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Acceleration of a compound pendulum

  1. Sep 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniform thin rod of mass M and length L is supported horizontally by two supports, one at each end. The acceleration of gravity, g, is constant and in the downward direction. At time t=0 the left support is removed.

    Find the downward acceleration of the center of mass at t=0 in terms of M, L and g. Also, find the angular acceleration and the force exerted by the remaining support.


    2. Relevant equations
    [tex] \vec{\tau} = \vec{r} \times \vec{F} [/tex]
    [tex] |\tau| = I \alpha [/tex]



    3. The attempt at a solution


    To calculate the downward acceleration, I first calculate the torque.

    [tex] \vec{\tau} = \frac{-L}{2}\hat{y} \times (-M g) \hat{z} = \frac{LMg}{2}\hat{x} [/tex]

    Equate this torque with the moment of inertia times the angular acceleration, and solve for angular acceleration.

    [tex] \alpha = \frac{\tau}{I} = \frac{\frac{L}{2} M g}{\frac{1}{3} M L^2}=\frac{3}{2}\frac{g}{L} [/tex]

    Then, solve for the acceleration,
    [tex] a = r \alpha = \frac{L}{2} \cdot \frac{3}{2}\frac{g}{L} = \frac{3}{4} g [/tex]

    Find the force by F=ma,

    [tex] F = m a = M \frac{3}{4} g [/tex]


    Where have I gone wrong here? Thanks!
     
    Last edited: Sep 19, 2011
  2. jcsd
  3. Sep 20, 2011 #2

    ehild

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    Your force is wrong. The resultant force is equal to the mass times the acceleration of the CM. The forces are gravity and the force at the support.


    ehild
     
  4. Sep 20, 2011 #3
    Thanks, that makes sense. But Im still a little confused because the acceleration is at the center of mass, the force of gravity is at the center of mass but the support is at a different location. I can sum up the forces even though the support is causing a torque? Perhaps I should sum up the torques! I just now realized, I think that is what I should do.
     
  5. Sep 20, 2011 #4

    ehild

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    The torque refers to an axis and equals force times the distance of the line of force from the axis. You calculated the torque of gravity with respect to the supported end of the rod. If there are more forces, their torques add up, but what is the torque of the supporting force with respect to the support?

    In case of a rigid body, the resultant force is the vectorial sum of all forces acting. The forces can be shifted along their line. Parallel forces add up algebraically. No matter where the forces attack. You have two forces, one is gravity, the other is the supporting force. The acceleration of the CM multiplied by the total mass is equal to the resultant force.
    What you should do is to find out is the acceleration of the CM. For that, you need to know its position. The rod is homogeneous. Where is its CM? The rod sweeps a circle with one end as the centre. Every point of it moves along a circular path, with the same angular velocity and angular acceleration. So what is the linear acceleration of the CM?

    ehild
     
  6. Sep 20, 2011 #5
    I found the CM downward acceleration to be
    [tex] \frac{3}{4} g [/tex]

    If I understand you correctly, its as simple as doing,

    [tex] ma = M \frac{3}{4}(- g) = - M g + F_{\text{Support}} [/tex]

    Solve for the force and get,
    [tex] F_{\text{Support}} = \frac{1}{4} g [/tex]


    I still cant help but be skeptical though... The point at which the support is applying a force is not accelerating at all. I feel like I can only use forces shifted along a line if there is no rotation in the rigid body. Ill take it as it is for now though. Thanks
     
  7. Sep 21, 2011 #6

    ehild

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    You forgot the M, Fs=Mg/4.


    No matter how the individual points move. The centre of mass accelerates as if the sum of all the external forces attacked at it and its acceleration is equal to the vector sum of all external forces divided by the total mass.

    If two parallel/antiparallel forces act on a rigid body the resultant is a force equal to the algebraic sum of the forces (and its point of attack is that point for which the the torque of the two forces is equal and opposite).

    You can understand this by looking at the picture. It is a rigid rod and two forces act on it at different points, F1 and F2, parallel and of opposite direction, like in the problem Fs an -mg.
    You can add two equal and opposite horizontal forces (the green ones): They would cancel, so do not influence the system of forces. Add the horizontal forces to the vertical ones: You get two forces (red) which are not parallel any more, they can be shifted to a common point where the resultant can be drawn. This resultant force can be shifted againso its point of attack is on the rod.

    ehild
     

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  8. Sep 21, 2011 #7
    That picture does help, thanks.
     
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