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A ball rolling up a parabolic ramp

  1. May 12, 2016 #1
    1. Assume there is gravity and no external force acting on the system. A ball has an initial velocity of 5 m/s and climbs up a parabolic ramp, which is defined by y=(x^2)/3. If the ball rolls exactly along the path of ramp and energy of the ball is conserved, starting from (0,0), calculate the time taken for the ball to climb to a height of y=1.

    2. Ek+Ep=Einit, Vy=||v||sin(arctan(dy/dx))

    3. I made a differential equations using the formula above but the differential I got is only numerical integratable (results given by wolfram alpha). I'm looking for an easier way to solve this problem. So, guys, please help! Thanks!
     
    Last edited: May 12, 2016
  2. jcsd
  3. May 12, 2016 #2

    Simon Bridge

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    What did you need the trig for?
    Did you start with ##\frac{1}{2}m(\dot x^2+\dot y^2) + mgy = \frac{1}{2}mu^2 : u=5,\; y=\frac{1}{3}x^2## ?
     
  4. May 12, 2016 #3
    That seems pretty helpful! I haven't learnt to use and x and y things because I'm a highschool student, but I've done stage 1 college maths so I think I can cope with that. I'll give it a try! Thank you!
     
  5. May 12, 2016 #4

    Simon Bridge

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    Is this a problem set for high school?
     
  6. May 12, 2016 #5
    Well. It's not, I made this question up. It'd be nice if you can solve it using highschool knowledge only though.
     
  7. May 12, 2016 #6
    I'm stuck again...
     
  8. May 12, 2016 #7
    I might got it.. trying now
     
  9. May 12, 2016 #8
    I'm trying now, but still, can you show me how you would do it please?

    Guess what. I used the formula you gave me and that ends up the same differential equation I got last time using trig, which I need to use a computer to calculate it. Can you help me?
     
  10. May 12, 2016 #9

    Simon Bridge

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    Please answer the question?
     
  11. May 12, 2016 #10
    I answered above. Basically this is some extracurricular work.
     
  12. May 12, 2016 #11

    Simon Bridge

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    Sorry - misread.
     
  13. May 12, 2016 #12
    It's ok. So, right now I used your formula, and it ended up exactly the same as last time I did it using trig.
    I got dt/dy=0.5sqrt((3+4y)/(25y-2gy^2)) dy, which even wolfram alpha couldn't give an exact solution.
    I believe there is a better way to do this. Can you show me please?
     
  14. May 12, 2016 #13

    Simon Bridge

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    ... this is a specific case of an object sliding inside a parabolic bowl.
    If you google for that, you'll get a bunch of examples. i.e. See attachment below.

    Note: ##\dot x = v(x)## is where you get stuck.
    You can do $$T = \int_0^{\sqrt 3} \frac{dx}{v}$$ ... i.e. what you were doing but as a definite integral.
    The indefinite integral involves elliptical functions.

    It is not unusual for quite simple-seeming setups to be very difficult to solve - or even have no analytic solutions at all.
    ie. Try working out the exact equation of motion for a simple pendulum - or a puck sliding in a spherical bowl.
     

    Attached Files:

  15. May 12, 2016 #14
    Thank you! No problem. I need to sleep as well. I'll have a look tomorrow.
     
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