# Physics mechanics question
**URL:** https://boards.straightdope.com/t/physics-mechanics-question/201148
**Category:** Factual Questions
**Created:** [September 12, 2003, 6:07am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148 "2003-09-12T06:07:32Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@optimystique](https://boards.straightdope.com/u/optimystique)
#### Post date: [September 12, 2003, 6:07am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/1 "2003-09-12T06:07:32Z")
I have a ball (initially stationary) rolling down a concave curved slope (of say, equation y = x^2). Assume that the only force acting on the ball is downward gravity i.e. no air resistance or friction. How do I find the general expression for the time taken for the ball to reach the bottom of this slope? I can find the expression if the slope were not curved; the downward acceleration would then be constant and time taken can be found from s = 1/2 (mg)(sin^2 theta)( t^2) where s = vertical displacement from top of slope, m = mass of ball, f = acceleration of free fall, theta = angle of incline of slope, t = time. However, if the slope is curved, theta is not a constant and consequently the acceleration of the ball varies as well. So how can one find the general expression for the time taken for a ball to reach the bottom of a concave curved slope?
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### Author:  [@MonkeyMensch](https://boards.straightdope.com/u/MonkeyMensch)
#### Post date: [September 12, 2003, 6:26am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/2 "2003-09-12T06:26:13Z")
Just a wag: use an expression for acceleration relative to distance covered. At any given distance you know the angle and thus the acceleration. Then integrate for an expression with time as a soluble quantity.
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### Author:  [@optimystique](https://boards.straightdope.com/u/optimystique)
#### Post date: [September 12, 2003, 6:44am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/3 "2003-09-12T06:44:11Z")
i think i can derive an expression for the acceleration wrt the angle, but how do i bring in the time variable? What is the relationship between time and the angle?
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### Author:  [@biqu](https://boards.straightdope.com/u/biqu)
#### Post date: [September 12, 2003, 8:29am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/4 "2003-09-12T08:29:50Z")
The complication with this problem is that there is a constraint force that affects the acceleration of the mass. One way to get around it without explicitly solving for the constraint force is to use Lagrangian mechanics.
The system can be described in terms of one variable, x, since the height y is a function of x. The kinetic energy T is (m/2) ( x’ [sup]2[/sup] + y’ [sup]2[/sup] ), where the prime indicates differentiation with respect to time. Inserting y=x[sup]2[/sup], this expression simplifies to T = (m/2) ( x’ [sup]2[/sup] + ( 2 x x’ )[sup]2[/sup] ). The potential energy V is mgx[sup]2[/sup].
Thus the Lagrangian L = T - V is given by
L = (m/2) ( x’[sup]2[/sup] + 4 x[sup]2[/sup] x’ [sup]2[/sup] ) - mgx[sup]2[/sup].
The equation of motion is:
d/dt ( m x’ + 4 m x[sup]2[/sup] x’ ) = 4 m x x’[sup]2[/sup] - 2 mgx.
m x’’ + 4 m ( 2 x x’[sup]2[/sup] + x[sup]2[/sup] x’’ ) = 4 m x x’[sup]2[/sup] - 2mgx
m x’’ + 4 m ( x x’[sup]2[/sup] + x[sup]2[/sup] x’’ ) = -2mgx
Now that’s a second-order non-linear differential equation, which sometimes means we can’t find an exact non-trivial solution. Often we stumble upon the right solution just by guessing. The first guess we usually try is x = exp(kt).
m k[sup]2[/sup] exp(kt) + 4 m ( k[sup]2[/sup] exp(3kt) + k[sup]2[/sup] exp(3kt) ) = -2mg exp(kt)
After dividing through by m exp(kt) we get this equation.
k[sup]2[/sup] ( 1 + 8 exp( 2kt ) ) + 2 g = 0
If x = exp(k t) is going to be a solution to the differential equation, then some constant k has to satisfy this characteristic equation for all t. Now that’s not possible, since the exponential depends on both k and t, and to make it independent of t would require setting k=0, which violates the equality we’re trying to solve. So an exponential function is not going to solve our differential equation. But then neither will a sinusoidal function, because we would have been able to obtain those if a constant imaginary k could be found to satisfy the characteristic equation. The solution is not a damped oscillation either, since those can be obtained if a constant complex k can be found that satisfies the characteristic equation.
The last technique I would try before solving numerically would be to try finding a solution expressed as a power series. I won’t go through the details, but if you’ve followed me up to this point, you can probably figure it out on your own.
