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

[Oculatus]

I found this picture somewhere:

My question being, as in the picture, which ball gets to the end faster? I guess you could naively say that ball A reaches the end in less time, but then again, there is the force of gravity which would accelerate the ball going downwards (and deaccelerate it going upwards, respectively). In essence, the question really might be: does the gravitational pull accelerate the ball to a high enough velocity to compensate for the increased distance? I hope I'm going into the right direction with this, but I certainly lack the mathematical/physical knowledge to solve this (could the conservation of energy be used?).

I'd appreciate any input.

 

[phinds]

Can you visualize roughly what the velocity vs time graph would look like for B as it goes down the trough and then comes back up?

 

[berkeman]

I found this picture somewhere:

My question being, as in the picture, which ball gets to the end faster? I guess you could naively say that ball A reaches the end in less time, but then again, there is the force of gravity which would accelerate the ball going downwards (and deaccelerate it going upwards, respectively). In essence, the question really might be: does the gravitational pull accelerate the ball to a high enough velocity to compensate for the increased distance? I hope I'm going into the right direction with this, but I certainly lack the mathematical/physical knowledge to solve this (could the conservation of energy be used?).

I'd appreciate any input.

I used to think it was obvious -- ball B has a higher average velocity, since its velocity is always greater than or equal to ball A. But on the other hand, ball B has to travel farther. Hmm.

So try picking a couple simple dip geometries to calculate the time it takes for the ball to make it across. Show us your calcs -- we are all interested in the correct answer...

 

[Nugatory]

I used to think it was obvious -- ball B has a higher average velocity, since its velocity is always greater than or equal to ball A. But on the other hand, ball B has to travel farther. Hmm.

I figured that the increase in distance traveled is a linear function of the depth of the dip, as is the increase in the ball's energy as it drops that distance. Speed goes as the square root of kinetic energy, so the distance traveled should increase more rapidly than the speed and A wins the race.

 

[willem2]

The question is, can the horizontal component of the velocity v_h ever become smaller than v₀?
The only force that can reduce v_h is the normal force. On a down-slope the horizontal component of the normal force points forward, so v_h can only increase. If the hole was symmetrical, we would now be done, because v_h on the up-slope would have to be the same as on the corresponding point on the down-slope, so v_h >= v₀ everywhere.

It seems possible to think of a real steep up-slope, that could reduce v_{h[/SUB to smaller than v0, but if that happens, the vertical component of the velocity would become so large that the ball would jump up above the starting level.}

 

[bahamagreen]

Is there anything changing the horizontal component of either of the balls' velocites?

Unless there is, it looks like a tie, to me.

 

[A.T.]

Is there anything changing the horizontal component of either of the balls' velocites?

You don't think that ball B will have a higher horizontal velocity than V₀ at the bottom of the dent? See willem's post.

 

[Delta2]

Is there anything changing the horizontal component of either of the balls' velocites?

Unless there is, it looks like a tie, to me.

Yes there is, it is a component of the normal force (it is normal to the surface orientation and not normal to horizontal velocity) from the cavity surface that increases horizontal component $v_{h_B}$ to a velocity $v_{hmax}>v_0$. That is during the descent of ball B. During the ascend we have again the component of the normal force that decreases the horizontal velocity from $v_{hmax}$ back to $v_0$.

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 $v_0$, ball B wins the race.

 

[Lok]

I do not like the tie, as that would imply that a deeper hole will still end up in a tie, yet gravity is limited and a slightly deeper hole will clearly slow B regarding to A.

 

[willem2]

I do not like the tie, as that would imply that a deeper hole will still end up in a tie, yet gravity is limited and a slightly deeper hole will clearly slow B regarding to A.

There's a limit to how deep the hole can become without the ball B losing the contact with the ground. As long as the normal force points upwards it will also point forward on the down-slope, and the horizontal velocity can only increase. If the normal velocity points downwards, the horizontal velocity could decrease, but this can only happen if ball B can't leave the ground. (if it's in a tunnel for example). If the horizontal velocity of ball B can't decrease, ball B can never be slower than ball A.

 

[Lok]

On a better thought this problem can have 2 solutions (if the never breaks contact statement holds) affected by the "ratio" of vo and g.

A very low speed with normal g will favor B.

