I What's wrong with this approach to the ant on a rubber rope problem?

[greypilgrim]

Hi.

An ant walks with 1 cm/s along a rubber rope with initial length 10 cm that is stretched uniformly with 10 cm/s. Find its velocity at time $t$.
This is a a well-known problem, but I'd like to know what's wrong with the following alternative approach:

At time $t>0$, any point $x_0$ on the rope gets stretched to
$$x(t)=x_0\cdot (1+\frac{10\,\frac{\text{cm}}{\text{s}}}{10\,\text{cm}}\cdot t)=x_0\cdot (1+\frac{1}{\text{s}}\cdot t)\enspace.$$
This is equivalent to
$$x_0=\frac{x}{1+\frac{1}{\text{s}}\cdot t}$$

The given speed of the ant is with respect to the rope at any time $t$, so $\frac{dx}{dt}=1\,\frac{\text{cm}}{\text{s}}$. However, we want its velocity with respect to a fixed scale, such as the rubber band at $t=0$. So I get
$$v(t)=\frac{dx_0}{dt}=\frac{dx_0}{dx}\cdot \frac{dx}{dt}=\frac{1}{1+\frac{1}{\text{s}}\cdot t}\cdot 1\,\frac{\text{cm}}{\text{s}}\enspace.$$

This of course is terribly wrong, as it would mean that the ant is even slower than 1 cm/s seen from the outside. But what exactly is wrong?

 

[A.T.]

The given speed of the ant is with respect to the rope at any time $t$, so $\frac{dx}{dt}=1\,\frac{\text{cm}}{\text{s}}$. However, we want its velocity with respect to a fixed scale, such as the rubber band at $t=0$. So I get
$$v(t)=\frac{dx_0}{dt}=\frac{dx_0}{dx}\cdot \frac{dx}{dt}=\frac{1}{1+\frac{1}{\text{s}}\cdot t}\cdot 1\,\frac{\text{cm}}{\text{s}}\enspace.$$

Are you multiplying the velocities from stretching and walking instead of adding them?

 

[greypilgrim]

Actually, I'm only using the chain rule of differentiation.

 

[SchroedingersLion]

Hi.

An ant walks with 1 cm/s along a rubber rope with initial length 10 cm that is stretched uniformly with 10 cm/s. Find its velocity at time $t$.
This is a a well-known problem, but I'd like to know what's wrong with the following alternative approach:

At time $t>0$, any point $x_0$ on the rope gets stretched to
$$x(t)=x_0\cdot (1+\frac{10\,\frac{\text{cm}}{\text{s}}}{10\,\text{cm}}\cdot t)=x_0\cdot (1+\frac{1}{\text{s}}\cdot t)\enspace.$$

This is already wrong, I guess. Not all parts of the rope move with this velocity, only one of the end (while the other end stays at its position - equivalently, both ends move with half the velocity in opposite directions).

 

[greypilgrim]

Not all parts of the rope move with this velocity, only one of the end (while the other end stays at its position - equivalently, both ends move with half the velocity in opposite directions).

But this formula exactly means that the velocity of a point on the rope is proportional to its initial position $x_0$. The end with $x_0=0$ remains at $0$, the other end with $x_0=10\,\text{cm}$ moves with $10\,\frac{\text{cm}}{\text{s}}$, the center point with $x_0=5\,\text{cm}$ moves with $5\,\frac{\text{cm}}{\text{s}}$ and so on.

 

[A.T.]

Actually, I'm only using the chain rule of differentiation.

Which functions are you chaining and why?

 

[greypilgrim]

I'm chaining $x_0(t)=x_0(x(t))$. Why? As I wrote, as I interpret the problem, I seek $v(t)=x_0'(t)$ (the velocity of the ant with respect to a fixed frame, such as the rubber rope at $t=0$), but have given the velocity of the ant with respect to the stretching frame, $x'(t)=1\,\frac{\text{cm}}{\text{s}}$, but I know the relationship $x_0(x)$.

 

[A.T.]

...but have given the velocity of the ant with respect to the stretching frame, $x'(t)=1\,\frac{\text{cm}}{\text{s}}$, ...

That's the velocity of the ant with respect to a moving point. The velocity of that point is determined by the stretching. You combine these velocities by adding them.

