Rotation speed for a particular system under gravity

[RobertGC]

TL;DR Summary

Classical physics problem involving conservation of energy and conservation of angular momentum.

You have a rope hanging over a fixed support with a heavy weight at one end and a lighter weight at the other end. You set the end of the rope with the lighter weight spinning in a circle and let the heavy weight end fall under gravity. As the heavy end falls the length of the rope that is spinning will decrease. By conservation of angular momentum the rotation speed of the light weight at the one end will increase. (The familiar example of the spinning skater whose speed increases as he draws his arms in.)

But you have also the issue of the conservation of energy of the whole system taking into account the heavy weight falling under gravity. The question is what will be the speed of the two weights taking into account both these conservation principles? You may assume the rope weightless and doesn’t twist up and kink from the rotation.


Robert Clark

 

[jbriggs444]

The question is what will be the speed of the two weights taking into account both these conservation principles?

Are you assuming that the trajectory will be circular? For most initial conditions, I would not expect it to be. Nor would I expect a simple closed orbit -- though I cannot immediately rule that out.

It seems straightforward to compute the radial and tangential velocities of the spinning weight associated with any radial position for that weight, given the initial conditions. A numerical prediction of its trajectory should then be simple. A prediction for the periapsis and apoapsis (low point and high point of the orbit) is just a matter of some simple algebra. (Probably solutions to a quadratic).

If we take the problem as stated -- circular path... then the heavy weight must remain motionless, the tension in the cord will be $Mg$ and we can set this equal to the centripetal force, $\frac{mv^2}{r}$, and immediately read off the required initial velocity for the light weight.

 

[RobertGC]

Yes. I was assuming it would remain circular. A more complex physics problem if it did not!

I don’t think the heavy weight would always remain fixed. For instance suppose the heavy weight was, say, a hundred times that of the lighter weight and you only rotated the lighter weight at an initial small speed. The heavier weight falling under gravity would seem to dominate.

It would be an interesting question to test out!

Also, I used the word “gravity” but the small body is not really in orbit. It’s just rotating around attached to a string.

Robert Clark

 

[jbriggs444]

For instance suppose the heavy weight was, say, a hundred times that of the lighter weight and you only rotated the lighter weight at an initial small speed.

Then the orbital path of the lighter weight will not be a circle.

It will still have a periapsis and an apoapsis. You can solve for those.

 

[Dale]

This would be a central force with a $r^0$ force instead of the usual $r^{-2}$ force

https://en.m.wikipedia.org/wiki/Classical_central-force_problem

 

[jbriggs444]

Let me try to outline the calculation.

You can split the total system kinetic energy into three parts:

1. The kinetic energy of the large weight: $\frac{1}{2}M{v_r}^2$
2. The kinetic energy corresponding to the radial motion of the small weight: $\frac{1}{2}m{v_r}^2$
3. The kinetic energy corresponding to the tangential motion of the small weight: $\frac{1}{2}m{v_t}^2$.

Meanwhile, one has the system's potential energy. If we use the initial position of the large weight as our reference then this is given by $Mg(r-r_0)$

The initial system energy is given by $\frac{1}{2}m{v_0}^2$.

Conservation of energy then allows us to write:$$\frac{1}{2}m{v_0}^2 + Mgr_0 = \frac{1}{2}(M+m){v_r}^2 + \frac{1}{2}m{v_t}^2 + Mgr$$I will assume that the initial velocity is purely tangential. Conservation of angular momentum then allows us to write:$$mv_0r_0 = mv_tr$$Now consider what happens at periapsis and apoapsis. In both situations, the radial velocity is zeroed out.

Applying angular momentum conservation, we then have that $v_t = \frac{v_0r_0}{r}$

Applying energy conservation [edit: and dividing by m] we have that $$\frac{1}{2}{v_0}^2 + \frac{Mg}{m}r_0 = \frac{1}{2}v_t^2 + \frac{Mg}{m}r = \frac{1}{2}{(\frac{v_0}{r_0})}^2r^2 + \frac{Mg}{m}r$$The only free variable here is $r$. This is a quadratic in $r$. The left hand side is a constant depending only on the initial conditions. The right hand side has one term which is a constant multiple of $r^2$ and one which is a constant multiple of $r$.

One solution to the quadratic will be $r=r_0$. That will be either apoapsis (if initial velocity is low) or periapsis (if initial velocity is high). The other solution will fall out from the quadratic formula and will reveal the height of the other orbital extreme. If the initial conditions are just right, you'll wind up with a duplicated root and a circular orbit. [You could probably also extract the other root with synthetic division, but I am old and lazy and my algebra skills are not quite up to par].

