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Particles moving freely on a vertical/horizontal loop

  1. Apr 30, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    1) A mouse of mass ##m## runs around the inner circumference of a vertical circle which is free to rotate about its centre. The mouse starts at the bottom of a stationary wheel. Let ##\phi## be the angle that the radius vector of the mouse makes with the downward vertical at time ##t## and let the distance run by the mouse around the circumference at time ##t## be a known function ##s(t)##. Show that the Lagrangian for ##\phi(t)## is $$L = \frac{1}{2}ma^2\dot{\phi}^2 + \frac{1}{2}I\left(\frac{\dot{s}}{a}-\dot{\phi}\right)^2 + mga\cos \phi$$
    (only need help with this part - but the rest of the question is attached for clarity (AttachmentQ2))

    2)In attachment Q1 Last part: If the particle has speed ##u## when passing through O, show that its speed relative to the wire when passing through the point P at the other end of the diameter is ##\sqrt{u^2 + 4a^2 w^2}##. (Rest of question provided for clarity again)

    2. Relevant equations
    Kinetic energies, angular velocity.

    3. The attempt at a solution
    1)The first term in the Lagrangian is the translational kinetic energy of the mouse relative to an inertial frame and the third term is its potential energy. Introducing ##\theta## as the angle through which a point on the surface of the wheel has rotated, assuming there is no slip between the mouse and the inner circumference, then ##\dot{\theta} = \dot{s}/a##. Let ##w## be the angular velocity of the mouse wrt an inertial frame. How do we obtain ##w = \dot{s}/a - \dot{\phi}##? Why isn't ##\dot{\phi}## enough?

    2) I think to get the speed at P relative to the wire then I need to solve ##v_p = 2a \dot{\phi}##. From a previous part of the question, I got that the equation of motion for ##\phi## incorporating the constraint ##\dot{\theta}=w##, is $$\ddot{\phi} + w^2 \sin \phi = 0$$ I could multiply this equation by ##\dot{\phi}##, integrate to get d/dt (expression) = constant and use the initial conditions for the velocity at O to get the constant. This would give me ##\phi## but the integral was non-elementary. Even so, if I could solve it, it would give me ##\phi(t)## and I would not know the time taken for the particle to get to P.

    Many thanks.
     

    Attached Files:

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    Last edited: Apr 30, 2014
  2. jcsd
  3. Apr 30, 2014 #2

    collinsmark

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    I might be able to help with the first question.

    [itex] s [/itex] is the distance traveled by the mouse, relative to the circumference of the wheel. Don't confuse that with the angular displacement of the wheel. Normally, you would expect them to be similar, but they are at least subtly different, and can be quite different if the mouse suddenly latches onto the wheel while spinning, as mice sometimes do (I've seen it!). So [itex] \frac{\dot s}{a} [/itex] gives an angular velocity, but it's not quite the angular velocity of the wheel. It's the angular velocity of the wheel with respect to the mouse, not with respect to the vertical. To find the angular velocity of the wheel with respect to the vertical (such that you can use that to find the rotational kinetic energy of the wheel), you need to subtract off the angular velocity of the mouse with respect to the vertical. That gives you the angular velocity of the wheel with respect to the vertical. And that's where the [itex] \frac{\dot s}{a} - \dot \phi [/itex] comes into play in the second term.
     
  4. May 1, 2014 #3

    CAF123

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    Hi collinsmark,

    Let ##v (\omega_v)## be the tangential (angular) velocity of the wheel wrt the mouse. Let ##u (\omega_u)## be the tangential (angular) velocity of the mouse wrt the vertical. Let ##V (\omega_V)## be the tangential (angular) velocity of the wheel wrt the vertical. Then I believe these are all related by the equation ##\mathbf{V} = \mathbf{v} + \mathbf{u}##. I believe to get the angular velocity of the wheel wrt the vertical, you can then say that ##\omega_V = \omega_u + \omega_v = \dot{\phi} + \frac{\dot{s}}{a}##. so I got a + where you got a -. Why is it we are interested in the rotational kinetic energy of the wheel? Are you we not looking for the rotational kinetic energy of the mouse since the question asks for the Lagrangian for ##\phi##, which by definition describes the motion of the mouse relative to the vertical?

    Thanks.
     
  5. May 1, 2014 #4
    Angular velocity w.r.t. a non-inertial frame? Avoid :)

    ##v## is the velocity of the mouse w.r.t. in an inertial frame.

    ##u## is the velocity of the of the wheel's point of contact with the mouse in the inertial frame.

    (not using vectorial notation here because the velocities are obviously parallel)

    Then ##v - u = \dot s## is the velocity of the mouse w.r.t. the wheel. Or, ##w = - \dot s = u - v ## is the velocity of the wheel's point of contact w.r.t. the mouse. So ## u = v - \dot s ## and the wheel's angular velocity is ## \omega = \frac u a = \frac v a - \frac {\dot s} a = \dot \phi - \frac {\dot s} a ##.
     
