think torque are the same. When we consider torque , we need to consider weight as well.Weights are not the same.But light rod angular accelaration is greater. Think!!!!
     
    Last edited: Jun 21, 2010
  6. Jun 22, 2010 #5

    Doc Al

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    There's no need to complicate things by including the weight. After all, we don't know the orientation of the rod or the location of of the axis. Keep it simple: Assume that the applied force is the only force acting.
     
  7. Jun 22, 2010 #6
    Firstly he has mentioned in the question that same force is applied at same distance from center so that torque for both rods is same. Since Torque T=F*R.
    And second point is that here we may/should assume that both rods are kept on a smooth horizontal surface like a table so need to take into account the weights of the rods.
     
  8. Jun 22, 2010 #7
    I think we need to consider two forces for each rod.Weight and applied force. Torque are not the same because of the weight.
    eg.
    Let we consider fixed point (axis) at the end for both rods.
    tau(1)=m1g+F(0.5l +x)
    tau(2)=m2g+F(0.5l +x)

    Take m2>m1
    tau(2)>tau(1)
    I2alpha(2)>I1alpha(1)
    I=ml2/3
    I2>I1
    m2 alpha(2)>m1 alpha(1)
    alpha(1)>alpha(2)
     
  9. Jun 22, 2010 #8

    Doc Al

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    Staff: Mentor

    You seem to be assuming that the rods are rotating in a vertical plane about an axis at one end and that they are oriented horizontally. Why?
     
  10. Jun 22, 2010 #9
    we can consider applied force for each rod.
    eg.
    If we consider fixed point (axis) at the the centre of the rod.
    tau(1)=F(x)
    tau(2)=F(x)

    Take m2>m1
    tau(2)=tau(1)
    I2alpha(2)=I1alpha(1)
    I=ml2/12
    I2>I1
    alpha(1)>alpha(2)
     
    Last edited: Jun 22, 2010
  11. Jun 22, 2010 #10
    Actually we can consider axis at any point because problem didn't mention where the axis is. If we consider at the centre, it is easier for us.
     
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