Finding the maximum elongation

[Saitama]

Homework Statement

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)##)

Homework Equations

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)?

 

[Tanya Sharma]

Hi Pranav

The answer doesn't seem right .Its dimensionally incorrect .

 

[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.

 

[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.

 

[barryj]

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

 

[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.

 

[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.

 

[barryj]

Where is the center of mass for the system?

 

[Saitama]

Where is the center of mass for the system?

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

 

[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?

 

[Saitama]

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?

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?

 

[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?

 

[Saitama]

What will be the velocities of the masses at maximum elongation?

If I knew, I would have solved the problem.

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

 

[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.

 

[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?

EDIT : Already answered above me.

 

[Saitama]

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.

You mean something like this:
[tex]\frac{1}{2}m\frac{v_0^2}{4}+\frac{1}{2}m\frac{v_0^2}{4}=\frac{1}{2}kx^2[/tex]

 

[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.

 

[Saitama]

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.

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,
[tex]\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)[/tex]
Solving for ##v'##,
[tex]v'=\frac{v_0l_0}{2(l_0+2x)}[/tex]
From conservation of energy,
[tex]\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[/tex]
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?

 

[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.

 

[Saitama]

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.

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

 

[voko]

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

You assume that ##v'##, whatever that means, is orthogonal to the separation. I do not know whether that is a valid assumption.

I suggest you follow the approach I outlined.

 

[dreamLord]

I think he has resolved the velocity at time t into 2 components, v' being the velocity along y-axis, going by the kinetic energy expression. Is that right Pranav? But then you would have more terms in the angular momentum equation, so I don't know.

But I agree with voko, follow his approach (and use polar coordinates, it will make your job easier).

 

[Saitama]

I think he has resolved the velocity at time t into 2 components, v' being the velocity along y-axis. Is that right Pranav?

I resolved the velocity into components, one along the length of spring and the other perpendicular to the length of spring. Is that wrong?

But I agree with voko, follow his approach (and use polar coordinates, it will make your job easier).

Can't we simply use Cartesian coordinates here? I am not familiar with polar coordinates.

I still don't understand how to start and what is wrong with my approach?

 

[dreamLord]

In that case, you must use (half of) the full length of the spring, not just it's length along the x-axis when finding out the final angular momentum of the masses.

 

[Saitama]

In that case, you must use (half of) the full length of the spring, not just it's length along the x-axis when finding out the final angular momentum of the masses.

I just tried something which gives me the right answer but I am not sure if its correct so here it goes.
Rewriting the energy equation,
[tex]\frac{mv_0^2}{2}=2mv'^2+2mv_x^2+4kx^2[/tex]
At maximum elongation, ##dx/dt=0 \Rightarrow v_x=0## i.e
[tex]\frac{mv_0^2}{2}=\frac{mv_0^2l_0^2}{2(l_0+2x)^2}+4kx^2[/tex]
As ##x<<<l_0##,
[tex]\frac{1}{(l_0+2x)^2}=\frac{1}{l_0^2}\left(1-\frac{4x}{l_0}\right)[/tex]
[tex]\Rightarrow \frac{mv_0^2}{2}=\frac{mv_0^2}{2}-\frac{2mv_0^2x}{l_0}+4kx^2[/tex]
Solving for ##x##,
[tex]x=\frac{mv_0^2}{2kl}[/tex]
Since maximum elongation is ##2x## hence final answer is ##\displaystyle \frac{mv_0^2}{kl_0}##. This matches with the given answer if we consider ##l_0^2## as a typo in the given answer. Does this look correct?

 

[TSny]

Solving for ##x##,
[tex]x=\frac{mv_0^2}{2kl}[/tex]
Since maximum elongation is ##2x## hence final answer is ##\displaystyle \frac{mv_0^2}{kl_0}##. This matches with the given answer if we consider ##l_0^2## as a typo in the given answer. Does this look correct?

Looks good to me.

 

[Saitama]

Looks good to me.

Thank you TSny for the check. :)

 

[barryj]

How did you get this?

As x<<<l0,
1(l0+2x)2=1l20(1−4xl0)

 

[Saitama]

How did you get this?