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### Author:  [@zut](https://boards.straightdope.com/u/zut)
#### Post date: [September 12, 2003, 11:17am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/5 "2003-09-12T11:17:44Z")
That looks right for a falling particle that follows the y = x[sup]2[/sup] constraint curve. I suspect that answers the spirit of the OP’s question. However, I wanna point out that if you’re actually looking at a _rolling ball_, the rotational inertia needs also to be accounted for, so the kinetic energy would have an additional (1/2)I[symbol]w[/symbol][sup]2[/sup] term. The final KE would be T = [(m + I/r[sup]2[/sup])/2] ( x’ [sup]2[/sup] + ( 2 x x’ )[sup]2[/sup] ).
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 12, 2003, 11:45am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/6 "2003-09-12T11:45:21Z")
> [@](#):
>
> \*Originally posted by amore ac studio \*
> m x’’ + 4 m ( x x’[sup]2[/sup] + x[sup]2[/sup] x’’ ) = -2mgx
>
> Now that’s a second-order non-linear differential equation, which sometimes means we can’t find an exact non-trivial solution. Often we stumble upon the right solution just by guessing. The first guess we usually try is x = exp(kt)…
>
> The last technique I would try before solving numerically would be to try finding a solution expressed as a power series. I won’t go through the details, but if you’ve followed me up to this point, you can probably figure it out on your own.
A power series for a non-linear equation? I don’t think I’d try that. While it’s often rather silly to bang your head against the wall trying to solve a differential equation, it is actually easy to make _some_ progress on this one. First get rid of that m:[ul]x’’ + 4 (x x’[sup]2[/sup] + x[sup]2[/sup] x’’) + 2gx = 0[/ul]This can be simplified by replacing x’ with a function on x, so let x’ = v(x), and then x’’ = v dv/dx. This simplifies to:ul v dv/dx + 2x (g + 2v[sup]2[/sup])[/ul]This is a first-order differential equation, and a separable one at that:[ul]v dv / (g + 2v[sup]2[/sup]) = -2x dx / (1 + 4x[sup]2[/sup])[/ul]Integrating and simplifying gives:[ul]2v[sup]2[/sup] + g = C / (1 + 4x[sup]2[/sup])[/ul]We can leave C as it is, or use the boundary condition that at the origin (x = 0), v = dx/dt = sqrt(2gy[sub]0[/sub]), by conservation of energy. Simplifying again gives:[ul]v[sup]2[/sup] = 2g (x[sub]0[/sub][sup]2[/sup] - x[sup]2[/sup]) / (1 + 4x[sup]2[/sup])[/ul]Since v = dx/dt, this is another separable first-order equation. However, I’m afraid that here the fun ends. I get the integral of:[ul]((1 + 4x[sup]2[/sup])/(x[sub]0[/sub][sup]2[/sup] - x[sup]2[/sup]))[sup]1/2[/sup]dx[/ul]According to [The Integrator](http://www.integrals.com/), this is solved using Elliptic integrals. So, I doubt I’d be able to find its inverse function. It looks pretty ugly too.
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### Author:  [@biqu](https://boards.straightdope.com/u/biqu)
#### Post date: [September 12, 2003, 12:21pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/7 "2003-09-12T12:21:26Z")
> [@](#):
>
> _Originally posted by Achernar_
> A power series for a non-linear equation? I don’t think I’d try that.
I would agree that it’s probably not the most efficient way of solving the problem. On the other hand, I had a combinatorics professor who seemed to passionately enjoy formal manipulation of power series, so perhaps some of that enthusiasm rubbed off on me.
> [@](#):
>
> According to The Integrator, this is solved using Elliptic integrals. So, I doubt I’d be able to find its inverse function.
See section 6.19.5 and subsequent sections of the CRC Standard Mathematical Tables, 30th edition. There the inverse functions of elliptic integrals are defined, their properties are listed, and their series expansions are provided.
> [@](#):
>
> _Originally posted by zut_
> However, I wanna point out that if you’re actually looking at a _rolling_ ball, the rotational inertia needs also to be accounted for.
Good catch, **zut**. I haven’t yet done the calculation using the KE you provided. Maybe taking into account the rotational KE will yield a differential equation that is solvable with elementary functions.
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 12, 2003, 12:35pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/8 "2003-09-12T12:35:06Z")
> [@](#):
>
> \*Originally posted by amore ac studio \*
> See section 6.19.5 and subsequent sections of the CRC Standard Mathematical Tables, 30th edition. There the inverse functions of elliptic integrals are defined, their properties are listed, and their series expansions are provided.
