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Kinetic and potential energy in circular and rotational motion

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a bead of mass m that is confined to move on a circular hoop of radius r. The axis of symmetry of the hoop is horizontal, and the hoop is rotating about a vertical axis at a uniform rate [itex]\omega[/itex]. Neglect friction and assume a constant gravitational acceleration of magnitude g. Let the angle [itex]\phi[/itex] specify the position of the bead on the hoop, with [itex]\phi = 0[/itex] corresponding to the bead lying at the bottom of the hoop.

    Derive the following Lagrangian:

    [tex] L = T-V = \frac{1]{2}mr^2(\dot{\phi}^2 + \omega^2\sin^2\phi) + mgr\cos\phi [/tex]

    Since the Lagrangian doesn't depend explicitly on time, there is a conserved quantity for this system, [itex]E = \dot{\phi}\frac{\partial L}{\partial \dot{\phi}}[/itex]. Compute E and show that it is constant. Is the constant the total energy of the system T+V?

    2. Relevant equations

    Most relevant equations are there

    3. The attempt at a solution

    I thought I computed the Lagrangian just fine. My logic was that there were two sources of kinetic energy: that of the bead moving in circular motion ([itex]mr^2\dot{\phi}^2/2[/itex]) and that of the bead rotating. My thinking was that as the bead rotated, if it was at an angle [itex]\phi[/itex], then it was tracing a circle of radius [itex]\r\sin\phi[/itex], so its kinetic energy would be [itex]mr^2\omega^2\sin^2\phi/2[/itex].

    I think the potential energy is purely gravitational. If h=0 corresponds to the height of the center of the hoop, then the potential energy is [itex]-mgr\cos\phi[/itex]. This gives the correct Lagrangian.

    However, it does not give the correct energy. When I compute E, it comes out to be [itex]1/2 mr^2(\dot{\phi}^2 - \omega^2\sin^2\phi - mgr\sin\phi[/itex]. That middle term has the wrong sign, but I can't figure out why. Moreover, this version of E is not conserved. The only conclusion I can reach is that somehow, what I took as rotational kinetic energy is actually potential energy, but that doesn't make sense. Any thoughts? Thank you!
     
  2. jcsd
  3. Nov 10, 2011 #2
    I don't understand how your Rωsinθ is negative since Rsinθ is just the radial length from the axis of rotation to a point on the hoop

    Potential NRG should be mgR(1-cosθ) because it is at a max when cosθ = 0 and that is perependicular to the axis of rotation and this is when your tangential velocity is a max

    The lagrangian should be

    .5mR2((dθ/dt)2 + ω2sin2θ) -mgR(1-cosθ)

    and i know this is correct since i did this problem for an assignment a month ago
     
  4. Nov 10, 2011 #3
    My R[itex]\sin\omega[/itex] is positive. If you mean in the E computation, it's because in computing E, you subtract the Lagrangian, hence the negative sign.

    As for the potential energy, I like yours much better than what I have; however, the first part of the assignment required matching the form shown in the OP, which didn't come through for some reason:

    [tex]L = T-V = \frac{1]{2}mr^2(\dot{\phi}^2 + \omega^2\sin^2\phi) + mgr\cos\phi[/tex]
     
  5. Nov 10, 2011 #4
    I'm also beginning to wonder if you would even expect that quantity E to be the total energy of the system, since I think there's also an effective potential in there somewhere
     
  6. Nov 14, 2011 #5
    Its the pot that changes sign not E. what u were given is wrong. My prof gave us the text book solution for the non inertial frame. And this question is an example in taylor.

    Edit talking the derivative of anything will give u force not energy
     
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