A variation of the ant-on-rubberband problem

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In summary: For the balloon context, let's say the ant starts at (r,0) and the balloon center is at (0,0), and the balloon is growing at rate v=1, so its radius after t seconds is rt=t. Then the ant's position at time t is at (r0 + t, 0). From the Pythagorean theorem, the distance from the ant to the balloon center at time t is s(t) = √(r0 + t)^2 + (0)^2 = √(r0 + t)^2 = r0 + t. So we need to find the time that s(t) = 10, and then the distance traveled will be r0 +
  • #1
bobie
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Homework Statement


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. Homework Equations ?
[tex]\frac{v_b}{v_c}\log{\left(\frac{c+vT}{c}\right)}=1 [/tex]
[tex]T_x=\frac{D}{v_b}\left(e^{v_c/v_b}-1\right) = 1* e^1 -1 = 1.7 (/1.2 )[/tex]
 
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  • #2
bobie said:
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.
At time 10 s , at distance r = 10v a bullet is shot from A at vb = vc toward B. (airfriction=0)
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 traveling 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?
 
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  • #3
haruspex said:
Perhaps you don't mean ... fired in whichever direction is necessary for it to reach B?
Surely, haruspex it was loosely expressed, as I knew you'd understand; I have edited it for future readers.
Thanks for your help!
 
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  • #4
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?
 
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  • #5
haruspex said:
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?

You mean sort of motion of car B
y=√3x+5?
 
  • #6
bobie said:
You mean sort of motion of car B
y=√3x+5?
No, I mean consider the y-component velocities of A and B. How do they compare? What about their y-coordinates?
 
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  • #7
haruspex said:
Let's take the y coordinate as the bisector of the A and B paths.
haruspex said:
No, I mean consider the y-component velocities of A and B.
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?
 
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  • #8
bobie said:
If y is the bisector they are the same
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?
 
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  • #9
haruspex said:
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?
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.

bobie said:
At time 10 s , at distance D =10v a bullet is shot from A at vb = vc
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
bobie said:
The bullet will never hit car B both if vb is muzzle velocity and if it ground speed ( the actual velocity ).
Please confirm.
Thanks.

I agree.
 
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  • #11
Thank you for your attention, haruspex, :smile:
 
  • #12
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 traveled by the ant? how do you integrate the original formula on space?
 
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  • #13
bobie said:
Could we go back to the ant (original) problem?

Suppose that now the angle is one radiant,
Angle? There is no angle in the ant problem.
 
  • #14
haruspex said:
Angle? There is no angle in the ant problem.
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
bobie said:
The radius of the balloon r is growing at 1 cm/s and the and is going at 1 cm/s.
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
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:
bobie said:
2. Homework Equations ?
[tex]\frac{v_b}{v_c}\log{\left(\frac{c+vT}{c}\right)}=1 [/tex]
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
[tex]T_r=\frac{rad}{v_a}\left(e^{v_e/v_a}-1\right) = 10/1* e^1 -1 = 17.2 s [/tex]
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
 
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  • #17
bobie said:
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
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
haruspex said:
it looks like we have an ant moving at the same speed as the balloon is expanding, right?
A radian is an angle,
the speed in the first denominator is the rate of expansion, not the speed of the ant/bullet:
but I guess you meant 10(e-1).
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?
 
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  • #19
bobie said:
It 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 right name: logarithmic spiral, what is its equation?
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.
 
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  • #20
haruspex said:
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.

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??
 
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  • #21
bobie said:
THanks, please let me know when you find it.
The next step is for you to post an attempt at the appropriate differential equation.
 
  • #22
haruspex said:
The next step is for you to post an attempt at the appropriate differential equation.
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

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

17.12/cos45° , LA,B' = 24.3 cm

Thanks for your attention, anyway
 
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  • #23
bobie said:

I do not agree we need a new differential equation:
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
haruspex said:
But the path is not a logarithmic spiral, it's an arithmetic spiral. .

:confused:
 
  • #25
bobie said:
:confused:
Hmm, yes, not surprised that confused you. I meant to write 'geometric'.
 

1. What is the ant-on-rubberband problem?

The ant-on-rubberband problem is a classic physics problem that involves an ant crawling along a stretched rubber band. The rubber band is gradually stretched at a constant rate, causing the ant to move along the band. The question is, how far will the ant travel before the rubber band snaps back to its original length?

2. What is the purpose of studying this problem?

The ant-on-rubberband problem is a simple and fun way to introduce concepts of elasticity and potential energy. It also helps to develop critical thinking skills and problem-solving abilities.

3. How does the mass of the ant affect the distance it travels?

The mass of the ant does not affect the distance it travels. This problem is an idealized scenario and assumes that the ant has no mass and the rubber band has no weight. In reality, the mass of the ant would play a role in the distance it travels, but for the purposes of this problem, it is negligible.

4. What are the key factors that affect the distance the ant travels?

The key factors that affect the distance the ant travels are the initial length of the rubber band, the rate at which the rubber band is stretched, and the elastic properties of the rubber band. These factors determine the amount of potential energy stored in the rubber band, which is converted into kinetic energy as the rubber band snaps back and propels the ant forward.

5. Is the ant-on-rubberband problem a realistic scenario?

No, the ant-on-rubberband problem is not a realistic scenario. It is a simplified version of a larger concept known as the "Hooke's Law of Elasticity," which states that the force needed to extend or compress a spring is directly proportional to the distance it is stretched or compressed. However, this problem is a great way to introduce basic concepts of physics and can be used as a starting point for more complex problems.

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