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Equation for motion with variable acceleration

  1. Sep 20, 2011 #1
    I'm stuck on the solution of a problem. I'm looking for the general equation for the motion of a body with respect to time as follows:

    A mass slides down a frictionless slope. The slope is not linear, but has the function y = -x^2 so that as the value of x increases, the acceleration (downward) increases.

    The formula I came up with for the acceleration at any given (positive) value of x is:
    a(x) = x * g * sqrt(1/(x^2 + 1/4)) where g is acceleration due to gravity.

    I know that when acceleration is constant, velocity with respect to time is the integral of acceleration. Also, that position is the integral of velocity with respect to time.

    What I'm having trouble with is the fact that velocy here depends not only on time, but also on position, since the slope is not linear.

    Any help on this would be greatly appreciated.
    Thanks
     
  2. jcsd
  3. Sep 20, 2011 #2

    Hootenanny

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    That is a fairly nasty non-linear differential equation. It would perhaps be easier to start from scratch and use Lagrangian mechanics (energy considerations) to derive the equations of motion.
     
  4. Sep 21, 2011 #3
    Could you give me an idea of how to get started with this? I'm not familiar with Lagrangian mechanics.

    It seems to me that if I could come up with an initial equation for acceleration as a function of time instead of as a function of position, it would just be a matter of integration to get velocity and position as functions of time.
     
  5. Sep 21, 2011 #4

    chiro

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    Hey Wanderbiker and welcome to the forums.

    Do you have any constraints or formulations for the velocity and acceleration? If your velocity depends on time and position you will have a function that looks like v(x,t) which will probably depend on acceleration a will probably be in the form a(x,t) if acceleration depends on position as well. If x is not a scalar, then it will be a vector if it depends on a position in three dimensional space.

    If you can give hints about the form of a(x,t) and/or v(x,t) then we could give you some concrete suggestions.
     
  6. Sep 22, 2011 #5

    Hootenanny

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    Essentially, this boils down to considering conservation of energy (we'll try and stay away from Lagrangian formalism). The trick here is to not treat [itex]v=v(x,t)[/itex], but [itex]v = (v\circ x)(t)[/itex] and [itex] y = (y\circ x)(t)[/itex].

    So, at the top of the parabola the gravitational potential energy is [itex]V = mg y_0[/itex]. For simplicity, let us assume that the particle starts from rest as some position [itex]x = x_0 > 0[/itex]. The total energy of the particle is then [itex]E = mg y_0[/itex], where [itex]y_0 = y(x_0)[/itex]. The height of the ramp at position [itex]x[/itex] is then described by the equation [itex]y(x) = y_0 + x_0^2 - x^2[/itex].

    Since the total energy of the system must remain constant

    [tex]mg y_0 = mg y(x) + \frac{1}{2}mv^2[/tex]

    or

    [tex]v = \sqrt{2g[y_0 - y(x)]} = \sqrt{2g[x^2 - x_0^2]}[/tex]

    More conveniently,

    [tex]v^2 = 2g[x^2 - x_0^2]\;.[/tex]

    Taking the derivative with respect to time

    [tex]2v\frac{\partial v}{\partial t} = 4g x\frac{\partial x}{\partial t}[/tex]

    [tex]v\frac{\partial^2 x}{\partial t^2} = 2gxv[/tex]

    Hence, for non-trivial velocities we obtain the following initial value problem

    [tex]\frac{\partial^2 x}{\partial t^2} - 2gx(t) = 0\;,[/tex]

    [tex]x(0) = x_0 \text{ and } x'(0) = 0\;.[/tex]

    Does that help? Can you take it from here?
     
  7. Sep 22, 2011 #6
    Thanks so much Hootenanny,

    I apologize for my ignorance.

    I've been trying to teach myself differential equations, and it is VERY slow going. I learn best by applying what I am learning to real-world situations. I thought this problem would be a straight forward one to work on.

    The approach you presented here is clear to me. I will take a stab at trying to solve this differential equation.
     
  8. Sep 22, 2011 #7
    Taking the derivative with respect to time

    2 v∂/v∂t = 4gx ∂x/∂t

    v ∂2x/∂t2 = 2gxv

    Can you please show me how you make this last step to the final second order diff. equation/

    Thanks
     
  9. Sep 22, 2011 #8
    I get a solution to x'' - 2gx = 0
    or (since g=32ft/sec^2)

    x'' - 64x = 0

    of x(t) = C1e8t + C2e-8t

    Using the initial conditions, I get C1 = C2 = xo/2
    and get as a final equation:

    x(t) = xo/2(e8t + e-8t)

    I've checked this several times, and It appears to me to be correct.
    However, that means that at an initial position of 1 foot and time of 1 second, the mass would have moved 1492 feet. This, of course, is way too large a number. Even a falling mass would only have fallen 16 feet in a second.
     
