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Thread: Movement of an object inside a bowl

  1. #1 Movement of an object inside a bowl 
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    Hey there,

    There is a homework I have been struggling with recently, and after days of fruitless reflection I turn to the pros at thescienceforum.com So, here is the quesiotn : I have to establish a set of 2 differential equations relevant to a round object moving inside a bowl. The bowl can be considered as a 2D curve, of equation y=(x1)^2. I have to find 2 differential equations such as (x1)'=x2, and (x2)'=f(...). The function f must be dependent on the previous values of x1 and x2, as well as the weight of the object, the gravitational acceleration and a constant of friction K. I really have no idea how to do this, a push in the right direction would be extremely helpful.

    Thank you!!


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  3. #2  
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    So the movement is on a 2D parabola not a 3D bowl correct? First off you know the ball is constrained to stay on the parabola.

    You know that movement is going to be a function of mass, position, velocity, and force right? You also know the conservation laws.

    Now find the force acting on the ball and come up with an equation of force acting parallel to the ball's constrained path given the instantaneous position of the ball. Then we will move on.


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  4. #3  
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    Thank you for your reply! The forces on the ball are due to the gravitational acceleration, the friction, and the slope of the trajectory (relevant to the angle of the current slope). The vector of the friction is opposite to that of the speed, times the constant K. The gravitational pull is towards the ground. As for the slope of the trajectory, I have no idea how I can put that into equation... that's sort of where I'm blocked...
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  5. #4  
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    Must you consider friction? If so ok... We can get to that later.

    You can find a formula for the slope of the ball trajectory at any point right? Then the force vector can be computed by trigonometry based on the slope at any given point right?
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  6. #5  
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    You can find a formula for the slope of the ball trajectory at any point right?
    If by that you mean the derivative of y=x^2 at a given x, then yes.

    Then the force vector can be computed by trigonometry based on the slope at any given point right?
    Yes, but there is my problem, I don't know how to put all the different forces together to find the final force vector. I assume if the system of differential equations is :
    (x1)'=x2
    (x2)'=f(),
    then (x2)' would give the acceleration of the object, which I should find as a function of all the different forces.
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  7. #6  
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    OK. So you have the slope of the bowl as the derivative at any point. If we define ∅ as the angle from horizontal. Then:
    tan∅= slope. Do you agree?
    And what is the formula for the tangential portion of gravity as a function of the angle?
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  8. #7  
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    Yes, I agree that tan∅ is the slope, it's the ratio sin∅/cos∅, which gives the rate of increase/decrease of height.

    As for gravity, as it's always directed vertically towards the ground, ∅ is always 90. So tan∅=sin∅, right?
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  9. #8  
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    Quote Originally Posted by Walker
    Yes, I agree that tan∅ is the slope, it's the ratio sin∅/cos∅, which gives the rate of increase/decrease of height.
    Ok good.

    As for gravity, as it's always directed vertically towards the ground, ∅ is always 90
    .

    Correct. So the force tangential to the direction the ball is traveling is what?
    F = f(g) where:

    g

    '➘F

    Edit: Actually it seems that you inferred it correctly but wrote it wrong.

    So tan∅=sin∅, right?
    No try again... you're on the right track though.
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  10. #9  
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    For gravity, I would say the weight of the object times the sin of ∅ : gravitational force= wsin∅.

    The horizontal component of the movement would be the weight time the cos of ∅ = wcos∅. Is that right?
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  11. #10  
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    Quote Originally Posted by Walker
    For gravity, I would say the weight of the object times the sin of ∅ : gravitational force= wsin∅.

    The horizontal component of the movement would be the weight time the cos of ∅ = wcos∅. Is that right?
    Yes. Now express the force in terms of x instead of ∅. f=gsin(arctan(??)) where ?? is the slope which you correctly identified as the derivative. Then simplify and then covert force to acceleration.

    Next you need one for velocity as a function of x, then you combine those to the general position formula. Right? Finally if you need to consider friction....
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  12. #11  
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    OK. So using the following formula : sin(arctan(x)) = x/(1+x^2), the equation for f becomes :
    f=gsin(arctan(??))
    f=g(??)/(1+(??)^2).
    There is no way so simplify this any more to my knowledge.

    As f=ma, I can say that
    ma=g(??)/(1+(??)^2), so
    a=[g(??)/(1+(??)^2)]/m.

    Is this any close to being correct?
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  13. #12  
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    Quote Originally Posted by Walker
    OK. So using the following formula : sin(arctan(x)) = x/(1+x^2), the equation for f becomes :
    f=gsin(arctan(??))
    f=g(??)/(1+(??)^2).
    There is no way so simplify this any more to my knowledge.

    As f=ma, I can say that
    ma=g(??)/(1+(??)^2), so
    a=[g(??)/(1+(??)^2)]/m.

    Is this any close to being correct?
    Yes that is correct except you dropped the square root on the denominator I think. But don't forget to plug in the derivative of (X^2) for the slope (??). Now you need a formula for position vs. time. You might notice that we are doing this using X as our reference so we using horizontal position as our reference frame. This could be done vertically also ... Not sure which is easier doubt it makes much difference but... just a thought. Carry on. Let me know if you get stuck again.
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  14. #13  
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    One thought is to express velocity first in terms of Y since energy is conserved, the velocity will always be a simple function of starting position, starting velocity and conversion of potential energy into motion energy. Finally you can again use trigonometry to then determine the position verses time function.
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  15. #14  
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    Thank you!

    So if x(t) is the position of the ball at time t, it's next position would be :
    x(t+1) = x(t) + v . (dt), where dt is the diffrence in time between x(t+1) and x(t).
    But as the velocity is the integral of acceleration, I could say:
    x(t+1) = x(t) + integral(a) . (dt)
    And as we have already found the acceleration, I can substitute it into the integral. Am I right? This does seem a little too messy...

    Edit : didn't see your second message when I wrote this...
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  16. #15  
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    Yes that is the general Idea. No it is not too messy.

    Again though one could also find velocity as a function of y and then substitute x^2 after the fact. velocity at any Y would be the same as if you simply dropped the ball from Yo. Either way would work . Ultimately you you are looking for x as a function of time correct? Or is it sufficient to have two equations but not solve it completely?
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  17. #16  
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    Thank you very much for your help! This is what I was looking for!

    For the moment this is enough, and I can work with this. If I have any firther questions I will post them here, but not today, I am very tired!

    Once again thak you very much, this has been extremely helpful and made me understand a lot of things!

    Walker
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