A very high speed with normal g will favor A. As the hole will act as a bump (never breaks contact).

 

[A.T.]

A very high speed with normal g will favor A. As the hole will act as a bump (never breaks contact).

"Act as a bump" is not a explanation. You have to justify this using forces that act to reduce the horizontal veloctiy of B.

 

[Lok]

"Act as a bump" is not a explanation. You have to justify this using forces that act to reduce the horizontal veloctiy of B.

As willem2 said above, at high speed (or very low g) ball B would lose contact to the ground, yet having a "does not break contact" constrain that does not necessarily imply a relationship between gravity and speed the hole would only act to create a bigger path.

While I know this is not the scope of the problem, it is there.

 

[jbriggs444]

As willem2 said above, at high speed (or very low g) ball B would lose contact to the ground, yet having a "does not break contact" constrain that does not necessarily imply a relationship between gravity and speed the hole would only act to create a bigger path.

While I know this is not the scope of the problem, it is there.

As I read the picture, the balls are rolling on the surface of the ground. Accordingly, the normal force can never be negative. Together with the "does not break contact" constraint, this limits the set of possible paths and starting velocities and limits them differently depending on gravity.

Unstated, however, is an important assumption that the balls roll without slipping.

Edit: My mistake on the rolling without slipping. "Negligible friction" covers this.

 

[A.T.]

As I read the picture, the balls are rolling on the surface of the ground. Accordingly, the normal force can never be negative. Together with the "does not break contact" constraint, this limits the set of possible paths and starting velocities and limits them differently depending on gravity.

That's how I understand it too. The assumption that B never breaks contact doesn’t imply an actual mechanical constraint that prevents that. It just limits the cases to be considered. And in these cases A can never be faster.

 

[olivermsun]

Unstated, however, is an important assumption that the balls roll without slipping.

Edit: My mistake on the rolling without slipping. "Negligible friction" covers this.

It's kind of tangential to the problem, but I think rolling without slipping is the opposite case from negligible friction. Otherwise, should the ball start rolling faster when it accelerates in translation? It could just continue rotating at whatever rate while also sliding along the slope.

 

[Delta2]

Er are we talking for ways that the horizontal velocity of B can become smaller than $v_0$? The only way for this to happen is that the normal force is negative (i.e pointing towards the ground) which simply cannot happen, not in this case.

 

[bahamagreen]

Yes there is, it is a component of the normal force (it is normal to the surface orientation and not normal to horizontal velocity) from the cavity surface that increases horizontal component $v_{h_B}$ to a velocity $v_{hmax}>v_0$. That is during the descent of ball B. During the ascend we have again the component of the normal force that decreases the horizontal velocity from $v_{hmax}$ back to $v_0$.

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 $v_0$, ball B wins the race.

If I drive a car off a cliff, the horizontal component of motion remains constant.

But, if I roll the car down a long flat incline, it will increase... because a normal force is present.

I think I see that...

So, since the horizontal component of the distance is the same for both balls, but it is ball B that spends part of its travel in the segment of the dip, within which its horizontal component velocity is always greater than ball A (because ball B's entry speed into the dip equals its exit speed, but is above Vo throughout the dip)... so ball B advances over ball A as B enters the dip and maintains the advantage it holds when exiting the dip to the end, finishing before ball A.

 

[voko]

Let's simplify this. Say $V_0$ is very close to zero. What happens then?

 

[haruspex]

If I drive a car off a cliff, the horizontal component of motion remains constant.

But, if I roll the car down a long flat incline, it will increase... because a normal force is present.

I think I see that...

So, since the horizontal component of the distance is the same for both balls, but it is ball B that spends part of its travel in the segment of the dip, within which its horizontal component velocity is always greater than ball A (because ball B's entry speed into the dip equals its exit speed, but is above Vo throughout the dip)... so ball B advances over ball A as B enters the dip and maintains the advantage it holds when exiting the dip to the end, finishing before ball A.

Yes, this is the critical observation: that during ascent the horizontal component cannot increase, yet on completing the ascent it must be back at Vo.

Here are some more interesting questions:
What shapes of dip (a) minimise, (b) maximise the time difference?

 

[voko]

Yes, this is the critical observation: that during ascent the horizontal component cannot increase

How is that so?

