# Homework Help: A variation of the ant-on-rubberband problem

1. Jun 16, 2014

### bobie

1. The problem statement, all variables and given/known data
You probably know that intriguing problem: http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope.

Now, suppose 2 cars A, B are speeding at vc on two roads at 60° (in a vacuum) so that their distance is at any time rt = Tt*vc. Road A points North (edit: A 0°, B 300°).
At time 10 s , at distance D =10v a bullet is shot from A at vb = vc toward (Edit: the future position of) B. (airfriction=0)
- after what time Tx the bullet will hit B? (≈14.2 s)
- what is the length L covered by the bullet? (≈21.05v )
- the shot must be fired at what angle? (≈35.65°)
- what is the right equation?

The problem is slightly different since the bullet has inertia radially but there is no rubber rope to pull it tangentially toward car B; the direction of the inertia (0°) is rougly at ≈84.35° from the direction of the bullet (≈275.65°), so the effective speed is > √2v (≈1.48v) and therefore Tx should be shorter.

2. Relevant equations ?
$$\frac{v_b}{v_c}\log{\left(\frac{c+vT}{c}\right)}=1$$
$$T_x=\frac{D}{v_b}\left(e^{v_c/v_b}-1\right) = 1* e^1 -1 = 1.7 (/1.2 )$$

Last edited: Jun 17, 2014
2. Jun 17, 2014

### haruspex

So, the two cars start at the intersection of the roads at time 0, right?
Is 'toward B' relative to the rest frame or relative to A's motion?
If it's relative to the rest frame then it is travelling towards where car B was at the moment it was fired, so will miss it completely, no?
If it is relative to A then relative to the rest frame it will move parallel to B, so will never reach it.
Perhaps you don't mean that it is fired 'toward B', but rather, fired in whichever direction is necessary for it to reach B? The question then is, what is the speed relative to?

3. Jun 17, 2014

### bobie

Surely, haruspex it was loosely expressed, as I knew you'd understand; I have edited it for future readers.
Thanks for your help!

Last edited: Jun 17, 2014
4. Jun 17, 2014

### haruspex

Let's take the y coordinate as the bisector of the A and B paths.
What is the relative motion of A and B in the y direction?
What is the relative position of A and B in the y direction?
What, therefore, must the relative motion of bullet be to A in the y direction?

5. Jun 18, 2014

### bobie

You mean sort of motion of car B
y=√3x+5?

6. Jun 19, 2014

### haruspex

No, I mean consider the y-component velocities of A and B. How do they compare? What about their y-coordinates?

7. Jun 20, 2014

### bobie

If y is the bisector they are the same 8.6-5. How can this lead to the solution?.
The key factor is the sum of the 2 vectors acting on the bullet, isn't it?

Last edited: Jun 20, 2014
8. Jun 20, 2014

### haruspex

Right. So go on to answer the other two questions I posed in post #4:
How do the y-components of their co-ordinates compare?
What does that tell you about the y component of the bullet's velocity relative to A?

9. Jun 23, 2014

### bobie

Have you found a solution, haruspex? do you think there is one solution at all?
I was carried away in OP by the analogy with the ant, but there is no analogy.

The bullet will never hit car B both if vb is muzzle velocity and if it ground speed ( the actual velocity ).
Please confirm.
Thanks.

10. Jun 24, 2014

### haruspex

I agree.

11. Jun 24, 2014

### bobie

Thank you for your attention, haruspex,

12. Jun 27, 2014

### bobie

Could we go back to the ant (original) problem?

Suppose that now the angle is one radiant, so that distance A,B = radiant = 10, and vant = vinflation =1,
according to the formula, which I assume correct, time T required for the ant to reach point B is 10*1.72 = 17.2 seconds, right? 27.2 seconds after inflation started.

Could you now please tell me how to calculate space L travelled by the ant? how do you integrate the original formula on space?

Last edited: Jun 27, 2014
13. Jun 27, 2014

### haruspex

Angle? There is no angle in the ant problem.

14. Jun 27, 2014

### bobie

One radiant is the distance the ant must go. When the ant starts off from point A, point B is one r away; 10 cm.The radius of the balloon r is growing at 1 cm/s and the and is going at 1 cm/s.According to the formula it will reach point B after 17.2 s when the distance A',B' is 27.2 cm.
We want to find the distance A,B' the ant has travelled.
Thanks

15. Jun 28, 2014

### haruspex

The link you posted deals with a linear elastic string, not a balloon, right? It can be posed in either context, but let's stick to one to avoid confusion. And let's use symbols not numbers (yet). You refer to 'the formula', but I don't know which one you mean.
For the balloon context, if that's what you want:
Start point A, end point B, ant's position at time t = X(t).
Centre of balloon is O.
Angle AOX at time t is θ(t).
Radius at time t = r0+rt.
Now, what is your equation?