As x<<<l0,
1(l0+2x)2=1l20(1−4xl0)

Using binomial theorem.
[tex]\frac{1}{(l_0+2x)^2}=\frac{1}{l_0^2}\left(1+\frac{2x}{l_0}\right)^{-2}[/tex]
Do you know the expansion for ##(1+y)^{-2}##?

 

[WannabeNewton]

I get the same result using energy and angular momentum conservation. I'm guessing the book just had a typo. Maybe the author(s) should have checked for units :p

 

[barryj]

Pran, This is not what you have in post 25.

 

[WannabeNewton]

His use of the binomial approximation was fine. ##(l_0 + 2x)^{-2}= l_0^{-2}(1 + 2\frac{x}{l_0})^{-2}\approx l_0^{-2}(1 - 4\frac{x}{l_0})##.

 

[voko]

In a center-of-mass frame, the displacement of one mass from the center of mass is ## \vec{x} ## and the displacement of the other is ## - \vec{x} ##. Conservation of energy yields ## mx'^2 + k(2x - l_0)^2/2 = mx_0'^2 ##. Since, as given, ## e = (2x - l_0)/l_0 ## is very small, ## k(2x - l_0)^2/2 = k (el_0)^2/2 \approx kel_0^2 ##, so, at max elongation, ## kel_0^2 = mx_0'^2 ##, thus ## e_{max} = \frac {mx_0'^2} {kl_0^2} ##.

This looks very much like the given answer, but ## x_0' = v_0/2 ##, so ## e_{max} = \frac {mv_0^2} {4kl_0^2} ##, which is four times smaller.

Any idea anyone?

 

[WannabeNewton]

Assuming ##x'## is the vectorial velocity of each endpoint mass in the CM frame, this is not entirely zero in the CM frame at the instant of maximal elongation. Only the component of ##x'## along the length of the spring is zero; there is still a component perpendicular to the spring. Recall that in the CM frame, the two endpoint masses are both rotating about the CM and translating outwards from the CM due to the spring elongation. At the instant of maximal elongation the endpoint masses will remain still along the length of the spring but they will still have an orbital component about the CM.

Note that if you assume ##x'## vanished identically at the instant of maximal elongation, then the angular momentum at that instant will vanish identically as well which cannot be because the system had an initial angular momentum ##L = 2m(\frac{l_0}{2})(\frac{v_0}{2})\hat{z}## and the angular momentum is conserved (the only force in the system, the spring force, acts radially).

 

[voko]

I should have followed my own advice literally, then :)

Still there is a problem.

Conservation of angular momentum implies ## x \tau = h = \text{const} ##, so ## \tau = h/x ##, where ##\tau## is azimuthal velocity (the component of velocity perpendicular to ## \vec{x} ##). The square of total velocity is ## \dot{x}^2 + \tau^2 = \dot{x}^2 + h^2/x^2 ##, where ## \dot{x} ## is the time-derivative of the length of ## \vec{x} ##. Conservation of energy: $$ m(\dot{x}^2 + \frac {h^2} {x^2}) + k(2x - l_0)^2/2 = m \frac {h^2} {x_0^2} $$ Now ##x_0 = l_0 /2 ## and ## \tau_0 = v_0 / 2 ##, then ## h = l_0 v_0 / 4 ##, so at max elongation $$ m \frac {(l_0 v_0)^2} {16 x^2} + k(2x - l_0)^2/2 = m \frac {(l_0 v_0)^2} {4 l_0^2} $$ ## 2x = (e + 1)l_0 ##, so ## m \frac {(l_0 v_0)^2} {16 x^2} = m \frac {v_0^2} {4(e + 1)^2} \approx mv_0^2/4 - m v_0^2e/2##, thus conservation of energy becomes ## mv_0^2/4 - mv_0^2e/2 + kl_0^2e^2/2 = mv_0^2/4 ##, giving $$ e = \frac {mv_0^2} {kl_0^2} $$ which is the desired result.

The problem is that the result was obtained by retaining only linear terms in kinetic energy, while keeping the entire quadratic term in potential energy. This is not consistent. To be consistent, we either have to linearize potential energy, or have to retain quadratic terms in kinetic energy. Either way, the answer is different from the one given.