 

[greypilgrim]

Ok, I think I understand now why I'd need to add the velocity of the moving point. What I still don't understand is why you can just take $1\,\frac{\text{cm}}{\text{s}}$ as velocity with respect to the moving point and don't need to correct for the fact that the respective coordinate system stretches as time passes. At any point in time, we could write a cm scale on the rubber rope to measure the velocity of the ant with respect to the rope, but this scale would be wrong at any other time, even an infinitesimal time later. So why can we just use $x'(t)=\lim_{h\rightarrow 0}\frac{x(t+h)-x(t)}{h}$, when $x(t+h)$ is already scaled differently than $x(t)$?

I think I was "inspired" by the transformation law for tensors, which also includes first derivatives of the coordinate transformation.

 

[A.T.]

Ok, I think I understand now why I'd need to add the velocity of the moving point. What I still don't understand is why you can just take $1\,\frac{\text{cm}}{\text{s}}$ as velocity with respect to the moving point and don't need to correct for the fact that the respective coordinate system stretches as time passes.

If you want to get the total speed in cm/s, you can just add the 1cm/s as given.

At any point in time, we could write a cm scale on the rubber rope to measure the velocity of the ant with respect to the rope

Not in cm/s. A cm doesn't change just because some rope stretches. Your method would measure the speed of the ant with respect to the rope in 10ths-of-the-current-rope-length/s.

 

[Chestermiller]

This is more easily visualized in terms of a material coordinate system that is permanently inscribed onto the rope. To do this, we use the coordinates of the material points along the rope at time zero $x_0$ as labels to keep track of the deformation. These material points get farther apart as time progresses, but their material coordinate labels remain unchanged. At time t, the differential spatial distance between two material points that were at $x_0$ and $x_0+dx_0$ at time zero is now $$dx=\lambda(t)dx_0$$where $$\lambda(t)=1+\frac{vt}{L_o}$$If, during the time interval between t and t+dt, the ant moves $dx=v_adt$ relative to the rope, then, since the rope has been stretched, the material coordinate of the ant changes by much smaller amount $$dx_{0}=\frac{v_adt}{\lambda(t)}$$So, at any time, the ant's material coordinate location is $$x_0(t)=v_a\int_0^t {\frac{dt'}{\lambda(t')}}=\frac{v_aL_0}{v}\ln{\left[1+\frac{vt}{L_0}\right]}$$When $x_0(t)=L_0$, the ant has reached the far end of the rope.

 

[greypilgrim]

Ok, that seems to be exactly what my calculation gives, though I interpreted it wrongly.

How would I now use this result to calculate the ant's velocity measured from an outside, resting observer?

 

[A.T.]

How would I now use this result to calculate the ant's velocity measured from an outside, resting observer?

You convert the material coordinate location to cm using λ(t) and differentiate that with respect to t.

 

[greypilgrim]

Ok, we have
$$x(t)=x_0(t)\cdot \lambda(t)+x_0(0)=\frac{v_aL_0}{v}\ln{\left[1+\frac{vt}{L_0}\right]}\cdot \left(1+\frac{vt}{L_0}\right)+x_0(0)$$
and hence
$$x'(t)=v_a+v_a\cdot\ln{\left[1+\frac{v}{L_0}\cdot t\right]}\enspace .$$

Is this the addition of velocities you have been referring to? However, it still contains a logarithm, so integration must have taken place somewhere. So is there no way to find an expression for the velocity measured from the outside without integrating or solving a differential equation?

 

[Chestermiller]

Ok, we have
$$x(t)=x_0(t)\cdot \lambda(t)+x_0(0)=\frac{v_aL_0}{v}\ln{\left[1+\frac{vt}{L_0}\right]}\cdot \left(1+\frac{vt}{L_0}\right)+x_0(0)$$

Actually, the ant's location at time zero $x_0(0)$ is equal to zero.

 

[Chestermiller]

$$x'(t)=v_a+v_a\cdot\ln{\left[1+\frac{v}{L_0}\cdot t\right]}\enspace $$

If you combine this with the final equation of post #11, you get $$x'(t)=v_a+v\frac{x_0}{L_0}=v_a+v\frac{x}{L}$$which is what one would expect.