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Kinetic Energy of rod with lower end in contact with smooth surface

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I've got a rigid rod, length [itex]2a[/itex], held at an angle [itex]\alpha[/itex] above a smooth horizontal surface. Its end is in contact with the ground. My two generalized coordinates are [itex]x[/itex], the horizontal position of the center of the rod and [itex]\theta[/itex] the angle between the rod and the vertical.

    I'm trying to find the kinetic energy of the rod. I'm not sure if I'm doing it right.


    2. Relevant equations



    3. The attempt at a solution
    As of now, this is what I've done:
    [tex]KE = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}I\dot{\theta}^2 = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}(\frac{1}{3}m(2a)^2)\dot{\theta}^2 = \frac{1}{2}m\dot{x}^2 + \frac{2}{3}ma^2\dot{\theta}^2[/tex]

    Something doesn't seem right though, because I think that [itex]\dot{x}[/itex] should be zero. Carrying out the Lagrangian doesn't seem to show this.
     
  2. jcsd
  3. Feb 6, 2009 #2
    [tex] \alpha [/tex] is the azimuthal angle? Is the rod held to a point in ground? If not, the speed of it's center of mass won't be zero.
    Also, I don't think you've got the formula right. K.E can be broken into a translational and a rotational term when you take the rotation about the center of mass. In this case the rotation is about a point in the ground. Perhaps you could use x and z?
     
  4. Feb 6, 2009 #3
    All motion takes place in the vertical plane. Alpha is just the angle the rod is at when it's initially released. The end of the rod is touching, but NOT attached, to the ground.


    I specified the KE of the rod by the translational motion in the x direction and by the rotational motion about the point touching the ground (even though it is slipping backwards). I'm almost certain this is incorrect but I couldn't come up with much better.
     
  5. Feb 6, 2009 #4
    Ok, so there's no force here along the horizontal. The 2 forces, gravity and normal are both along the vertical. Consequently we should not have any motion along the horizontal. So we can use just one gen coordinate - the angle. Then we can rate of change of the angle ([tex] \dot(\theta) [/tex] ) . Then the speed of any infinitesimal portion at a distance r from the point of contact is just [tex]r \dot(\theta) [/tex] . So you can integrate along the length and find the KE.
     
  6. Feb 6, 2009 #5
    The point of the problem is to use [itex]x[/itex] and [itex]\dot{x}[/itex] with the lagrangian and prove that the CM stays in place
     
  7. Feb 7, 2009 #6
    Oh. To know that you don't actually need to know the KE : the potential term is mgy and the constraint condition [tex] \dot z [/tex] =0. None of which contains x, ergo [tex] \dot x = 0 [/tex] .

    I think the KE may be found this way: Let x, y be the co-ordinates of the CM. Then [tex] y= a\sin\theta [/tex] ( This is using the condition that the end keeps touching the ground)Then the co-ordinates of any other point on the rod :
    [tex] (X,Y) = ( x - r\cos\theta , a\sin\theta - r\cos\theta) [/tex] . Now you can take derivative, square and integrate along r (from -a to a) to find the KE.
     
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