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Homework Help: Particle sliding on a thin rod

  1. Aug 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is free to slide on a thin rod. The rod rotates in a plane about one end at constant angular velocity ##\omega##. Show that the motion is given by ##r=Ae^{-\gamma t}+Be^{+\gamma t}##, where ##\gamma## is a constant which you must find and A and B are arbitrary constants. Neglect gravity.

    Show that for a particular choice of initial conditions [that is, r(t=0) and v(t=0)], it is possible to obtain a solution such that r decreases continually in time, but that for any other choice r will eventually increase. (Exclude cases where the bead hits the origin.)


    2. Relevant equations



    3. The attempt at a solution
    I guess I have to use polar coordinates. In polar coordinates,
    [tex]\textbf{a}=(\ddot{r}-r\dot{\theta}^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta}) \hat{ \theta }[/tex]
    Here, ##a=0## and ##\ddot{\theta}=0##. Hence
    [tex]0=(\ddot{r}-r\omega^2)\hat{r}+2\dot{r}\omega\hat{\theta}[/tex]
    This gives,
    ##\ddot{r}-r\omega^2=0## and ##2\dot{r}\omega=0##.
    Solving the first equation gives a solution of the form presented by the question and I get ##\gamma=\sqrt{\omega}## but the second equation gives ##\dot{r}=0##. This doesn't look right. :confused:

    Any help is appreciated. Thanks!
     

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  3. Aug 12, 2013 #2

    BruceW

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    It says 'the particle is free to slide on the rod'. What direction is it free to move in? And what does that imply about the other direction?
     
  4. Aug 12, 2013 #3
    It is free to move in the radial direction.

    What is the other direction? Perpendicular to rod? :confused:
     
  5. Aug 12, 2013 #4

    BruceW

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    yeah. I'm sure you'll be like 'oh of course' when you realize why equation 2 is not equal to zero.
     
  6. Aug 12, 2013 #5
    Actually, I still don't realise why second equation should not be zero. :uhh:
     
  7. Aug 12, 2013 #6

    WannabeNewton

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    If it were zero, then what is keeping the particle constrained to move along the rod?
     
  8. Aug 13, 2013 #7

    ehild

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    If the total acceleration is zero the particle moves along a straight line with constant velocity. (Newton's First Law. )
    Your formula for the acceleration refers to an inertial frame of reference, using polar coordinates.

    ehild
     
  9. Aug 13, 2013 #8
    This problem is easiest in the co-rotating frame. What forces act on the particle? What forces does the rod neutralize?
     
  10. Aug 13, 2013 #9
    Normal force from the rod? :confused:

    Sorry, I don't know about the co-rotating frame.
     
  11. Aug 13, 2013 #10

    BruceW

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    hehe, I think you have a mind-block on this problem. You are saying that the normal force from the rod keeps the particle constrained to move along the rod and you are saying there is zero force acting on the particle. These two things are inconsistent.

    edit: except it is not inconsistent in the special case when omega is zero.
     
  12. Aug 13, 2013 #11

    ehild

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    The particle slides freely along the rod, but it must stay on the rod. Assume it is a bead with a hole in the middle and the rod goes across the hole. The road exerts a normal force on the bead.


    ehild
     
  13. Aug 13, 2013 #12
    This is the frame in which the rod is stationary. It will have fictitious (inertial) forces. The frame rotates uniformly, so these forces are well known.

    But even if you do not know them, these forces will have two components: along the rod and perpendicular to it. Do you care about the perpendicular component? Why? What about the parallel component?
     
  14. Aug 13, 2013 #13
    I meant that the force ##2m\dot{r}\omega## (coriolis force?) is balanced by the normal force exerted by the rod on the particle or bead.

    The particle is constrained to move along the length of rod. The normal force balances the perpendicular component. The parallel component is responsible for the motion of particle along the rod.
     
  15. Aug 13, 2013 #14

    BruceW

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    as you said, [itex]\ddot{\theta}[/itex] is zero, therefore [itex]2m\dot{r}\omega[/itex] is the net force on the particle in the angular direction. This is not a fictional force, or anything like that. When you add up all the real forces in the angular direction, they must equal [itex]2m\dot{r}\omega[/itex]
     
  16. Aug 13, 2013 #15

    ehild

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    You mix the two frames -the inertial and the rotating ones. There is no Coriolis force in the inertial frame of reference, but there is normal force from the rod.

