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Magnitude of the angular acceleration about the pivot point?

  1. Mar 24, 2008 #1
    [SOLVED] Magnitude of the angular acceleration about the pivot point?

    1. The problem statement, all variables and given/known data

    A uniform horizontal rod of mass 2.7 kg and length 2.4 m is free to pivot about one end as shown. The moment of inertia of the rod about an axis perpendicular to the rod and through the center of mass is given by I=ml^2/12

    If 4 N force at an angle of 64 degrees to the horizontal acts on the rod as shown, what is the magnitude of the resulting angular acceleration about the pivot point? The acceleration of gravity is 9:8 m/s^2 : Answer in units of rad/s^2

    2. Relevant equations

    I=(2.7*(2.4)^2)/12

    F=4N=2.7*a


    3. The attempt at a solution

    I=(2.7*(2.4)^2)/12=1.296

    F=4N=2.7*a

    How do I figure out the magnitude of a? or [tex]\alpha[/tex]?

    I just cant understand how to relate the F and the I, I would think the magnitue of a was sin(64).
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2008 #2

    Doc Al

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    Staff: Mentor

    That's the moment of inertia through the center of the rod; what you need is the moment of inertia about the pivot point.
    That would be true if 4N were the net force acting on the rod, but it's not. For this problem, what you need to find is the torque that this force produces.

    Find the net torque on the rod about the pivot point. I assume you need to take into account the force of gravity acting on the rod as well as the applied force.

    Then you can use Newton's 2nd law for rotation to find the angular acceleration.
     
  4. Mar 24, 2008 #3
    So Net torque would = 2.4 *4(sin64)=8.628? or the net force is 9.81 down - 8.832 up? = -.968 Net force?

    How do I account for gravity?

    So the moment of Inertia would be (ML^2)/3 which is I=(2.7*(2.4)^2)/3 = 5.184, but how does it work into Newtons second law, and how does it help me solve angular acceleration?
     
  5. Mar 24, 2008 #4

    Doc Al

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    Staff: Mentor

    That's the counter-clockwise torque due to the 4 N force. But is that the only force acting on the rod?
    :confused:
    What's the weight of the rod? Where does it act?

    Newton's 2nd law for rotational motion:

    [tex]\tau = I \alpha[/tex]
     
  6. Mar 24, 2008 #5
    Ohhhhh, ok so its 8.628+-gravity/5.184 ? t/I? Would the force down due to gravity be mg(l/2)= force of gravity? if so do I subtract that from the 8.628 force up? so the force of gravity is 31.704 and the torque up is 8.628? Then would I have to convert that answer to radians?
     
  7. Mar 24, 2008 #6

    Doc Al

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    Staff: Mentor

    Right.
    That will be the clockwise torque due to gravity.
    You would subtract the clockwise torque from the counter-clockwise torque.
    Since you are using standard units (kg, N, m), the angular acceleration will be in rad/s^2. No need to convert.
     
  8. Mar 24, 2008 #7
    Awesome! Thank you so much.
     
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