 

[jbriggs444]

This would be a central force with a $r^0$ force instead of the usual $r^{-2}$ force

https://en.m.wikipedia.org/wiki/Classical_central-force_problem

This assumes that the light weight is negligibly massive compared to the heavy weight. Otherwise, I believe that the kinetic energy associated with the motion of the heavy weight interferes dangling weight is negligibly massive compared to the spinning weight. Otherwise, the tension in the cord will not be closely approximated by a constant $F=Mgh$. See my working above.

My initial intuition was the same as yours.

 

[Dale]

This assumes that the light weight is negligibly massive compared to the heavy weight. Otherwise, I believe that the kinetic energy associated with the motion of the heavy weight interferes dangling weight is negligibly massive compared to the spinning weight. Otherwise, the tension in the cord will not be closely approximated by a constant $F=Mgh$. See my working above.

My initial intuition was the same as yours.

Oh, right. It isn’t quite a central potential because the force depends on the second derivative of the radial coordinate.

 

[RobertGC]

Yes. Because the length of the rotating string is shrinking the path of the smaller weight can not remain circular.
I have a question about the calculation in post #6 however. It would seem there should be a solution for r depending on time. But the equation provided only has 2 or only 1 solution.

Robert Clark

 

[jbriggs444]

Yes. Because the length of the rotating string is shrinking the path of the smaller weight can not remain circular.

The length of the rotating portion of the string may be shrinking initially, but in the absence of frictional losses it will reach a minimum and then grow again.

I have a question about the calculation in post #6 however. It would seem there should be a solution for r depending on time. But the equation provided only has 2 or only 1 solution.

I was solving for the maximum $r$ and the minimum $r$. Apoapsis and periapsis. Naturally there are two solutions, one for each. Or just the single solution in the case of a perfectly circular orbit.

As I mentioned in post #2 above, the same idea can be used to compute $v_r$ and $v_t$ for any permissible value of $r$.

You know $v_t$ immediately from angular momentum conservation. Divide angular momentum by $r$ and out pops $v_t$.

With $v_t$ in hand, you can then compute $v_r$ from energy conservation. You know the original total energy. You know potential energy from $r$. You know tangential kinetic energy from $v_t$. Whatever is left over is radial kinetic energy. Solve $E_r = \frac{1}{2}(M+m){v_r}^2$ for $v_r$.

Note that you cannot calculate the sign of $v_r$ from energy conservation alone. Only its absolute value. Similarly, you can only determine the absolute value of the trajectory's angle from the horizontal in this way. You cannot tell whether the angle is upward or downward. The fact that the trajectory is smooth helps you out in this respect. The trajectory angle can only change sign at periapsis and apoapsis.

While the equations do not give you position as a function of time, they do give you a differential equation. You could attempt to solve that. Or you could just iterate forward in time with a computer program. You can even use Excel for this.

Note also that you only have to simulate one half of one orbit. The second half will be the mirror image of the first. Then the next half will be the same as the original half. And so on.

If you turn friction on, you get a new term in the equations. There will be an energy drain from the system at a non-constant rate and a possibly lop-sided in-spiral. While Excel could be used to model this, it lacks the graphic capability to display a proper x-y plot in polar coordinates.

Interesting question... With friction, is the resulting inspiral finite or infinite? e.g. an Archimedean spiral is finite but an exponential spiral is infinite.

Assume a smooth in-spiral. The tangential velocity is responsible for centripetal force approximately equal to $Mg$. The required velocity and, accordingly, the energy loss rate will scale with the square root of $r$. The orbital energy will scale directly with $r$.

So we have a pool of energy depleting at a rate proportional to its square root. $\frac{dE}{dt} = -c\sqrt{E}$. What function has a first derivative that is its own square root? Let us try something of the form $t^k$. Like $t^2$. Oooh, that guess works out well. So we would expect energy and orbital radius to follow a parabolic arc to the bottom while tangential velocity follows a straight line graph to zero in finite time.

Or more directly without the differential equation, if the descending weight is approaching the rope limit, we have an orbitting weight with velocity $v$ subject to a retarding acceleration of magnitude $\mu g$ so the orbitting weight slows to zero speed in finite time given by $\frac{v}{\mu g}$.

 

[RobertGC]

If it does reach a minimum radius and start growing again, then presumably once it reaches again a maximal radius it will start decreasing again. Then the appearance should be of the heavy weight periodically going up and down, in concert with the smaller weight going in and out, instead of the heavy weight undergoing an expected continual descent due to gravity.

That would be quite interesting to observe experimentally!