I’m sure it would be easy if it was _just_ an Elliptical function, but [here’s what it actually is](http://www.integrals.com/graphics.cgi?format=StandardForm&FontSize=Medium&expr=Sqrt%5B%281%2B4x%5E2%29/%28x0%5E2-x%5E2%29%5D&x=0&y=0). You think the inverse function of _that_ is in CRC?
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### Author:  [@pjd](https://boards.straightdope.com/u/pjd)
#### Post date: [September 12, 2003, 1:13pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/9 "2003-09-12T13:13:45Z")
> [@](#):
>
> So how can one find the general expression for the time taken for a ball to reach the bottom of a concave curved slope?
=t
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 12, 2003, 1:25pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/10 "2003-09-12T13:25:06Z")
> [@](#):
>
> \*Originally posted by amore ac studio \*
> Good catch, **zut**. I haven’t yet done the calculation using the KE you provided. Maybe taking into account the rotational KE will yield a differential equation that is solvable with elementary functions.
No such luck. The only difference that gets made in the equation of motion is equivalent to replacing the constant g with the constant g / (1 + I / mr[sup]2[/sup]). For a ball, this is equal to 5g/7. It’s also equivalent to replacing the independent variable t with sqrt(7/5) t. So the only effect it’ll have on the final solution is stretching time by a factor of 1.183.
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### Author:  [@biqu](https://boards.straightdope.com/u/biqu)
#### Post date: [September 12, 2003, 6:16pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/11 "2003-09-12T18:16:07Z")
> [@](#):
>
> _Originally posted by Achernar_
> Since v = dx/dt, this is another separable first-order equation. However, I’m afraid that here the fun ends. I get the integral of:
>
> ((1 + 4x[sup]2[/sup])/(x[sub]0[/sub][sup]2[/sup] - x[sup]2[/sup]))[sup]1/2[/sup] dx
>
> According to The Integrator, this is solved using Elliptic integrals. So, I doubt I’d be able to find its inverse function. It looks pretty ugly too.
I don’t think you need to find the explicit expression for the inverse function in order to solve for the time it takes for the particle to reach the origin. If I’m reading your notation correctly, the differential form quoted above is equal to sqrt(2 g) dt, when you separate the equation dx/dt = v(x).
Then you can integrate the side that depends on x from x = x[sub]0[/sub] to x = 0, and the side that depends on t from t = 0 to t = t. The definite integral with respect to x, divided by sqrt(2 g), gives t as a function of x[sub]0[/sub], which is what the OP wanted, even if it’s not expressed in terms of elementary functions.
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 13, 2003, 12:55am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/12 "2003-09-13T00:55:49Z")
Okay, you’re right. I was trying to find x(t) explicitly, but that’s not the way to do it. Additionally, I did the integral by hand and got a different answer than The Integrator. Here’s the solution to how long it takes the ball to reach the bottom of the parabola:[ul]T = ((1+4y[sub]0[/sub])/2g)[sup]1/2[/sup] E[(4y[sub]0[/sub]/(1+4y[sub]0[/sub]))[sup]1/2[/sup]][/ul]Here E[k] is the [Complete Elliptic Integral of the Second Kind](http://mathworld.wolfram.com/CompleteEllipticIntegraloftheSecondKind.html). Notice that for y[sub]0[/sub] very large, the argument in the brackets approaches 1, and E[1] = 1. Thus T approaches (2y[sub]0[/sub] / g)[sup]1/2[/sup], which is the time it would take just to be dropped straight down. This seems reasonable to me.
Consider the other limit, though, as y[sub]0[/sub] goes to 0. It’s actually not equal to 0 as I expected! As y[sub]0[/sub] approaches 0, T approaches pi/(8g)[sup]1/2[/sup]. It’s certainly counterintuitive!
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### Author:  [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)
#### Post date: [September 13, 2003, 1:14am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/13 "2003-09-13T01:14:15Z")
Surely you must know the equation of the parabola before you can even start to talk about the period etc.
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 13, 2003, 1:58am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/14 "2003-09-13T01:58:04Z")
Actually, oddly enough, I just went back and derived the first-order differential equation using only conservation of energy. I guess I’m not used to using Lagrangian mechanics the way it’s supposed to be used. This makes my previous analysis a lot simpler.