16. Jun 28, 2014

### bobie

This is the equation from wiki, I supposed an inflating balloon is equivalent to a streching rubberband, so to keep the analogy with the bullet. If D must be T*v, then the angle should be 57° instead of 60°
The difference is that the bullet moves on a straight line and the ant on a curve, a spiral:
And this should be the equation to find the time it takes the ant to reach point B', to cover an original distance D of 1r = 10cm
We want now to calculate what is the space L actually covered by the ant , which is of course > 17.2 cm
I guess it must be ≈24.3245, but I'd like to learn how to find it by calculus, which I do not know.
Is there also a formula to define the spiral described by the path of the ant?

Thanks for your attention

Last edited: Jun 28, 2014
17. Jun 28, 2014

### haruspex

This is the ant on the rubber band/balloon model, not your bullet and two cars model. But you seem to be using numbers more appropriate to the bullet model. So it looks like we have an ant moving at the same speed as the balloon is expanding, right?
Your equation $T_r=\frac{rad}{v_a}\left(e^{v_e/v_a}-1\right)$ doesn't seem right. A radian is an angle, not a distance, and I think the speed in the first denominator is the rate of expansion, not the speed of the ant/bullet:
$T_r=\frac{D}{v_e}\left(e^{v_e/v_a}-1\right)$
You then wrote 10/e-1, but I guess you meant 10(e-1).

Anyway, you are now asking a different question, and we need to derive the equation from scratch. In time dt, how far does the ant move in space?

18. Jun 28, 2014

### bobie

Sorry, I made a couple of slips, let's see if I can write better:
TA,B' = $\frac{r}{v_a}\left(e^{v_e/v_a}-1\right) = 10/1 * (1.72) = 17.2s$
The speed in the denominator is the speed of the ant: distance /speed = time elapsed to cover DA,B (1 r) if space were not stretching (= 10 s), since it is, then TA,B is multiplied by 1.72 : TA,B' 17.2s
TA,B' = $\frac{10}{1}\left(e^{1/1}-1\right) = 17.2 s$
It is the same question, I think:

what is the actual distance DA,B' covered by the ant in 17.2 s ?
Do we need integration (on space), do we need a new equation or can we calculate directly the length of the spiral? what is its rigth name: logarithmic spiral, what is its equation?

Last edited: Jun 29, 2014
19. Jun 29, 2014

### haruspex

Yes, it's the same question in a sense, but the solution will be rather different. There is no simple way to adapt the existing solution. You need to start from scratch with a new differential equation. I suspect it will be harder to solve. Until it is solved, we won't know what kind of spiral it is.

20. Jun 29, 2014

### bobie

THanks, please let me know when you find it. My guess is ≈24.3 cm
Is this a suitable representation?:http://de.wikipedia.org/wiki/Logarithmische_Spirale#mediaviewer/Datei:Logarithmic_spiral.svg
is there any difference with this?http://en.wikipedia.org/wiki/Rhumb_line#mediaviewer/File:Loxodrome.png

Could you tell me what is the length of that spiral if the distance of the ending point from o (on the x axis) x= r is 10cm , and we proceed for 1 radian??

Last edited: Jun 29, 2014
21. Jun 29, 2014

### haruspex

The next step is for you to post an attempt at the appropriate differential equation.

22. Jun 29, 2014

### bobie

I do not know calculus and this is not, of course, homework. School is over, anyway, haruspex
I do not agree we need a new differential equation:

the equation for time is simply a log spiral: T = e1*Θ

$$\arccos \frac{\langle \mathbf{r}(\theta), \mathbf{r}'(\theta) \rangle}{\|\mathbf{r}(\theta)\|\|\mathbf{r}'(\theta)\|} = \arctan \frac{1}{b} = \phi.$$
b= 1, 1/b=1, $\varphi$= tan-1 1 , pitch should be 90°-45°= 45°

17.12/cos45° , LA,B' = 24.3 cm

Thanks for your attention, anyway

Last edited: Jun 29, 2014
23. Jun 29, 2014

### haruspex

Sorry, you're right, no new DE is needed to find the path. But the path is not a logarithmic spiral, it's an arithmetic spiral. You will see this if you eliminate time to find the relationship between the radius and the angle traversed.

24. Jun 30, 2014

### bobie

25. Jun 30, 2014

### haruspex

Hmm, yes, not surprised that confused you. I meant to write 'geometric'.

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