Robert Clark

 

[RobertGC]

Let me try to outline the calculation.
…

Applying angular momentum conservation, we then have that $v_t = \frac{v_0r_0}{r}$

Applying energy conservation [edit: and dividing by m] we have that $$\frac{1}{2}{v_0}^2 + \frac{Mg}{m}r_0 = \frac{1}{2}v_t^2 + \frac{Mg}{m}r = \frac{1}{2}{(\frac{v_0}{r_0})}^2r^2 + \frac{Mg}{m}r$$The only free variable here is $r$. This is a quadratic in $r$. The left hand side is a constant depending only on the initial conditions. The right hand side has one term which is a constant multiple of $r^2$ and one which is a constant multiple of $r$.

One solution to the quadratic will be $r=r_0$. That will be either apoapsis (if initial velocity is low) or periapsis (if initial velocity is high). The other solution will fall out from the quadratic formula and will reveal the height of the other orbital extreme. ..

There may be a problem with this last equation. Using $v_t = \frac{v_0r_0}{r}$, then in the last equation the r² and the r₀² should be in the reversed positions.

Robert Clark

 

[jbriggs444]

They may be a problem with this last equation. Using $v_t = \frac{v_0r_0}{r}$, then in the last equation the r² and the r₀² should be in the reversed positions.

Yes indeed, you are correct. That messes up a nice theory with a pesky fact.

 

[jbriggs444]

If it does reach a minimum radius and start growing again, then presumably once it reaches again a maximal radius it will start decreasing again. Then the appearance should be of the heavy weight periodically going up and down, in concert with the smaller weight going in and out, instead of the heavy weight undergoing an expected continual descent due to gravity.

You should only expect a continual descent if you expect friction.

If you had continual descent to the rope limit together with angular momentum conservation then you would wind up with the orbitting weight moving infinitely fast and exerting infinite force.

If you had a continual descent only to a circular orbit where angular momentum is conserved and a force balance is achieved then you would find yourself violating energy conservation.

You would also find yourself violating macroscopic reversibility. In the absence of friction the laws of physics are time reversible. You can run the film backward and get a valid physical scenario. You cannot run a stable circular orbit backward to an unstable beginning. So you cannot run an unstable beginning forward to a stable circular orbit.

 

[RobertGC]

Thanks for that. The set up for your equations looks valid. Only, the issue with the r and r₀ would mean it’s a cubic in r. Rather unexpected. I would love to see an experimental demonstration of how it looks.

Robert Clark

 

[jbriggs444]

Thanks for that. The set up for your equations looks valid. Only, the issue with the r and r₀ would mean it’s a cubic in r. Rather unexpected. I would love to see an experimental demonstration of how it looks.

Right you are. I was already working through that. I have an explanation for how it plays out and a plan for how to solve the cubic. You can get some good thinking done on a long walk. However, it is a struggle against laziness.

A cubic with two real solutions will normally have three real solutions. You cannot have only two solutions except in edge cases where the graph of the cubic goes tangent to the x axis at some point. Otherwise, if you have two real solutions, you must have a third. [I think that this is a theorem by Galois. Any complex solutions for a polynomial with real coefficients will come in pairs, complex conjugates of each other]

At first glance, this would seem to make a cubic implausible. Surely something must have gone wrong with the anaysis. But wait... What if one of the real solutions is unphysical? For instance, what if one of the solutions has $r < 0$?

That would correspond to the dangling weight dropping below the rope limit and the orbitting weight spinning on the opposite side of the origin at the end of a rope segment with negative length.

The plan to solve the cubic is to multiply the corrected equation through by $r^2$ (to turn it into a cubic in the first place) and then since $r=r_0$ is a known solution, carry out synthetic division by $(r-r_0)$ to obtain a quadratic. The quadratic would then be expected to have two solutions, one positive and one negative.

Right now, the laziness is winning. Having a plan is enough. Carrying it out is too much effort.

Joke...

An experimental physicist is in a hotel room. There is a pitcher of water on the night stand. A fire erupts in the room. The physicist pours out a glass of water precisely sized to douse the blaze, extinguishes the fire and goes back to sleep.

An engineer is in a similar hotel room in a similar circumstance. When the fire erupts, the engineer pours from the pitcher directly on the blaze, extinguishing it and then goes back to sleep.

A mathematician in the same circumstance wakes up, sees the fire, declares that a solution exists and returns to sleep.

Ask me some time how a mathematician captures a lion in the savannah.

 

[A.T.]

You should only expect a continual descent if you expect friction.

If the friction is only opposing the motion of the hanging mass, but not the motion of the spinning mass, then I would expect it to settle in a circular path.

 

[RobertGC]

Let me try to outline the calculation.

…

Applying energy conservation [edit: and dividing by m] we have that $$\frac{1}{2}{v_0}^2 + \frac{Mg}{m}r_0 = \frac{1}{2}v_t^2 + \frac{Mg}{m}r = \frac{1}{2}{(\frac{v_0}{r_0})}^2r^2 + \frac{Mg}{m}r$$The only free variable here is $r$. This is a quadratic in $r$. The left hand side is a constant depending only on the initial conditions. The right hand side has one term which is a constant multiple of $r^2$ and one which is a constant multiple of $r$.

