Finding the maximum elongation

[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

[itex]\ddot r-r{\dot {\theta}}^2=- \frac{k}{\mu}(r-Lo)[/itex]

From conservation of angular momentum [itex]r^2 \dot {\theta}=A[/itex] (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: [tex]\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)[/tex]
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: [tex]\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)[/tex]
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

[itex]\ddot r-r{\dot {\theta}}^2=- \frac{k}{\mu}(r-Lo)[/itex]

From conservation of angular momentum [itex]r^2 \dot {\theta}=A[/itex] (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: [tex]\ddot r-\frac{A^2}{r^3}=- \frac{k}{\mu}(r-Lo)[/tex]
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 = [tex]\frac{\frac{mv_0^2}{kl_0^2}}{1+1.5\frac{mv_0^2}{kl_0^2}}[/tex]

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
[tex] \ddot u-\frac{v_0^2/L_0^2}{(1+u)^3}=-\frac{2k}{m} u[/tex]

If u<<1 he equation can be approximated by

[tex] \ddot u-\frac{v_0^2}{L_0^2}(1-3u)=-\frac{2k}{m} u[/tex]

that is

[tex] \ddot u+(\frac{2k}{m}+3\frac{v_0^2}{L_0^2})u=\frac{v_0^2}{L_0^2}[/tex].

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 [tex]u=B(1-cos(ωt))[/tex], where

[tex] \omega^2=\frac{2k}{m}+3\frac{v_0^2}{L_0^2}[/tex]

and

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

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]

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

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 [itex]3v_0^2/L_0^2 << 2k/m[/itex], allowing the [itex]3v_0^2/L_0^2[/itex] 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.