  6. May 1, 2014 #5

    CAF123

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    Hi voko,

    I see. Let me break down the Lagrangian into pieces: The first term is the translational kinetic energy of the mouse wrt an inertial frame. The second term is the rotational kinetic energy of the wheel to an inertial frame. Why do we not have here only the rotational kinetic energy of the mouse? Are we not only interested in the motion of the mouse, since the question says find the lagrangian for ##\phi(t)## which describes the motion of the mouse to an inertial frame?

    I would have expected something like total kinetic energy = trans. kinetic energy of mouse to inertial frame + rotational kinetic energy of mouse to inertial frame.
     
  7. May 1, 2014 #6

    collinsmark

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    Actually, it's the other way around. The mouse has no rotational kinetic energy. That's because the mouse is treated as a point particle (well, per the instructions given in the problem statement). Spinning (yet translationally stationary) point particles have no kinetic energy, simply due to their spinning.

    The [itex] \frac{1}{2}ma^2 \dot \phi^2 [/itex] term is attributed to the mouse's translational kinetic energy, not its rotational energy.

    What is the mouse's translational speed, as a function of [itex] \dot \phi [/itex]? It should be clear after answering that. :wink:
     
    Last edited: May 1, 2014
  8. May 1, 2014 #7

    collinsmark

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    Mice are awesome, by the way!

    https://www.youtube.com/watch?v=N633gnUxJ50
     
  9. May 1, 2014 #8
    First, see the comment by collinsmark.

    Second, the mouse and the wheel are clearly coupled in the problem via s(t). That alone means it has to enter the Lagrangian.
     
  10. May 1, 2014 #9

    CAF123

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    Okay, I think the words 'Show that the Lagrangian for ##\phi(t)##..'' put me off for some reason. The Lagrangian given embodies the kinetic energy of the whole system, that of the translational kinetic energy of the mouse (no rotational kinetic energy of the mouse, as was clear after viewing the video posted by collinsmark), the rotational kinetic energy of the wheel and the potential energy of the mouse.

    Can you help at all with my other question? See OP, second question.

    Thanks!
     
  11. May 1, 2014 #10
    I have not checked whether the equation you obtained is correct, but assuming it is, it results in some ## \dot \phi = f(\phi) ##. Consider ##f(0)## and ##f(\pi)##.
     
  12. May 1, 2014 #11

    CAF123

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    Multiplying the equation by ##\dot{\phi}##, I get that $$\frac{d}{dt} \left(\frac{{\dot\phi}^2}{2} -w^2\cos\phi\right) =0$$ so ##\frac{{\dot\phi}^2}{2} -w^2\cos\phi = \text{const} = A##

    Now, ##\dot{\phi}/a = u \Rightarrow \dot{\phi}^2 = a^2u^2## where ##\phi = \pi## meaning that ##A = a^2 u^2/2 + w^2##

    So now I would rearrange for ##\dot{\phi}## and find the value when ##\phi = 0##. Then ##v_p = 2a\dot{\phi}|_{\phi=0}##
     
    Last edited: May 1, 2014
  13. May 1, 2014 #12
    That does not address #10 fully.
     
  14. May 1, 2014 #13

    CAF123

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    Yes, ofcourse and it had an error. My internet connection lapsed. I have corrected it now and added more detail. Thanks.
     
  15. May 1, 2014 #14
    I wonder why you could not finish the analysis. Once you get ##A##, there are no unknowns in ##f(\phi)##. What is ##f(0)##?
     
  16. May 1, 2014 #15

    CAF123

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    I didn't get the right result, so I wanted to check my method. ##f(0) = \sqrt{4w^2 + a^2u^2}## Then ##v_p = 2a\sqrt{4w^2 + a^2u^2}##, but it is not quite the same as the answer.
     
  17. May 1, 2014 #16
    How did you get that? Note that addends under the radical are incompatible dimensionally.

    Where does the ##2## come from?
     
  18. May 1, 2014 #17
    I see that the wrong dimensions were present already in #11, though it was not so apparent as in #15.
     
  19. May 2, 2014 #18

    CAF123

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    Okay with the corrections I obtain ##\dot{\phi}|_{\phi =0} = \sqrt{u^2 + 2a^2w^2}##. My expression for v_p came about because relative to O, P is a distance 2a away.
     
    Last edited: May 2, 2014
  20. May 2, 2014 #19
    But is the rotation due to ##\phi## relative to O?
     
  21. May 2, 2014 #20

    CAF123

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    No, it is relative to the radius as in the sketch. So, that would mean that ##v_p = a \dot{\phi}|_{\phi=0} = a\sqrt{u^2 + 2a^2w^2}## but I seem to still be missing a factor and that is no longer dimensionally consistent.
     
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