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Rotating rod

  1. Mar 10, 2005 #1
    Hi, I have a problem here I was hoping someone could help me out with. The first part of the problem I think I've got, but I would appreciate it if someone could just look through what I did and tell me if I'm way off.

    The problem, first part: A solid rod of length L and mass m is attached horizontally in both endpoints (A to the left, and B to the right). Point A is then released and the rod rotates freely about point B until the point A is directly below point B (i.e. it starts out horizontally and finish vertically) at which point it is released also in point B.

    It starts out horizontally, and ends vertically at time [tex] t = 0 [/tex]. So we're asked to find the angular velocity and the center-of-mass velocity at [tex] t = 0 [/tex].

    Here's what I've done: I used conservation of energy and chose to assign zero potential energy to the center of mass (CoM) when the rod is vertical, that is, the CoM is at L/2 below point B. Then it will start out horizontally with a potential energy of [tex] E_p = mg(L/2) [/tex] which will all be kinetic when the rod is vertical, i.e. [tex] mg(L/2) = (1/2)I \omega^2 = (1/6)mL^2\omega^2[/tex] where I've used the moment of inertia of a rod about it's end. That eventually leads to [tex] \omega = \sqrt{3g/L}[/tex]. To get the velocity of the CoM I then multiply by L/2 to get [tex]v = \sqrt{3gL/4} [/tex].

    Now to the next part. The rod is vertical, attached in it's top point (B), with angular velocity as found in the previous part. The rod is then released from point B and falls freely.

    I'm now assuming that the rod continues, rotating about it's CoM with the same angular velocity as found in the last part. We are then to introduce (generalized) coordinates to describe the motion of the rod during the free fall, and find the Lagrangian.

    Here's what I've done: I let [tex]\vec{r} [/tex] be the position vector of the CoM, and [tex]\theta [/tex] be the angle between the y-axis and the position vector. My coordinate system has it's origin in point B. This then gives [tex] x = r\sin \theta[/tex] and [tex] y = x \cos \theta[/tex] where [tex] r = \arrowvert \vec{r} \arrowvert[/tex].

    This now gives me kinetic energy: [tex]T = (1/2)I\omega^2 + (1/2)m\dot{x}^2 [/tex] where the moment of inertia this time is about the rod's CoM, [tex]I = (1/12)mL^2 [/tex] and [tex]\omega^2 = 3g/L [/tex]. For the potential energy it seems reasonable to choose the x-axis as the level with zero potential energy. Then I get that [tex]V = -mgy [/tex]. Putting these two into the Lagrangian then gives me: [tex] L = T - V = (mgL/8) + (1/2) m\dot{x}^2 + mgy[/tex].

    So, that's what I got. Now I would like to hear what you guys say, am I way off, or at least in the right neighborhood?
     
  2. jcsd
  3. Mar 10, 2005 #2
    Part 1.

    Total Energy = potantial E. + Rotational K.E. + Translational K.E.

    From Conservation of Energy,

    [tex] mg(L/2) = (1/2)I \omega^2 + (1/2)m v^2[/tex]

    Do you agree?
    ========================================

    Part 2

    There are certain things I don't agree in your writing of the lagrangian. L should be in generalized coordinates.

    However, you have used the initial rotational KE in writing lagrangian. One can use the initial condition later when solving differential equations. I would write the lagrangian as follows.

    [tex] L = T - V = (1/2) I\dot{\theta}^2 + (1/2) m\dot{r}^2 + mgrcos\theta[/tex]
     
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