  10. Sep 23, 2011 #9

    Hootenanny

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    I've just realized that there is an error in my derivation. I have used that [itex]v^2 = \dot{x}^2(t)[/itex], when in actual fact, it is [itex]v^2 = \dot{x}^2(t) + \dot{y}^2(t)[/itex]. Apologies for the mix up - it was early when I typed up the solution and I hadn't had my coffee! So, if we start from here
    We have that (by the chain rule, which should also answer your earlier question about this step).
    [tex]2v\frac{\partial v}{\partial t} = 2g[x^2 - x_0^2][/tex]

    [tex]2v\frac{\partial }{\partial t}\left(\left[\frac{\partial y}{\partial t}\right]^2 + \left[\frac{\partial x}{\partial t}\right]^2\right)^{1/2} = 4gx\frac{\partial x}{\partial t}[/tex]

    And unfortunately, this will yield an equally nasty non-linear differential equation. Apologies again for wasting your time!

    At the moment, I can't see any straightforward method other than using Lagrange formalism to solve this problem. I'm afraid I'm going on vacation this afternoon and so won't be able to help you further for a week or so, but I'm sure someone else will drop in. In the meantime, take a look at this thread: https://www.physicsforums.com/showthread.php?t=74631, which solves a similar problem.

    Apologies for my earlier slip!
     
    Last edited: Sep 23, 2011
  11. Sep 23, 2011 #10
    No worries for the slip. It was a good exercise for me. I was hoping that the solution would be fairly straightforward. Since this is getting much deeper than I am comfortable with, I am going to leave it at this point.

    Thank you much for your help.
     
  12. Sep 23, 2011 #11

    Ben Niehoff

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    Since your "inverted half-pipe" is frictionless, it should be easy to treat the x and y components separately.

    To solve this problem, you just need four facts:

    The constraint [itex]y = -x^2[/itex]

    The potential energy [itex]V = mgy[/itex]

    The relationship between force and potential energy [itex]\vec F = -\vec \nabla V[/itex]

    Newton's 2nd law [itex]F = ma[/itex].

    First, we can use the constraint to make the problem effectively 1-dimensional. Let's choose to get rid of y and keep x. Then using the constraint, the potential energy becomes

    [tex]V = -mgx^2[/tex]
    and hence the force is

    [tex]F_x = -\frac{dV}{dx} = 2mgx[/tex]
    Now using the second law gives us

    [tex]a_x = 2gx[/tex]
    or

    [tex]\frac{d^2 x}{dt^2} - 2gx = 0[/tex]
    This is just a linear ordinary differential equation with constant coefficients. The solution will be a linear combination of exponentials

    [tex]x = C_1 e^{\sqrt{2g} t} + C_2 e^{-\sqrt{2g} t}[/tex]
    which is exactly what you had written earlier.

    If you look carefully, though, you'll notice the units are inconsistent. This is because the constraint equation [itex]y = -x^2[/itex] has inconsistent units. You can't set a length equal to a length squared! What you really need is an equation like [itex]y = - Ax^2[/itex], where the constant A has units of inverse length.
     
  13. Sep 24, 2011 #12
    OK, I'm with you. So if I start with [itex]y = - Ax^2[/itex] instead of [itex]y = -x^2[/itex], I should end up with a solution of [tex]x = C_1 e^{\sqrt{2Ag} t} + C_2 e^{-\sqrt{2Ag} t}[/tex]. Right? How would I find a value for A?
     
  14. Sep 26, 2011 #13

    vela

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    This first part won't be much help to Wanderbiker, but I found the Lagrangian is given by
    [tex]\mathcal{L} = \frac{1}{2}m\dot{x}^2(1+4x^2) + mgx^2[/tex]which leads to the equation of motion
    [tex]\ddot{x}=g\frac{2x}{1+4x^2}-\frac{\dot{x}^2}{1+4x^2}[/tex]If [itex]\theta[/itex] is the angle the slope makes with the horizontal, you have [itex]\tan\theta=2x[/itex]. With a little bit of work, you can see the first term corresponds to horizontal component of the tangential acceleration. The other term turns out to represent the horizontal component of the normal acceleration.

    You can also derive this result using Newton's second law using the fact that the http://en.wikipedia.org/wiki/Curvature" [Broken] for a point on the path is given by[tex]\rho=\frac{1}{2}(1+4x^2)^{3/2}[/tex]so that the components of acceleration are
    \begin{align*}
    a_t &= g\sin\theta \\
    a_n &= \frac{v^2}{\rho} = \frac{2v^2}{(1+4x^2)^{3/2}}
    \end{align*}
    where [itex]v^2 = v_x^2 + v_y^2[/itex].

