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Object sliding down smooth quarter circle as a function of time

  1. Apr 17, 2010 #1
    I can't find anything on the internet about this...
    How is it done?


    I've got a smooth curve given by these parametric equations:
    y = 5cos( theta )+5; x = 5sin( theta )
    taking g = 9.81

    how can I model the position of the ball as a function of time?
    Or how can I model it so that i can write it in multiples of deltatime?
    (e.g. new_pos = current_pos + velocity*dt)
    how can I estimate, or determine the time it takes to slide down it?
     
  2. jcsd
  3. Apr 17, 2010 #2
    Mate...

    1) Show your work.

    2)Post it in the homework's section...
     
  4. Apr 17, 2010 #3
    I don't know where to start since the acceleration is constantly changing.

    I did, but this is an extension to the work.
     
  5. Apr 17, 2010 #4
    Your "work".. That is your effort in doing the sum...And not the work done by/on the ball...lol... ;)

    Anyways.. Use vectors... Diff y and x wrt time. And remember do not forget the "d(theta)/dt".. You cant write it as some random omega cause that aint make any sense.

    Instead think like this. You have got a curve which is nothing but a circle with centre at y=5. So now take the normal forces and the g effects. Resolve and now find the omega. And plug that omega as d(theta)/dt...
     
  6. Apr 17, 2010 #5
    I know, I attempted it and failed hard :P

    Surely there must be an easy way?
     
  7. Apr 17, 2010 #6
    Look at it this way. The trajectory is the same as a pendulum with the pivot at the center of the circle with radius a="5" (say 0.5 m), so the time for the object to slide from top to bottom is exactly a quarter period of a pendulum swinging from + 90 degrees to - 90 degrees. This is exactly the problem of a pendulum of arbitrary amplitude.

    Bob S
     
  8. Apr 18, 2010 #7
    But doesn't that forumula 2pi*sqrt(L/g) only work for small amplitudes

    this is a fairly large one at pi/2 radians :P
     
  9. Apr 18, 2010 #8
    I've got this now, but how do I get them as functions of time?
    http://mathbin.net/45802 [Broken]
     
    Last edited by a moderator: May 4, 2017
  10. Apr 18, 2010 #9
    If am getting your point, your task is to find the equation of the motion of a particle on a smooth curve.
    Just try to use the Lagrangian.
    But you should first show us how far you have tried!
    Good luck
     
  11. Apr 18, 2010 #10
    If am getting your point, your task is to find the equation of the motion of a particle on a smooth curve.
    Just try to use the Lagrangian.
    But you should first show us how far you have tried!
    Good luck
     
  12. Apr 18, 2010 #11
    I did, it's here
    http://mathbin.net/45802 [Broken]
     
    Last edited by a moderator: May 4, 2017
  13. Apr 18, 2010 #12
    I worked it on separate paper because i could not write it properly using this Latex codes. Please refer to attached its pdf file View attachment prob.pdf .

    Just let me know if you still have any question.

    Here is the latex version

    Here is my attempt:
    As the mass is slide on a frictionless curve, i can use Lagrange's equation for a conservative force. That is,
    [tex]\frac{d}{dq}[/tex]([tex]\frac{dL}{d\dot{q}}[/tex])-[tex]\frac{dL}{dq}[/tex]=0
    where L is the lagrangian and L=Kinetice energy - potential energy = T-V
    q is a generalised coordinate, the dot over q means the time derivative ([tex]\dot{q}[/tex]=[tex]\frac{dq}{dt}[/tex]

    For that curve,
    [tex]\dot{x}[/tex]=5[tex]\dot{\theta}[/tex]cos[tex]\theta[/tex]
    [tex]\dot{y}[/tex]=-5[tex]\dot{\theta}[/tex]sin[tex]\theta[/tex]
    Hence, kinetic energy T=[tex]\frac{25}{2}[/tex]m[tex]^{\dot{\theta}}[/tex]
    and the potential V= mgy=mg(5cos[tex]\theta[/tex]+5)

    L=[tex]\frac{25}{2}[/tex]m[tex]^{\dot{\theta}}[/tex]+mg(5cos[tex]\theta[/tex]+5)
    [tex]\frac{dL}{d\dot{\theta}}[/tex]=25m[tex]\dot{\theta}[/tex]
    [tex]\frac{d}{dq}[/tex]([tex]\frac{dL}{d\dot{q}}[/tex])=25m[tex]\ddot{\theta}[/tex]
    [tex]\frac{dL}{d\theta}[/tex]=5mgsin[tex]\theta[/tex]
    The lagrangian equation (which is the equation of motion) becomes
    25m[tex]\dot{\theta}[/tex]-5mgsin[tex]\theta[/tex]=0
    5[tex]\dot{\theta}[/tex]-gsin[tex]\theta[/tex]=0
     
    Last edited: Apr 18, 2010
  14. Apr 18, 2010 #13
    how would I get the lagrangian equation as a function of time though?

    n/m
    I used Runge-Kutta method

    thanks :D
     
    Last edited: Apr 18, 2010
  15. Apr 19, 2010 #14
    Alright, you can use any method to solve that ODE.
    Remember your initial conditions that at point (-5,5), its speed is 0.
    Actually if you are allowed to use matlab, it would be very easy!
    Good luck then!
     
  16. Apr 19, 2010 #15
    just out of interest, apart from euler and the improved euler methods?
     
  17. Apr 19, 2010 #16
    Frankly i havent studied lagrangians yet...

    Lets make it simple... The mass feels the Normal force the mg and the centripetal force...

    Make a free body diagram and then write down all the forces in the x and the y direction and then solve the diff equations.. I guess this shall work...
     
  18. Apr 19, 2010 #17
    If you want to use the decomposition methods. I think that the given equations of the system are like constraints of that motion. Then they will help to find the resultant force (or acceleration). If you can decompose the whole motion equate the found equations to those of resultant force, i think you can get the same results.
    The Lagrange method is the simplest actually
     
  19. Apr 19, 2010 #18
    The solution is a special case of the solution for a pendulum of arbitrary amplitude, as I pointed out in an earlier post. Neither Lagrangians nor Hamiltonians are needed. First, because the system is frictionless, the kinetic energy is simply related to the initial elevation, so the equation for (dθ/dt)2 vs cos(θ) can be written down. Then take the square root, and with several substitutions on the right had side, an exact solution can be found. You might want to look at the solution for a pendulum of arbitrary amplitude:

    http://en.wikipedia.org/wiki/Pendulum_(mathematics)#Arbitrary-amplitude_period

    You will find that the large-amplitude pendulum period (for a quarter circle) deviates less than ~20% from the small-angle pendulum period (or transit time).

    Bob S
     
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