What Is the Maximum Elongation of the Spring Between Two Moving Discs?

[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:
\frac{1}{2}m\frac{v_0^2}{4}+\frac{1}{2}m\frac{v_0^2}{4}=\frac{1}{2}kx^2

 

[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,
\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?

 

[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,
\frac{mv_0^2}{2}=2mv'^2+2mv_x^2+4kx^2
At maximum elongation, $dx/dt=0 \Rightarrow v_x=0$ i.e
\frac{mv_0^2}{2}=\frac{mv_0^2l_0^2}{2(l_0+2x)^2}+4kx^2
As $x<<<l_0$,
\frac{1}{(l_0+2x)^2}=\frac{1}{l_0^2}\left(1-\frac{4x}{l_0}\right)
\Rightarrow \frac{mv_0^2}{2}=\frac{mv_0^2}{2}-\frac{2mv_0^2x}{l_0}+4kx^2
Solving for $x$,
x=\frac{mv_0^2}{2kl}
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 $x$,
x=\frac{mv_0^2}{2kl}
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.
\frac{1}{(l_0+2x)^2}=\frac{1}{l_0^2}\left(1+\frac{2x}{l_0}\right)^{-2}
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.

 

[WannabeNewton]

Perhaps I'm misunderstanding you because I don't immediately see the issue at hand. The final orbital velocity of each mass is $v_{f} = \frac{l_0v_0}{2l_0 + 4x}$ hence $\frac{mv_0^{2}}{4} = \frac{1}{2}k(2x)^{2} + \frac{ml_0^{2}v_0^{2}}{[2l_0(1 + \frac{2x}{l_0})]^{2}} = 2kx^{2} + \frac{mv_0^{2}}{4}(1 + \frac{2x}{l_0})^{-2}$. The dimensionless quantity $(1 + \frac{2x}{l_0})^{-2}$ can be Taylor expanded to 1st order in $\frac{x_0}{l_0}$ to give $\frac{mv_0^{2}}{4} = 2kx^{2} + \frac{mv_0^{2}}{4}(1 - \frac{4x}{l_0})$ from which the result follows. We have retained both the squared velocity in the kinetic energy and the squared displacement in the spring potential while throwing out the terms $O([\frac{x}{l_0}]^{2})$.

 

[TSny]

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.

That's an interesting point. At first it does appear that the kinetic energy should be expanded to order $e^2$. But, suppose we take your exact equation $$m\frac {v_0^2} {4(1+e)^2} + k l_0^2e^2/2 = m \frac {v_0^2} {4} $$ and rewrite it as $$ \frac{mv_0^2}{2kl_0^2}(1-\frac{1}{(1+e)^2}) = e^2 $$
I think you can argue that the factor $\frac{mv_0^2}{2kl_0^2}$ on the left must be "small" (of order e). The spring force $k(l_0e)$, when the spring stretches by $l_0e$, should be on the order of the "centrifugal" force $mv_0^2/l_0$. So, $\frac{mv_0^2}{kl_0^2}$ should be on the order of $e$.

So, in the equation $\frac{mv_0^2}{2kl_0^2}(1-\frac{1}{(1+e)^2}) = e^2$, you only need to expand $\frac{1}{(1+e)^2}$ to first order in $e$ in order to have the left hand side of the equation accurate to order $e^2$.

 

[haruspex]

I think you can argue that the factor $\frac{mv_0^2}{2kl_0^2}$ on the left must be "small" (of order e).

Yes, but you don't need to consider the physics to arrive at that. In any equation of the form f(x) = g(x), the first order terms of the two sides are necessarily of the same order as each other. If for small x f(x) = kx + 3x² and g(x) = x² - x³ then kx ≈ x². That the first order terms of the two sides should be different orders of x is not an inconsistency. Of course, if k is not small c.w. 1 it would mean that the assumption that x is small was wrong.

 

[TSny]

Yes, but you don't need to consider the physics to arrive at that. In any equation of the form f(x) = g(x), the first order terms of the two sides are necessarily of the same order as each other. If for small x f(x) = kx + 3x² and g(x) = x² - x³ then kx ≈ x². That the first order terms of the two sides should be different orders of x is not an inconsistency. Of course, if k is not small c.w. 1 it would mean that the assumption that x is small was wrong.

Yes, I see what you are saying. Mathematical consistency alone requires that if $e$ is small then the quantity $mv_0^2/kl_0^2$ must be small and of the same order as $e$. Good.

 

[voko]

After some good sleep it is all clear to the point that it is not clear why I did not do the exact same thing TSny did in #37 which should have convinced me. Oh well. Thanks to all for enlightening me.

