A variation of the ant-on-rubberband problem

[bobie]

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 v_c on two roads at 60° (in a vacuum) so that their distance is at any time r_t = T_t*v_c. Road A points North (edit: A 0°, B 300°).
At time 10 s , at distance D =10v a bullet is shot from A at v_b = v_c toward (Edit: the future position of) B. (airfriction=0)
- after what time T_x 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 T_x should be shorter.

2. Homework 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 )

 

[haruspex]

suppose 2 cars A, B are speeding at v_c on two roads at 60° (in a vacuum) so that their distance is at any time r_t = T_t*v_c. Road A points North.
At time 10 s , at distance r = _10v a bullet is shot from A at v_b = v_c 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?

 

[bobie]

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!

 

[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?

 

[bobie]

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?

 

[haruspex]

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?

 

[bobie]

Let's take the y coordinate as the bisector of the A and B paths.

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?

 

[haruspex]

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?

 

[bobie]

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.

At time 10 s , at distance D =10v a bullet is shot from A at v_b = v_c

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

 

[haruspex]

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

I agree.

 

[bobie]

Thank you for your attention, haruspex,

 

[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 v_ant = v_inflation =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?

 

[haruspex]

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.

 

[bobie]

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

 

[haruspex]

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?

 

[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:

2. Homework Equations ?
\frac{v_b}{v_c}\log{\left(\frac{c+vT}{c}\right)}=1

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

T_r=\frac{rad}{v_a}\left(e^{v_e/v_a}-1\right) = 10/1* e^1 -1 = 17.2 s

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

 

[haruspex]

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?

 

[bobie]

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:
T_{A,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 D_A,B (1 r) if space were not stretching (= 10 s), since it is, then T_A,B is multiplied by 1.72 : T_A,B' 17.2s
T_{A,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 D_A,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?

 

[haruspex]

It the same question, I think:
what is the actual distance D_A,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.

 

[bobie]

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??

 

[haruspex]

THanks, please let me know when you find it.

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

 

[bobie]

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 = e^1*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° , L_A,B' = 24.3 cm

Thanks for your attention, anyway

 

[haruspex]

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.

 

[bobie]

But the path is not a logarithmic spiral, it's an arithmetic spiral. .

 

[haruspex]

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