**MC Master of Ceremonies** , you’re right for the general case. But for illustration, we were assuming the function given in the OP, y = x[sup]2[/sup]. However, I went through and did it assuming a general function of y = y(x), and you can still get pretty far:[ul]T = (2g)[sup]-1/2[/sup] Integral((1 + (dy/dx)[sup]2[/sup]) / (y[sub]0[/sub] - y))[sup]1/2[/sup]dx[/ul]The integral runs from 0 to x[sub]0[/sub]. Whether that integral is easily solvable depends on y(x), but you could even do it numerically if you want.
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### Author:  [@g8rguy](https://boards.straightdope.com/u/g8rguy)
#### Post date: [September 13, 2003, 2:01am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/15 "2003-09-13T02:01:20Z")
> [@](#):
>
> \*Originally posted by Achernar \*
> \*\*Consider the other limit, though, as y[sub]0[/sub] goes to 0. It’s actually not equal to 0 as I expected! As y[sub]0[/sub] approaches 0, T approaches pi/(8g)[sup]1/2[/sup]. It’s certainly counterintuitive! \*\*
Hmm… for an answer that counterintuitive, I would expect that it’s probably just wrong, somehow, wouldn’t you? In fact, it looks off hand to be not even dimensionally correct; that 1 in the 1 + 4y[sub]0[/sub] worries me a bit.
That’s not really a cast-iron argument, of course, since y = x[sup]2[/sup] is also not dimensionally correct if we take y to be a height and x to be a distance, but it is suggestive, eh?
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 13, 2003, 2:36am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/16 "2003-09-13T02:36:51Z")
No, the dimensionality being off doesn’t mean anything. I noticed that very early on, but you are right that we abandoned dimensionality in **amore ac studio** ’s second paragraph with “Inserting y = x[sup]2[/sup]”. No equation after that point is dimensionally correct.
As for whether the result is wrong, yeah it could be. But I’ve been trying to think of a way to disprove it since I first noticed it, and I haven’t come up with anything yet. Any ideas?
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### Author:  [@g8rguy](https://boards.straightdope.com/u/g8rguy)
#### Post date: [September 13, 2003, 3:23am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/17 "2003-09-13T03:23:37Z")
Since:[list=a][li]x[sub]0[/sub] is positive,[_]x is non-negative, and [_]x \<= x[sub]0[/sub],[/list] that complicated ugly mess you found simplifies quite a bit, doesn’t it? I’m not sure that helps any, though.[/li]
Just to help get the dimensionality correct, I’ll write y = x[sup]2[/sup]/L and carry L as a parameter. The first order equation becomes… hmm
1/L dx/dt = sqrt(2g/L) sqrt((x[sub]0[/sub][sup]2[/sup] - x[sup]2[/sup])/(L[sup]2[/sup] + 4x[sup]2[/sup]))
if I haven’t any errors. That looks right, anyway.
And at the end of the day, I am getting the same result as you, ignoring the existence of L for the moment. Not that this shocks me or anything, but I just don’t understand it, which makes me wonder if we’ve managed to omit a term somehow.
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### Author:  [@g8rguy](https://boards.straightdope.com/u/g8rguy)
#### Post date: [September 13, 2003, 4:02am UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/18 "2003-09-13T04:02:50Z")
Hmm, I actually get, with Maple, sqrt(L/(2g)) E(2 _i_ x[sub]0[/sub]/L). I don’t know if that winds up being the same thing or not; I’m not up on my elliptic integrals. I do get the same behaviors for small and large x[sub]0[/sub], though, as of course the basic issue remains. Odd. It might actually be worth solving for y(t) in the more general case, to see what’s going on.
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### Author:  [@Achernar](https://boards.straightdope.com/u/Achernar)
#### Post date: [September 13, 2003, 12:32pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/19 "2003-09-13T12:32:19Z")
:smack:
Of course it has to be right! At very small displacements, it’s the same as a classical pendulum. And we all know the period of a pendulum does not approach 0 as the displacement approaches 0.
And I get, after about five lines of algebra and trig, that:[ul]E[ik] = (1+k[sup]2[/sup])[sup]1/2[/sup] E[k/(1+k[sup]2[/sup])[sup]1/2[/sup]][/ul]
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### Author:  [@g8rguy](https://boards.straightdope.com/u/g8rguy)
#### Post date: [September 13, 2003, 8:34pm UTC](https://boards.straightdope.com/t/physics-mechanics-question/201148/20 "2003-09-13T20:34:24Z")
:smack: is right! Thanks for clearing that up; it was driving me nuts.