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

[WannabeNewton]

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

I accidentally deleted my comment when trying to edit it :[ but if you want you can go ahead and give your reasoning.

 

[dreamLord]

I'll just re-post what WBN said in short ; torque is not zero since at any instant of time since the radius vector w.r.t lab frame of any particle is NOT parallel to the spring force acting on the mass. Hence angular momentum is not conserved.

 

[Tanya Sharma]

WNB,I beg to differ with your view .With respect to a fixed origin,the angular momentum of the CM remains constant .Again,Angular momentum of the masses w.r.t CM remains constant.The total angular momentum of the system comprising spring and masses remains constant w.r.t fixed origin .

Again, where is the force applying net torque on the system ?

 

[WannabeNewton]

Never mind, I agree with Tanya. One can show that if the net force on a system of particles is zero then the net torque on the system is the same about all origins (proof on request). The net force on the spring+disks system must vanish in any inertial frame hence the net torque about the lab fame origin should be the same as the net torque about the CM i.e. zero. If there is a flaw in this argument then maybe ehild or someone else can point it out. What I was thinking of before was the torque about the lab frame origin on a single disk not vanishing, as opposed to the net torque about the lab frame origin on the system.

 

[dreamLord]

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

In the C.M frame, at the instant of maximum elongation, we will have both masses moving only tangentially with a velocity, and a radial force acting on them. How do we prove that this does not result in a circular motion with a constant radius? Will we have to show that $mv^2/(l/2 + x/2) \neq k(l+x)$ meaning the spring force does not provide the required centripetal force?

 

[Tanya Sharma]

There are far more twists and turns in the movie than what I anticipated .

WannabeNewton,I really appreciate your patience and knowledge .

I am feeling pretty embarrased asking so many doubts.But,still I have few more :shy: .

 

[WannabeNewton]

Why would the positions of the disks relative to the CM remain at constant radii from the CM after maximal elongation? The spring force doesn't suddenly switch off, it will continue to pull the disks back inwards. Perhaps I am misunderstanding what you are asking.

 

[ehild]

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) ?

.

During that maximum elongation, the spring is stretched. So it pulls the disks further inward. During that phase of motion the disks gain inward radial velocity, and after the spring regains its original length, the disks will move further inward and compress the spring. The motion is rotation and vibration at the same time. Two-atomic molecules do such motion and you can see nice rotational-vibrational spectrum of them in the infrared domain: transition between rotational levels superimposed to a vibrational transition. ( I saw many of them in my active life working with an IR spectrophotometer.)

ehild

 

[dreamLord]

In any uniform circular motion, the force acts perpendicular to the instantaneous direction of motion, which is what is happening here (at the time when elongation is maximum). My question is - can we show that the spring force, which pulls the disks inward as you said, does not change the magnitude of the velocity, but only it's direction, like in circular motion? I'm not saying that the force switches off - I'm asking if the force is such that it only changes the direction of the velocity vector.

 

[WannabeNewton]

My god ehild you and your pretty diagrams :) Is there a badge for best diagram provider?

 

[ehild]

In any uniform circular motion, the force acts perpendicular to the instantaneous direction of motion, which is what is happening here (at the time when elongation is maximum). My question is - can we show that the spring force, which pulls the disks inward as you said, does not change the magnitude of the velocity, but only it's direction, like in circular motion? I'm not saying that the force switches off - I'm asking if the force is such that it only changes the direction of the velocity vector.

I think it depends on the parameters, k, Vo, m, and lo. It can happen that the spring force provides the centripetal force for rotation with constant angular speed, but it can go on both with rotation and vibration. We should examine it ...

ehild

 

[dreamLord]

So ultimately will we have to prove what I wrote in post #65? In other words, is that a 'test' for any force acting always perpendicular to the velocity vector?

 

[WannabeNewton]

The spring force will change both the magnitude and direction of the radial velocity. As for the angular speed of the rotation about the CM, this will be given in magnitude by $\dot{\theta} = \frac{L}{mr^{2}(t)}$. The only time varying quantity is $r(t)$.

 

[Tanya Sharma]

If instead of a spring,we had a massless ,inextensible string connecting the two masses,how would the situation have been different ?

First,an impulsive tension would act on masses,and afterwards,the system would have translated and rotated about the CM,just as we have in this problem.The tension after a while would have been constant .Right?

 

[voko]

I think it depends on the parameters, k, Vo, m, and lo. It can happen that the spring force provides the centripetal force for rotation with constant angular speed, but it can go on both with rotation and vibration. We should examine it ...

ehild

If the spring is relaxed initially, as in the this problem, circular motion will never result. In circular motion the spring has to be uniformly extended everywhere, and the extension must be just right to provide const centripetal acceleration for the given constant speed.

Any kind of central force field admits circular motion, but initial conditions must be fine tuned.

 

[ehild]

Well, do it in polar coordinates, substituting the two particles with a single one with reduced mass μ=m/2, and r as their distance.

The radial acceleration is

$\ddot r-r{\dot {\theta}}^2=- \frac{k}{\mu}(r-Lo)$

From conservation of angular momentum $r^2 \dot {\theta}=A$ (A is a constant)
Note that conservation of angular momentum allows simultaneous change of both the radius and angular speed.