    I'm not exactly certain about where the problem in Ben's approach lies, but I'm pretty confident it is incorrect because it would lead to the relationship
    [tex]\frac{1}{2}mv_x^2 = mgx^2[/tex]whereas a straightforward application of conservation of energy to the original problem yields
    [tex]\frac{1}{2}m(v_x^2+v_y^2) = mgx^2[/tex]
    The problem, I think, is that you can't say Fx = -dV/dx when x and y aren't independent.
     
    Last edited by a moderator: May 5, 2017
  15. Oct 1, 2011 #14
    I've done a bit of numerical analysis on this problem

    I used a ramp in the shape of the curve y = x^2 that is 16 feet high and 4 feet wide (some real world numbers)

    The mass is released at the top of the ramp and slides to the bottom. The question is how long does it take for the mass to slide down the ramp (frictionless)

    Using microsoft excel I ran a series of calculations involving 1600 calculation steps

    The y axis was divided into 1600 intervals of Δy and all x values for the necessary equations were derived from the y axis values. I obtained 1.0651 seconds for the mass to slide down the ramp which is slighty longer than if it had fallen straight down. This result should be accurate to 3 decimal places - I hope. (I used a value of 32.2 for g)

    How does this result compare with the equations so far?
     
    Last edited: Oct 1, 2011
  16. Oct 2, 2011 #15
    This thread is about variable acceleration so
    variableacceleration.jpg

    The following definite integral should calculate the time

    [tex] t= \int_0^4 \frac{ \sqrt {4x^2+1} }{ \sqrt {2g(16-x^2)} } \,\, dx[/tex]
    I'm not sure if this is correct because I have not yet solved it Any ideas on how to solve that integral?
     
    Last edited: Oct 2, 2011
  17. Oct 3, 2011 #16
    One thing for sure, that integral is not easy to solve. I ran it through wolfram integrator and the solution used appell hypergeometric functions.

    The obvious thing to do is put the 2g constant outside the integration sign:

    [tex] t= \frac{1}{ \sqrt {2g} } \int_0^4 \frac{ \sqrt {4x^2+1} }{ \sqrt {(16-x^2)} } \,\, dx[/tex]
    That looks a little better .. but where to go from here??
    If it was not for that "16-" in the denominator then it would be easy to solve. It will take some trials to see if there might be a way around that. The denominator gives the tangent velocity at any y value and the numerator gives the infitesimal arc length as respects x. The tangent velocity of the mass is proportional to it's position at some vertical distance from the top of the curve at y =16 so that is why the "16-" is there.
     
  18. Oct 4, 2011 #17
    My first numerical analysis was wrong. I looked over the equations in my excel spread sheet and found that I had squared a value when it should have been a square root.
    That mistake was found when I did an approximation on the above integral - and the numbers did not jive. This gives it a bit more credibility. The approximations so far indicate the time it takes for mass to slide down the curve should be close to 1.03 seconds (with g = 32.2)

    Then another attempt at solving the integral is with x^2 replaced by y and the limits changed. Another approx. in excel should verify if that is allowable or not. Anyway the integral looks much easier to solve



    [tex] t= \frac{1}{ \sqrt {2g} } \int_0^{16} \frac{ \sqrt {4y+1} }{ \sqrt {(16-y)} } \,\, dy[/tex]
     
  19. Oct 5, 2011 #18
    No, that last integral was wrong.................................


    It should be

    [tex] t= \frac{1}{ \sqrt {2g} } \int_0^{16} \frac{ \sqrt {\frac {1}{4y}+1} }{ \sqrt {(16-y)} } \,\, dy[/tex]
    The following two integrals seem to be equal but not proven yet



    [tex] t= \frac{1}{ \sqrt {2g} } \int_0^4 \frac{ \sqrt {4x^2+1} }{ \sqrt {(16-x^2)} } \,\, dx \,\,\,\,\,\,\,\,\,= \,\,\,\,\,\,\frac{1}{ \sqrt {2g} } \int_0^{16} \frac{ \sqrt {\frac {1}{4y}+1} }{ \sqrt {(16-y)} } \,\, dy [/tex]

    using numerical approximation I ran 40 million calculation steps on the left hand integral and got a time of 1.02765 seconds. Then I ran 16 million calculation steps on the right hand integral and got a time of 1.02752 seconds. So it looks like there is a definite convergence there.

    I have come up with yet a third integral, using a different method and again I got a time of 1.027 seconds with only 16,000 calculation steps

    Still no progress on solving the integrals yet
     
    Last edited: Oct 5, 2011
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