At a deepest point of the track, the entire velocity must be horizontal. If the deepest point is below the starting point, conservation of energy implies that the horizontal velocity is greater (greatest?) at the deepest point.

 

[A.T.]

maximise the time difference?

For low V₀ and low inertial moments of the ball, it becomes the brachistochrone:
http://en.wikipedia.org/wiki/Brachistochrone_curve

 

[voko]

... and we can always go to a low $V_0$ by choosing a reference frame ...

 

[ZetaOfThree]

For low V₀ and low inertial moments of the ball, it becomes the brachistochrone:
http://en.wikipedia.org/wiki/Brachistochrone_curve

Why does $V_0$ need to be low? As long as the ball stays in contact with the ground, $V_0$ shouldn't matter right?

 

[haruspex]

How is that so?

At a deepest point of the track, the entire velocity must be horizontal. If the deepest point is below the starting point, conservation of energy implies that the horizontal velocity is greater (greatest?) at the deepest point.

That accords with my statement. Maybe you misread it.

 

[haruspex]

For low V₀ and low inertial moments of the ball, it becomes the brachistochrone:
http://en.wikipedia.org/wiki/Brachistochrone_curve

No, that's the catch. The standard brachistochrone problem is in the context of, say, bead on a wire. In the present context there's only gravity to maintain contact. The initial descent cannot be steeper than the ballistic path.

 

[olivermsun]

One thing that I noticed:

If I start from the energy equation without rotation:
$$\frac{1}{2}mv^2 + mgy = \frac{1}{2}mv_0^2$$
then after some manipulation I get
$$\ddot{x} = -\dfrac{\dot{y}}{\dot{x}} (g + \ddot{y}).$$
Since $\ddot{y} \ge -g,$ it follows that $\ddot{x} \ge 0,$ with equality in the case of the ballistic trajectory $(\ddot{y} = -g).$

If I include rotation, however, say for a hollow sphere:
$$
\frac{1}{3}mv^2 + \frac{1}{2}mv^2 + mgy
= \frac{5}{6}mv^2 + mgy = \frac{5}{6}mv_0^2,
$$
then I get
$$\ddot{x} = -\dfrac{\dot{y}}{\dot{x}} \left(\frac{3}{5}g + \ddot{y} \right).$$
Now for $\ddot{y} = -g,$ the expression in the second parentheses is negative, implying that $\ddot{x} < 0$.

Am I missing something?

 

[haruspex]

$\frac{1}{3}mv^2 + \frac{1}{2}mv^2 + mgy = \frac{5}{6}mv^2 + mgy = \frac{1}{2}mv_0^2$

$= \frac 56 mv_0^2$?

 

[olivermsun]

$= \frac 56 mv_0^2$?

Thanks for catching that. Fixed in post above.

 

[voko]

That accords with my statement. Maybe you misread it.

Indeed I did. I misread "ascent" as something else :)

 

[haruspex]

then I get
$$\ddot{x} = -\dfrac{\dot{y}}{\dot{x}} \left(\frac{3}{5}g + \ddot{y} \right).$$
Now for $\ddot{y} = -g,$ the expression in the second parentheses is negative, implying that $\ddot{x} < 0$.

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.

 

[A.T.]

For low V₀ and low inertial moments of the ball, it becomes the brachistochrone:
http://en.wikipedia.org/wiki/Brachistochrone_curve

No, that's the catch. The standard brachistochrone problem is in the context of, say, bead on a wire. In the present context there's only gravity to maintain contact.

Would the standard brachistochrone path between two points on the same elevation ever cause a rolling ball to take off?

The initial descent cannot be steeper than the ballistic path.

That's why I said low V₀. It's the limiting case where the ballistic path would be almost vertically down.

 

[haruspex]

Would the standard brachistochrone path between two points on the same elevation ever cause a rolling ball to take off?

In the present case, the ball is initially moving horizontally. The usual brachistochrone solution (a cycloid) starts steeply downwards. Yes, the ball would take off.
And I'm not interested in the limiting case, where the initially velocity is effectively zero.

 

[A.T.]

... and we can always go to a low $V_0$ by choosing a reference frame ...

Why does $V_0$ need to be low? As long as the ball stays in contact with the ground, $V_0$ shouldn't matter right?

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 $V_0$.

 

[pleco]

My question being, as in the picture, which ball gets to the end faster?

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.