One solution to the quadratic will be $r=r_0$. That will be either apoapsis (if initial velocity is low) or periapsis (if initial velocity is high). The other solution will fall out from the quadratic formula and will reveal the height of the other orbital extreme. If the initial conditions are just right, you'll wind up with a duplicated root and a circular orbit. [You could probably also extract the other root with synthetic division, but I am old and lazy and my algebra skills are not quite up to par].

Putting r and r₀ in the right spots the equation becomes:
$$\frac{1}{2}{v_0}^2 + \frac{Mg}{m}r_0 = \frac{1}{2}v_t^2 + \frac{Mg}{m}r = \frac{1}{2}{(\frac{v_0}{r})}^2r_0^2 + \frac{Mg}{m}r$$
When you subtract the two kinetic energy terms from each other and two potential energy terms from each other you see the result is an equation where r - r₀ can be factored out. Then the equation simplifies to:
$$\frac{1}{2}v_0^2(\frac{r+r_0}{r^2}) = \frac{Mg}{m}$$
A quadratic equation in r. This will indeed have one positive solution and one negative solution, so only one physical solution.
But it may not oscillate as I suggested before. It may remain in equilibrium with the heavy weight in fixed position and the light weight rotating in a constant radius.


Robert Clark

 

[RobertGC]

Another question. Will the small weight also descend due to gravity while it is rotating?

If so, will angular momentum still be conserved in such a scenario? I’ve tried looking this up on the internet and not gotten a definitive answer on this.

If it does descend, then the equations will be more complicated, whether or not angular momentum is conserved.

Robert Clark

 

[jbriggs444]

Another question. Will the small weight also descend due to gravity while it is rotating?

Ahhh, a picture helps. I had imagined the upper puck on a something like an air hockey table. I

 

[RobertGC]

I’m informed in the conical pendulum case the total force of the tension force along the string and the gravitational force downwards is horizontal, so absent frictional and air drag forces the pendulum bob will rotate in a horizontal circle. Then your analysis should be valid.

Still, as far as experimental verification because you might need high initial speed of the light weight to get a (near) horizontal initial rotation path, you might want to use an air table for testing it at various speeds with the initial path still being horizontal.

Note, since the initial question was for the speed, using $$v_t = v_0r_0/r$$ we can derive:

$$v_t(v_t + v_0) = 2Mgr_0/m$$

For v₀ being relatively small, v_t should be significantly larger, so will about the square-root of the term on the right.

Note then for the heavy weight M being much larger than the small weight m, and r₀ being large, we can get a large speed for v_t


Robert Clark

 

[jbriggs444]

I’m informed in the conical pendulum case the total force of the tension force along the string and the gravitational force downwards is horizontal, so absent frictional and air drag forces the pendulum bob will rotate in a horizontal circle. Then your analysis should be valid.

That is nonsense, I am afraid. There is no assurance that the resultant of tension and gravity on the orbitting weight will sum to $\frac{mv^2}{r}$ in the horizontal direction and to zero in the vertical direction. There is also no guarantee that the tension that achieves the requisite balance will match $Mg$.

If it were so for one case, any modification to the mass of the dangling weight or to the initial velocity of the orbitting weight would make it no longer so.

Still, as far as experimental verification because you might need high initial speed of the light weight to get a (near) horizontal initial rotation path, you might want to use an air table for testing it at various speeds with the initial path still being horizontal.

Right. If you adjust the initial conditions so that $Mg\cos \phi = \frac{mv^2}{r}$ and $Mg \sin \phi = mg$ then the orbit will be circular and stable. But I thought the whole point of the original problem statement was that the dangling weight would descend and produce some interesting dynamic behavior. A stable circular orbit is boring.

The conical pendulum has quite a few degrees of freedom. Even if we discard the current rotation angle, $\theta$, I count five state variables: $r$, $\phi$ below the horizontal, $v_\phi$ pitching up and down, $v_\theta$ yawing along the orbit and $v_r$ reeling in and out.

 

[RobertGC]

If the light weight is descending under gravity then the analysis and the equations become more complicated than the case where it is assumed to travel in a horizontal plane.

Robert Clark

 

[A.T.]

If the light weight is descending under gravity then the analysis and the equations become more complicated than the case where it is assumed to travel in a horizontal plane.

Yes, but I expect the qualitative behavior to be similar: If the initial conditions don't give you a horizontal circular orbit, then you have an oscillation. Just that here the non-circuital orbits will not lie in the horizontal plane.