# Finding the maximum elongation

1. Jun 26, 2013

### Saitama

1. The problem statement, all variables and given/known data
Two small identical discs, each of mass $m$, lie on a smooth horizontal plane. The discs are interconnected by a light non-deformed spring of length $l_0$ and stiffness $k$. At a certain moment one of the discs is set in motion in a horizontal direction perpendicular to the spring with velocity $v_0$. Find the maximum elongation of the spring in the process of motion, if it is known to be considerably less than unity.

(Answer: $mv_0^2/(kl_0^2)$)

2. Relevant equations

3. The attempt at a solution
I don't really know how should I begin here. Writing down the differential equation for the motion seems to be complicated. A hint follows the given answer stating that the problem is easier to solve in frame of centre of inertia. I don't see how can I utilize this to solve the problem. I mean, I can find $v_{CM}$ at t=0, then I guess I have to use conservation of energy. But how do I calculate the initial energy here? Do I have to calculate the relative velocities of discs w.r.t CM to find the energy (this is a guess)?

2. Jun 26, 2013

### Tanya Sharma

Hi Pranav

The answer doesnt seem right .Its dimensionally incorrect .

3. Jun 26, 2013

### barryj

Some hints: Where is the center of inertia. Using that as a reference, could you redefine the problem as one disk moving in one direction and the other disk in an opposite direction at a different velocity.

The answer does look suspicious.

4. Jun 26, 2013

### andrien

maximum elongation will occur when the discs will have relative velocity zero.you can find the velocity using momentum conservation(velocity of both discs being equal also consider springs as internal).Then just use conservation of energy to find elongation and yes,there is problem with dimensions.

5. Jun 26, 2013

### barryj

How is elongation defined? Is it just the change in length or a percent thing?

6. Jun 26, 2013

### dreamLord

The answer is definitely wrong.

As for the question, try to figure out what the motion of the 2 masses looks like in the center of mass frame. (Hint : It is symmetric ; what 1 mass does, the other one does in an opposite direction). Then all you need is energy conservation to find out the elongation.

7. Jun 26, 2013

### Saitama

Sorry everyone, I should have checked the dimensions. Yes, it does look wrong but this is what stated in the answer key. Perhaps I shouldn't have used the parentheses. In the key, the answer is written as $mv_0^2/kl_0^2$ but this still looks wrong to me.

@barryj: Elongation is defined as change in length.

I still don't know how to begin though.

8. Jun 26, 2013

### barryj

Where is the center of mass for the system?

9. Jun 26, 2013

### Saitama

At t=0, it is located at the centre of line joining the two discs.

10. Jun 26, 2013

### barryj

Yes. Now,picture yourself positioned at the center of mass. hen the first disk moves, the center of mass will move, right? So using symetry, what will be the velocity of each mass relative to the center of mass? Will the center of mass move, that is if this is your reference point? What is the initial kinetic energy due to the masses. What has to be the stored spring energy when the two masses stop moving?

11. Jun 26, 2013

### Saitama

At t=0, $v_{CM}=v_0/2$ in the direction perpendicular to the length of spring.

The relative velocity w.r.t CM of the disc initially at rest is $v_0/2$ opposite in the direction to that of CM and the relative velocity of the other disc initially with v velocity is $v_0/2$ in the direction of $v_{CM}$. CM will remain at rest for the subsequent motion. I don't know what would happen to velocities at maximum elongation. Any hints for that?

12. Jun 26, 2013

### barryj

What will be the velocities of the masses at maximum elongation? What is the total kinetic energy in both masses at that time. Where did the KE go?

13. Jun 26, 2013

### Saitama

If I knew, I would have solved the problem. :uhh:

Won't the discs perform circular motion around CM with an increasing radius?

14. Jun 26, 2013

### barryj

As the masses move out from the center of mass, the velocity of the masses will slow because there is a spring between them. So, from the conservation of energy, if the velocity of the masses goes to zero, then the initial energy must go into the spring. I assume you remember the formula for the energyu stored in a spring as a function of k and d.

15. Jun 26, 2013

### dreamLord

You don't need to look at the motion of the masses. Just look at the final position. Each mass has some kinetic energy. The spring applies some force on both of them, and does work on them, thereby reducing their velocities. Where do you think all the kinetic energy goes?

16. Jun 26, 2013

### Saitama

You mean something like this:
$$\frac{1}{2}m\frac{v_0^2}{4}+\frac{1}{2}m\frac{v_0^2}{4}=\frac{1}{2}kx^2$$

17. Jun 26, 2013

### barryj

This is what I was thinking. However, you asked about circular motion. This might be a factor here. When at maximum elongation, there does seem to be a rotation and hence there will be some rotational KE associated with the rotation.

Maybe another expert will join us on this point.

18. Jun 26, 2013

### Saitama

I am not sure but here's what I am thinking.
Initially, both the discs are at a distance of $l_0/2$. Let at a time t, they move a distance x away from the CM i.e. their new distance from CM is $l_0/2+x$.
Conserving angular momentum about CM,
$$\frac{mv_0l_0}{4}+\frac{mv_0l_0}{4}=mv'\left(\frac{l_0}{2}+x\right)+mv'\left(\frac{l_0}{2}+x\right)$$
Solving for $v'$,
$$v'=\frac{v_0l_0}{2(l_0+2x)}$$
From conservation of energy,
$$\frac{1}{2}m\frac{v_0^2}{4}+\frac{1}{2}m\frac{v_0^2}{4}=\frac{1}{2}m(v'^2+v_x^2)+\frac{1}{2}m(v'^2+v_x^2)+\frac{1}{2}k(2x)^2$$
where $v_x$ is the velocity of discs along the length of spring.

But this $v_x$ is troubling me here. I don't see how can I proceed further. Should I write $v_x$ as $dx/dt$ solve the D.E?

19. Jun 26, 2013

### voko

Total momentum is conserved.

Total angular momentum is conserved.

Total energy is conserved.

It is almost completely the same as in the two-body problem, except that the law of interaction is a bit different.

The two-body problem can be reduced to a one-body problem, where the one body revolves around a fixed center.

Do the same thing here.

20. Jun 26, 2013

### Saitama

Can you have a look at my attempt in the previous post?