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

[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.

 

[Tanya Sharma]

Okay...Things look better if we work from the CM frame.Lets leave this frame for a while .

I had worked from the Lab frame in#41 .What modifications do we have to make in #41 when we are working the problem in Lab frame ?

Can't we capture the essence of the problem in the ground frame? I apolozise for my limitations in understanding the problem at hand.

 

[dreamLord]

In that case angular momentum will not be conserved, as there is a torque acting on the masses.

 

[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 .

The conservation of linear momentum and angular momentum is OK, but it is not, that the disks can only move along the spring. Initially, one disk is given a kick initially somehow, that was an external force. Imagine that the spring is rigid so it is straight during the motion, but under the initial impulsive force it can bend a bit, so all points of the spring get some velocity (fig 1a). After the kick, the CM will move horizontally with Vo/2. Figure b shows the situation just after the kick in the CM frame of reference. The disks move with Vo/2 and -Vo/2 horizontal velocities, so they stretch the spring. Therefore they experience the spring force (the green arrows), accelerated inward by it, changing the direction of the velocity. Their outward radial components decrease. At the end, at maximum elongation, the velocities become normal to the spring, and the whole system rotates about the CM as a rigid body at that instant (fig.c)

ehild

 

[Tanya Sharma]

In that case angular momentum will not be conserved, as there is a torque acting on the masses.

No torque acts on the system comprising spring and masses .Angular momentum is definitely conserved.

 

[WannabeNewton]

I'm not sure to what extent you want the lab frame to be used. What I mean by this is that even in the lab frame, we can describe the motion at the instant of maximal elongation as a translation of the CM (the CM starts at the origin with velocity $\frac{v_0}{2}$ and since $\frac{\mathrm{d} P_{\text{net}}}{\mathrm{d} t} = 0$, the CM moves in a straight line with constant velocity $\frac{v_0}{2}$ starting from the origin) plus a rotation of the disks about the CM. This allows us to write the kinetic energy at that instant, in the lab frame, as $T_f = \frac{1}{2}(2m)(\frac{v_0}{2})^{2} + \frac{1}{2}k(2x)^{2} + \frac{1}{2}I_{cm}[(\omega^2_{1})_{f} + (\omega^2_{2})_f]$ where $(\omega_{1})_f$ and $(\omega_{2})_f$ are the angular velocities of the disks about the CM at the instant of maximal elongation. We can find $(\omega_{1})_f$ and $(\omega_{2})_f$ using conservation of angular momentum in the exact same way as before and we will get the same answer as before after using conservation of energy (the initial kinetic energy is just $T_i = \frac{1}{2}mv_0^{2}$).

EDIT: ehild that is a perfect diagram. Why are you so awesome :)?

 

[ehild]

EDIT: ehild that is a perfect diagram. Why are you so awesome :)?

I do not have enough brains to understand complicated problems without imagining a picture

ehild

 

[dreamLord]

No torque acts on the system comprising spring and masses .Angular momentum is definitely conserved.

Yes, silly answer on my part. No external forces. Don't know what I was thinking.

 

[Tanya Sharma]

The conservation of linear momentum and angular momentum is OK, but it is not, that the disks can only move along the spring. Initially, one disk is given a kick initially somehow, that was an external force. Imagine that the spring is rigid so it is straight during the motion, but under the initial impulsive force it can bend a bit, so all points of the spring get some velocity (fig 1a). After the kick, the CM will move horizontally with Vo/2. Figure b shows the situation just after the kick in the CM frame of reference. The disks move with Vo/2 and -Vo/2 horizontal velocities, so they stretch the spring. Therefore they experience the spring force (the green arrows), accelerated inward by it, changing the direction of the velocity. Their outward radial components decrease. At the end, at maximum elongation, the velocities become normal to the spring, and the whole system rotates about the CM as a rigid body at that instant (fig.c)

ehild

Excellent explanation :)

Does that mean ,once the spring attains maximum elongation,the masses rotate about the CM ,while spring maintaining this maximum elongation (i.e there is no further change in spring length) ?

EDIT:I agree WNB,ehild is truly awesome.

 

[dreamLord]

That is what I was thinking as well. Torque on a system can exist even if external forces do not - is that right?

 

[WannabeNewton]

Excellent explanation :)
Does that mean ,once the spring attains maximum elongation,the masses rotate about the CM ,while spring maintaining this maximum elongation (i.e there is no further change in spring length) ?

No it will definitely change. The spring force will pull the disks back.