Optimal Paths: Comparing the Motion of Two Balls on Different Trajectories

[haruspex]

The standard brachistochrone assumes $V_0$ = 0. If that is not the case, you probably get a different cycloid.

Also, if we demand that B starts out horizontally, like shown on the picture, keeping contact during transition to downslope requires low $V_0$.

No, it's just that the value of Vo sets a limit on how quickly the slope can curve downwards.

 

[voko]

If we switch to a reference frame where $V_0 = 0$, we have a classical brachistochrone setup in that frame. The question is, is the duration of motion along the brachistochrone from zero velocity to zero velocity finite?

If it is infinite, it will will be infinite for any other curve.

 

[olivermsun]

But what do you think $\ddot{y} = -g$ means here? Don't forget you have rolling contact, so a y acceleration implies an angular acceleration. This is not going to be like free fall.

Ah good point. I was trying to imagine what were the implications of the maximal "falling" path with rotation. So the "contact" condition ensures that $\ddot{y} = -g$ has the greatest magnitude for $\ddot{x} = 0$, hence the part in the parenthesis is identically zero and we get the expected behavior.

I suppose this can be seen more easily if you change the frame so that $v_0 = 0$.

Cool.

 

[haruspex]

If we switch to a reference frame where $V_0 = 0$, we have a classical brachistochrone setup in that frame. The question is, is the duration of motion along the brachistochrone from zero velocity to zero velocity finite?

If it is infinite, it will will be infinite for any other curve.

Switching reference frame cannot suddenly make it adhere to a cycloid! If a ball sits on a level surface of a block at the top of a ramped edge, and I snatch the block away, the ball will not stay in contact with the ramp as it falls.

 

[haruspex]

Ball A, of course. Ball B will loose as much speed going up as it gained going down, and then it is left to travel over greater distance. But more distance also means more friction, so if you push them with sufficiently small speed ball B could end up in the ditch and never finish the race, or it could manage to get out but still stop short of reaching the goal.

Speed bump with the same shape upwards would produce the same result, it doesn't slow down much if the ball manages to pass on the other side, but it makes for a greater distance the ball has to travel over.

No, it's ball B. As noted by Delta2, while going through the dip ball B will have an increased horizontal component of velocity, but only has the same horizontal distance to cover.
You should ignore friction for this problem.
Going over a speed bump instead would make it take longer, since its horizontal speed would be reduced.

 

[voko]

Switching reference frame cannot suddenly make it adhere to a cycloid!

I am not sure what this "it" is. My reasoning is that if the unrestricted brachistochrone time is infinite, certainly any other time will also be infinite. Which, if true, settles the issue immediately in my opinion.

If it is finite, then one would need to check whether an inverted cycloid is a possible gravity-only track for a ball to draw further conclusions.

 

[pleco]

No, it's ball B. As noted by Delta2, while going through the dip ball B will have an increased horizontal component of velocity, but only has the same horizontal distance to cover.
You should ignore friction for this problem.
Going over a speed bump instead would make it take longer, since its horizontal speed would be reduced.

I'm not sure if surface decline and incline is valid example without friction. In any case the problem seems to be more complex than that or what I thought. I see now all three possible outcomes, depending on the angle and horizontal distance between the decline and incline. Here is an example where they reach the finish line in the same time:t= 0
--- start
A: x= 0, v= 1
B: x= 0, y= 0, v= 1 t= 10 seconds
-- B begins acceleration
A: x= 10, v= 1
B: x= 10, y=0, v= 1, a= 10t= 15 seconds
-- B ends acceleration
-- B begins deceleration
A: x= 15, v= 1
B: x= 15, y= 127, v= 51, a= -10t= 20 seconds
-- B ends deceleration
A: x= 20, v= 1
B: x= 20, y=0, v= 1If during acceleration phase ball B travels in horizontal direction less distance than ball A, and if deceleration phase begins right after acceleration, then ball A would win. Which I admit looks like an extreme case example, so for a ditch as illustrated in the OP you would be right and ball B would win after all.

 

[haruspex]

I am not sure what this "it" is. My reasoning is that if the unrestricted brachistochrone time is infinite, certainly any other time will also be infinite. Which, if true, settles the issue immediately in my opinion.

I don't see how the time is going to be unbounded. Take the width of the dip and the initial speed as givens, W, V, both > 0. Under the constraints that:
- the particle/ball stays in contact without the normal force going negative, and
- the dip is entirely below the horizontal
we have already seen that the time in the dip is always less than W/V.
One of my questions was the minimum time to cross the dip.
My other question I need to rephrase: of those dips for which the time is only infinitesimally shorter than W/V, what's the deepest possible?

If it is finite, then one would need to check whether an inverted cycloid is a possible gravity-only track for a ball to draw further conclusions.

As I keep pointing out, it is not possible for the path to consist of an inverted cycloid. The gradient near the ends of the dip must approach zero. A cycloid cannot do that. It is of course possible that part of the curve is a cycloid, but you have to negotiate getting into a descent first.

