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Homework Help: Find the magnitude of the initial acceleration of the rod's center of mass

  1. Mar 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of length 1.15m is attached to a frictionless pivot at one end. It is released from rest an angle theta=17.0 degrees above the horizontal.

    Find the magnitude of the initial acceleration of the rod's center of mass

    2. Relevant equations

    torque=moment of inertia*angular acceleration
    torque=length*force(perpendicular)

    I=1/3 ML^2

    3. The attempt at a solution

    I set the torque about pivot equal to the moment of inertia about pivot times the angular acceleration to find the angular acceleration. Then I found the linear acceleration by multiplying the found angular acceleration by the length of rod. So my work was....

    L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)
    3/2*(g/L)*cos17=(angular acceleration)
    12.22 rad/s^2= (angular acceleration)

    a=(angular acceleration)*L= 14.05 m/s^2

    However the answer I got is wrong. (Unfortunately, I do not have access to the correct answer) What step did I miss?
     
  2. jcsd
  3. Mar 14, 2007 #2

    andrevdh

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    ...... L/2*(Mg)cos17=1/3*ML^2*(angular acceleration)

    This should be the sin not the cos.

    The com will then experience only a tangential acceleration:

    [tex]a_t = r \alpha[/tex]
     
  4. Mar 14, 2007 #3
    I replaced the cos with the sin17 in my equation and I just tried to solve but I was told my answer was wrong. And I did exactly what you suggested. I wonder why that was wrong? Did I need to do something else?
     
  5. Mar 14, 2007 #4

    Mentz114

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    Have you used the correct angle ? 17 deg above the horizontal is 90 + 17 from the vertical.
     
  6. Mar 14, 2007 #5

    hage567

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    If you are trying to find the tangential acceleration of the centre of mass, wouldn't it be [tex]a_t = \alpha(L/2)[/tex], since the centre of mass will be at half the length of a uniform rod? Just a thought.
     
  7. Mar 14, 2007 #6
    Mentz114, thank you for mentioning that! I looked through some notes and saw I need to add 90 degrees. However, I just redid the problem, using the info you told me and what andrevdh told me, and oddly enough, I got the same answer!

    hage567, I will try to find the initial acceleration with L/2 instead of L.
     
  8. Mar 14, 2007 #7
    Thanks everyone, I got the right answer now.
     
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