    In the rotating frame of reference, you have fictitious forces: the centrifugal force and the Coriolis force. And you also have the real normal force from the rod. But the bead moves only radially in that system, so dθ/dt=0.

    ehild
     
  17. Aug 13, 2013 #16
    Okay, let's not talk about the other reference frames and solve it in the inertial frame. I haven't yet studied the Coriolis force so I think it would confuse me more if I try to work on it using the rotating frame.

    But how do I solve this in inertial frame now? Hmm...I think my acceleration vector is wrong. I did not take into account that the normal force also acts in the ##\hat{θ}## direction. Correct?
     
  18. Aug 13, 2013 #17

    ehild

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    You wrote the acceleration vector correctly, but did not wrote the equation F=ma. If you write it, you need to include the normal force to the left-hand side.


    ehild
     
  19. Aug 13, 2013 #18
    Correct, so you only care about the parallel (radial) force. Still pretending you do not know what it is, assume that the particle is glued to the rod at some radius. Then it is stationary in the rotating frame, consequently the radial force must be neutralized by the normal force toward the center. I am sure you do know what this normal force toward the center is - just consider that particle in the inertial frame now.
     
  20. Aug 13, 2013 #19
    Normal force = ##mr\omega^2 ##? :confused:

    I don't really understand where this discussion is heading. I asked about the wrong result produced but I don't think I am anywhere close to clearing my misconception. :(
     
  21. Aug 13, 2013 #20

    WannabeNewton

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    Go back to the inertial frame. The particle can slide freely along the length of the rod. This means it is moving freely in the radial direction but clearly there is a constraint force acting on the particle in the tangential direction that is keeping it constrained to move along the length of the rod.
     
  22. Aug 13, 2013 #21
    The normal force (N) in the tangential direction balances the force ##2m\dot{r}\omega##.

    Equating forces in the tangential direction, ##N=2m\dot{r}\omega \Rightarrow \dot{r}=N/(2m\omega)##. But the acceleration vector I wrote down gives ##\dot{r}=0##. :(
     
  23. Aug 13, 2013 #22

    WannabeNewton

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    As ehild already mentioned, you're mixing up the corotating frame and the inertial frame. In the inertial frame the only force in the tangential direction is the constraint (normal) force. There is nothing balanced in the tangential direction.
     
  24. Aug 13, 2013 #23

    ehild

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    Think about forces acting on the particle. It interacts with the rod, and with nothing else. The rod does not exert force along its length: "the particle slides freely along the rod". What is the radial component of the net force acting on the particle then?
    The force of constraint is perpendicular to the constraint: The normal force is perpendicular to the rod. That force causes the azimuthal acceleration.

    I suggest again that you write up the equation F=ma in radial and azimuthal components.

    ehild
     
  25. Aug 13, 2013 #24
    That would be the force felt by the glued particle toward the center, also known as the centripetal force. In the co-rotating frame, that force neutralizes the radial force toward the periphery - the centrifugal force. Now if the particle is no longer glued (but still has to slide along the rod), all it feels is the centrifugal force (the Coriolis force is checked by the reaction of the rod).

    Do you see what radial force acts on the particle?

    Do you see why there is zero net tangential force in the co-rotating frame?

    Do you in particular see that tangentially there is a fictitious force, which can be eliminated by switching to an inertial frame, and a real force, which is present in any frame?

    What is left of your misconception?
     
  26. Aug 13, 2013 #25
    I think I will have to learn more about these frames. I guess I should leave the problem of ##\dot{r}=0## for now and move on to the other parts of the question.

    In the first post, it is ##\gamma=\omega##. Sorry about the typo.

    The next part of the question states that r decreases continually with time. I am not sure how would I approach this but it looks like ##Be^{\omega t}## would increase too much with time so if we set B=0, we can have r which continuously decreases with time but how should I find the initial conditions? :confused:

    For radial direction,
    ##F_r=ma_r \Rightarrow r\omega^2=\ddot{r}##. The solution of this differential is asked in the question.

    For azimuthal direction,
    ##F_{\theta}=ma_{\theta} \Rightarrow N=2m\dot{r}\omega##. I am sorry if this wrong. Please have a look at my attempt on the other part of the given question.
     
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