 

[haruspex]

I'm not sure if surface decline and incline is valid example without friction.

No one said it was real-world. It is very common to presume no friction for the purpose of such exercises, and there's no reason it's invalid to do so in the present problem.

In any case the problem seems to be more complex than that or what I thought. I see now all three possible outcomes, depending on the angle and horizontal distance between the decline and incline. Here is an example where they reach the finish line in the same time:


t= 0
--- start
A: x= 0, v= 1
B: x= 0, y= 0, v= 1


t= 10 seconds
-- B begins acceleration
A: x= 10, v= 1
B: x= 10, y=0, v= 1, a= 10


t= 15 seconds
-- B ends acceleration
-- B begins deceleration
A: x= 15, v= 1
B: x= 15, y= 127, v= 51, a= -10


t= 20 seconds
-- B ends deceleration
A: x= 20, v= 1
B: x= 20, y=0, v= 1


If during acceleration phase ball B travels in horizontal direction less distance than ball A, and if deceleration phase begins right after acceleration, then ball A would win. Which I admit looks like an extreme case example, so for a ditch as illustrated in the OP you would be right and ball B would win after all.

You cannot assume that the numbers you have made up are actual solutions within the constraints of the problem.
There is a clear and convincing reason (see posts by Delta2 and myself) why B is always quicker. Please try to understand it, or find a flaw in it.

 

[pleco]

You cannot assume that the numbers you have made up are actual solutions within the constraints of the problem.

But I didn't make them up, I calculated them. Check it out.

There is a clear and convincing reason (see posts by Delta2 and myself) why B is always quicker. Please try to understand it, or find a flaw in it.

I think this is the argument you are referring to: "The total horizontal displacement is the same for both cases but because ball B has a time interval where its horizontal velocity is bigger than v0, ball B wins the race."

I generally agree, but according to my calculations there can also exist such decline/incline angle where ball B during its acceleration and deceleration will travel less horizontal distance than ball A, and if deceleration phase begins right after acceleration ball A will win. The ditch in this case is deep and 'v' shaped.

 

[olivermsun]

... according to my calculations there can also exist such decline/incline angle where ball B during its acceleration and deceleration will travel less horizontal distance than ball A, and if deceleration phase begins right after acceleration ball A will win. The ditch in this case is deep and 'v' shaped.

Is it exactly 'v' shaped? If so, such a path won't satisfy the condition that the ball doesn't leave the surface.

 

[haruspex]

But I didn't make them up, I calculated them. Check it out.

Sure, but as olivermsun points out, your model does not fit the constraints of the problem. The ball has to stay in contact with the surface under its own weight. The normal force from the surface can fall to zero, but it must not go negative. In particular, this means that the initial part of the curve must not descend more steeply than the ball would under gravity in a ballistic trajectory.

I think this is the argument you are referring to: "The total horizontal displacement is the same for both cases but because ball B has a time interval where its horizontal velocity is bigger than v0, ball B wins the race."

No, there's more to it than that.
During descent, the normal force can only act in the forward and up directions. The horizontal component of the velocity therefore can increase, but not decrease. So through the entire descent the horizontal component of the velocity must exceed Vo.
During ascent, the normal force can only act in the backward and up directions. The horizontal component of the velocity therefore can decrease, but not increase. Yet, once back at the original level, by conservation of work, it must again be traveling with the original horizontal speed. This means that through the entire ascent the horizontal component of the velocity must exceed Vo.
Thus, the entire time in the dip the horizontal component of the velocity exceeds Vo. Since the horizontal distances are the same, the time taken must be less.

 

[pleco]

Sure, but as olivermsun points out, your model does not fit the constraints of the problem. The ball has to stay in contact with the surface under its own weight.

That looks to me to be a matter of initial velocity. With slow enough horizontal speed, or high enough vertical gravity acceleration, a ball should stay in contact with the surface as long as decline/incline is less steep than completely vertical.

During descent, the normal force can only act in the forward and up directions. The horizontal component of the velocity therefore can increase, but not decrease. So through the entire descent the horizontal component of the velocity must exceed Vo.

I see my mistake. I bent initial velocity to point in the direction of the decline instead of to keep it horizontal.

 

[haruspex]

With slow enough horizontal speed, or high enough vertical gravity acceleration, a ball should stay in contact with the surface as long as decline/incline is less steep than completely vertical.

No, you have to consider the second derivative too. Even if it only slopes down at 5 degrees, if that's a sudden change from horizontal then the ball will leave the surface, no matter how slow the ball.

 

[voko]

Considering the case $V_0 = 0$.