 

[Tanya Sharma]

Applying conservation of momentum and using the condition that at maximum elongation the relative velocity of the disks will be zero.

mv_0=mv+mv
v=\frac{v_0}{2}

Applying conservation of energy ,

\frac{1}{2}m{v_0}^2 = 2(\frac{1}{2}mv^2) + \frac{1}{2}kx^2

\frac{1}{2}m{v_0}^2 = 2(\frac{1}{2}m(\frac{v_0}{2})^2) + \frac{1}{2}kx^2

\frac{1}{4}m{v_0}^2 = \frac{1}{2}kx^2

x = v_0\sqrt{\frac{m}{2k}}

The maximum elongation comes out to be l_0 + v_0\sqrt{\frac{m}{2k}}

Where am I getting it wrong ?

 

[ehild]

Applying conservation of momentum and using the condition that at maximum elongation the relative velocity of the disks will be zero.

The relative radial velocity is zero but the system rotates, both disks moves perpendicularly to the spring connecting them, in opposite directions.

ehild

 

[Tanya Sharma]

Hi ehild

The relative radial velocity is zero but the system rotates, both disks moves perpendicularly to the spring connecting them, in opposite directions.

ehild

I understand that the system rotates,but how does disks move perpendicularly to the spring connecting them.The disk initially at rest can move in the direction of applied force which will be along the spring.Yes,the disks will move such that both the linear momentum and angular momentum are conserved.

I am still unable to understand ,where I am getting it wrong with linear momentum and energy conservation .

 

[voko]

Applying conservation of momentum and using the condition that at maximum elongation the relative velocity of the disks will be zero.

This is the same mistake I made in #33, and which was pointed out in #34, and corrected in #35.

 

[Saitama]

I am still unable to understand ,where I am getting it wrong with linear momentum and energy conservation .

If you assume that final velocity is zero, the angular momentum of system vanishes which contradicts the fact of angular momentum being conserved. This is already pointed out by WBN. Check his post on page 2.

 

[haruspex]

I understand that the system rotates,but how does disks move perpendicularly to the spring connecting them.

If you take the frame of reference of the common mass centre as it moves after kick-off, and apply that to include the kick-off itself, the kick-off becomes reformulated as each disk being sent off at speed v0/2, in opposite directions. Does that help?

 

[Tanya Sharma]

Hi haruspex

I understand what you have explained.Still,I am having difficulty perceiving the situation.

Could you point out mistake in post #41.

 

[WannabeNewton]

Applying conservation of momentum and using the condition that at maximum elongation the relative velocity of the disks will be zero.

mv_0=mv+mv
v=\frac{v_0}{2}

There are two components of linear momentum for each disk in the CM frame. There is the component of momentum radial to the spring (which we will call the radial direction) and the component of momentum perpendicular to the spring (which we will call the tangential direction). The initial momentum of the system in the CM frame along the tangential direction is $P_{i} = m\frac{v_0}{2} - m\frac{v_0}{2} = 0$. Since there is no force in the tangential direction, the tangential momentum of the system is conserved thus at the instant of maximal elongation $P_{f} = m(v_1)_f + m(v_2)_f = 0$ i.e. $v_f \equiv (v_1)_f = -(v_2)_f$. The initial angular momentum in the CM frame is $L_0 = 2m(\frac{l_0}{2})(\frac{v_0}{2})$ and, after using the result obtained through conservation of linear momentum above, the final angular momentum at the instant of maximal elongation is $L_f = 2m(\frac{l_0}{2} + x)v_f$. Since angular momentum is conserved (no torque), we have that $v_f = \frac{l_0v_0}{(\frac{2l_0}{2} + 4x)}$. We then plug this into the conservation of energy equation.

The total radial momentum is also conserved in the CM frame (at any given instant the radial momentum of one disk is equal in magnitude and opposite in direction to the radial momentum of the other disk in the CM frame) but this won't really help us for the problem since both disks have vanishing initial radial momentum relative to the CM and vanishing radial momentum at the instant of maximal elongation relative to the CM.

 

[Tanya Sharma]

Hi WannabeNewton

Superb explanation!

What will be orientation of the spring and the disks at the point of maximum elongation (The spring and the masses are horizontal at the beginning)?

 

[WannabeNewton]

I think to find that, you would have to solve the equations of motion in polar coordinates for $\theta(t), r(t)$, use the fact that $\dot{r} = 0$ at the instant of maximal elongation to, in principle, solve for the time $t$ at which the maximal elongation happens and then plug that into $\theta(t)$ to find the angle subtended with respect to the horizontal. You could also obtain the first integrals of motion coming from the conserved quantities and use them to solve for $\theta(r)$ in principle and use the result from above that tells us the position $r$ of the disks, relative to the CM, at maximal elongation to find the angle subtended with respect to the horizontal, at that instant.

As voko mentioned earlier, this system is very similar to the two body problem from gravitation, except the interaction is through the spring force, so the same method used there (the standard method for solving the two body problem in gravitation also takes advantage of CM coordinates) may be of use here in finding the trajectories of the disks.

I can't think of any easier/slicker method for finding the angle subtended at this very moment. Hopefully someone else knows of such a method and can outline it here. Cheers and best of luck.