We arrive at a de for r: $$\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)$$
with initial condition r(0)=Lo and dr/dt(0)=0

ehild

 

[dreamLord]

The spring force will change both the magnitude and direction of the radial velocity.

How do we prove this?

 

[WannabeNewton]

How do we prove this?

The spring force acts along the radial direction in the CM frame so it will change the magnitude of the radial velocity and the change in direction happens periodically due to the spring wanting to restore itself back to the unstretched state. Also see post #76 above.

 

[voko]

The spring force will change both the magnitude and direction of the radial velocity.

The force will change the direction of the entire velocity vector. The radial velocity will change its direction without any force, provided azimuthal velocity is non-zero.

 

[haruspex]

We arrive at a de for r: $$\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)$$
with initial condition r(0)=Lo and dr/dt(0)=0

As usual, multiplying that by $\dot r$ allows it to be integrated to arrive at ... conservation of energy, and quite a nasty DE it is .

 

[dreamLord]

How does ehild's post #76 prove that the magnitude of velocity will necessarily change or in other words that circular motion is not possible with the initial conditions chosen?

 

[voko]

How does ehild's post #76 prove that the magnitude of velocity will necessarily change or in other words that circular motion is not possible with the initial conditions chosen?

If motion is circular, then $r \equiv l_0$, and $\dot{r} \equiv 0$. Plug these into the equation, and you get $$ - \frac {A^2} {l_0^3} = 0 $$ The only solution is $A = 0$, then $\dot{\theta} \equiv 0$, which means there is no rotation at all.

 

[Chestermiller]

Well, do it in polar coordinates, substituting the two particles with a single one with reduced mass μ=m/2, and r as their distance.

The radial acceleration is

$\ddot r-r{\dot {\theta}}^2=- \frac{k}{\mu}(r-Lo)$

From conservation of angular momentum $r^2 \dot {\theta}=A$ (A is a constant)
Note that conservation of angular momentum allows simultaneous change of both the radius and angular speed.

We arrive at a de for r: $$\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)$$
with initial condition r(0)=Lo and dr/dt(0)=0

ehild

I also analyzed this problem using polar coordinates (this problem begs to be solved in cylindrical coordinates), and got the same result that you did. I then linearized the term 1/r³ by expanding in a Taylor series and then solved the resulting 2nd order linear ODE. But, in the end, I got a slightly different final answer than that indicated by the OP:

elongational strain = $$\frac{\frac{mv_0^2}{kl_0^2}}{1+1.5\frac{mv_0^2}{kl_0^2}}$$

Of course, if the elongational strain is small, the second term in the denominator can be neglected.

 

[dreamLord]

Thanks voko! (And ehild, WBN and the others).

 

[ehild]

The relative change of the length of spring u=L/L₀-1 fulfils the differential equation
$$\ddot u-\frac{v_0^2/L_0^2}{(1+u)^3}=-\frac{2k}{m} u$$

If u<<1 he equation can be approximated by

$$\ddot u-\frac{v_0^2}{L_0^2}(1-3u)=-\frac{2k}{m} u$$

that is

$$\ddot u+(\frac{2k}{m}+3\frac{v_0^2}{L_0^2})u=\frac{v_0^2}{L_0^2}$$.

The initial conditions are: u(0)=0 and du/dt=0 at t=0, as initial velocity was perpendicular to the spring.

The solution is of the form $$u=B(1-cos(ωt))$$, where

$$\omega^2=\frac{2k}{m}+3\frac{v_0^2}{L_0^2}$$

and

$$B=\frac{v_0^2 /L_0^2}{2k/m+3v_0^2/L_0^2}$$.

The spring oscillates between the relaxed length Lo and its maximum elongated length. Circular motion is possible if the spring was initially stretched by the appropriate amount.

ehild

 

[Chestermiller]

This is exactly the result that I got, as expressed in post #83.

Chet

 

[ehild]

Sorry, I wanted to mention your post but I forgot ...

ehild

 

[haruspex]

$$B=\frac{v_0^2 /L_0^2}{2k/m+3v_0^2/L_0^2}$$.

I seem to have lost track of why this thread has continued. I thought it had already been established that the book answer was correct. The above may be more accurate, but from the provided information that B << 1 we can deduce that $3v_0^2/L_0^2 << 2k/m$, allowing the $3v_0^2/L_0^2$ term to be dropped.
Btw, I think all versions of the result can be reached more quickly using conservation of energy and of angular momentum. If x is the max elongation and v the speed of each mass at that time, and we use the suggested frame of reference:
Energy $= mv_0^2/4 = mv^2+kx^2$
Ang mom $= mv_0L = 2mv(L+x)$, whence $v = v_0(1+x/L)^{-1}/2$ (since the speed must be purely tangential at max elongation), etc.

 

[ehild]

There was a question what happens after reaching maximum elongation. It is not forbidden to think about the problem further. DreamLord suggested (#65) that the disks will continue to move along a circle of constant radius around the CM. I have shown an approximate solution, which is a rotational-vibrational motion.

ehild

 

[voko]

There was a question what happens after reaching maximum elongation. It is not forbidden to think about the problem further. DreamLord suggested (#65) that the disks will continue to move along a circle of constant radius around the CM.

His concern had been addressed earlier, using exact equations of motion.