Assuming potential energy is zero at the initial (and final) point, and $y$ growing downward, conservation of energy yields $$ v = \sqrt {2gy} .$$ The horizontal component of velocity is $$v_x = v \cos n,$$ where $n$ is the angle of the velocity with the horizontal, so $$ \tan n = y'(x), $$ giving $$v_x = {v \over \sqrt {1 + (y')^2}} $$ Assuming further $x = 0$ at the initial point and $x = a$ at the final point, the total time of travel is $$ T= \int\limits_0^a {dx \over v_x} = {1 \over \sqrt {2g}} \int\limits_0^a \sqrt {1 + (y')^2 \over y} dx $$ The integrand does not depend on $x$, so the Beltrami identity can be employed: $$ \sqrt {1 + (y')^2 \over y} - \frac {y'} {\sqrt y} \frac {y'} {\sqrt {1 + (y')^2} } = {1 \over \sqrt y \sqrt {1 + (y')^2} } = C ,$$ so $$ y' = \sqrt {\frac D y - 1}, $$ which is the differential equation of the inverted cycloid generated by a circle of radius $D$ (c.f. http://en.wikipedia.org/wiki/Brachistochrone_problem). Thus, $$ T = \sqrt {D \over 2 g} \int\limits_0^a {dx \over y} .$$ A canonical parametrization of the cycloid is $$ x = \frac D 2 (u - \sin u) \\ y = \frac D 2 (1 - \cos u) $$ (c.f. http://en.wikipedia.org/wiki/Cycloid#Area), where $u$ is the angle of rotation of the generating circle. In this problem one full rotation $u = 2 \pi$ must correspond to horizontal displacement $a$, so $$ D = {a \over \pi} ,$$ and the total time integral becomes $$ T = \sqrt {a \over 2\pi g} \int \limits_0^{2 \pi} {1 - \cos u \over 1 - \cos u } du = \sqrt {2\pi a \over g} .$$ Thus the minimal time of motion for a particle starting from rest and reaching the same height at a given distance, where minimization is over the set of arbitrary smooth curves, is finite and is realized when the curve is the inverted cycloid specified above.

A question remains, is such a motion realizable when the particle slides freely "on top" of the curve?

$$ v = \sqrt {\dot x^2 + \dot y^2} = \sqrt {1 + (y')^2 } \dot x = \sqrt {D \over y} \dot x= \sqrt {2 g y }, $$ hence $$ \dot x = \sqrt {2g \over D} y, $$ and, using the parametric equations above, $$ (1 - \cos u) \dot u = \sqrt {2g \over D} (1 - \cos u), $$ giving $$ u(t) = \sqrt {2 \pi g \over a} t. $$

Acceleration vector: $$ \ddot x = g \sin \sqrt {2 \pi g \over a} t \\ \ddot y = g \cos \sqrt {2 \pi g \over a} t. $$ The force gravity is $(0, mg)$, hence the normal force is $$ N_x = mg \sin \sqrt {2 \pi g \over a} t \\ N_y = mg \cos \sqrt {2 \pi g \over a} t - mg. $$ The vertical component is everywhere non-positive, i.e, directed upward, so a cycloidal track does not "pull" the particle downward; the horizontal component is positive during the descent, and negative during the ascent, again as is to be expect from a track that can be "freely slidden upon" by a particle, which I think settles the issue completely for the case $V_0 = 0$.

Turning to $V_0 \ne 0$ now. As I said earlier, by introducing a reference frame moving uniformly at $V_0$, the problem is reduced to the one just dealt with above. To which haruspex (and possibly others) objected, reasoning that an inverted cycloid cannot possibly be a path "freely slidden upon" by a particle whose initial horizontal velocity is not zero, because such a path is vertical initially. That argument is flawed, however, because the inverted cycloid is the minimal time path in the initially comoving frame; in the lab frame where $V_0 \ne 0$, the minimal time path is the trajectory of the particle following a cycloidal path in the initially comoving frame; that path is not a cycloid per se. Moreover, the width of the dip in the lab frame is not distance $a$ in the initially comoving frame.

Let the width of the dip be $A$. During time $T$, the initially comoving frame will cover $V_0 T$; in the initially comoving frame, the particle will cover $a$. So, $$ A = V_0 T + a = V_0 T + {g \over 2\pi} T^2 , $$ yielding $$ T = {\pi \over g } \left( -V_0 + \sqrt {V_0^2 + {2 A g \over \pi}} \right) .$$

I hope there are no mistakes in the above, but any corrections are welcome.

 

[voko]

It can be shown that the lab-frame equations, in some normalized units, are: $$ x = ct - \sin t \\ y = 1 - \cos t, $$ where $t \in [0, 2 \pi ]$, $c > 1$, and where the units of time and length depend on $A, g$ and $V_0$. Attached is a lab-frame trajectory that I plotted with $c = 10$.

 

[haruspex]

haruspex (and possibly others) objected, reasoning that an inverted cycloid cannot possibly be a path "freely slidden upon" by a particle whose initial horizontal velocity is not zero, because such a path is vertical initially. That argument is flawed, however, because the inverted cycloid is the minimal time path in the initially comoving frame; in the lab frame where $V_0 \ne 0$, the minimal time path is the trajectory of the particle following a cycloidal path in the initially comoving frame; that path is not a cycloid per se.

Apologies - I had not understood what you giving as the solution. Yes, very neat - thanks.