Falling Spool (no conveyor belts!)

[brossa]

Here’s a balance-of-forces problem that has bothered me for several weeks.

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Take a solid disk of mass m and radius** r** (thickness is not important). This disk has an infinitely long wire of zero mass and thickness wrapped around its circumference. One end of the wire is fixed to a spring scale in the ceiling. The disk is aligned in such a way that as it falls under the acceleration of gravity (g ), the disk begins to spin as the wire unwinds. Furthermore, the unspooling wire remains vertical, so the force vector of the wire is always vertical and tangent to the disk at its edge (I dunno how. Maybe it’s falling/sliding down a frictionless wall).

I know that the moment of inertia for the disk through its center is 1/2mr ^2, and through a point on its edge is 3/2mr ^2. I have been able to show to my own satisfaction that under the above conditions, the tension on the wire as the spool falls/spins up is 1/3mg , and that the center of mass of the disk falls at 2/3g.

However, what I would like to know is what the behavior of the disk is as different forces are applied to the wire. For example, my gut tells me that if a force of mg were applied to the wire the disk would not hang motionless; but I can’t seem to derive the relationship between the force on the wire and the motion of the disk’s center of mass. So, what happens when the force exceeds mg? Does the disk ever not fall, or does its linear acceleration only, say, approach zero as the force on the wire becomes infinite?

My faded memories of college physics have failed me. What say the Dopers? I fear a response that will make me slap my forehead and say “D’oh!”

[David_Simmons]

@brossa

However, what I would like to know is what the behavior of the disk is as different forces are applied to the wire. For example, my gut tells me that if a force of mg were applied to the wire the disk would not hang motionless; but I can’t seem to derive the relationship between the force on the wire and the motion of the disk’s center of mass. So, what happens when the force exceeds mg? Does the disk ever not fall, or does its linear acceleration only, say, approach zero as the force on the wire becomes infinite?

My faded memories of college physics have failed me. What say the Dopers? I fear a response that will make me slap my forehead and say “D’oh!”

An upward force of mg would result in the spool standing still.

Suppose you had the spool on a spindle sticking up from a horizontal frictionless surface and you pulled on the wire with a force of mg. There would be a reaction on the spindle of mg.

This is exactly the case with the falling spool. There is an upward force on the wire of mg and a downward force of mg on the center of the spool.

Other forces on the wire can be handled similarly. There is always a downward force of mg and varying upward forces on the wire.

[David_Simmons]

My analysis was incorrect. Give me some time. It’s been a long time for me too and I need things like this to keep from ossifying.

[treis]

brossa, I believe I have stumbled upon your problem here. I assume you use the following equations to solve this problem:

Ft-mg=Am
Ftr=alphaI
alphar=A

Where Ft=tension in rope, A=linear acceleration of the disk, m=mass of disk, alpha=rotational accleration of the disk and I=moment of inertia for the disk.

Now, lets first combine the last two equations and then plug in mg. Solving for alpha in the third equation and plugging it into the 2nd equation yields:

Ft-mg=AM
Ftr=A*I/r

Ok, now lets plug in mg for Ft. The first equation gives us mg-mg=Am. Therefore A must be 0. Plugging into the second equation gives us mgr=AI/r. Solving for A gives us mgr[sup]2[/sup]/I=A. Since m, g, r and I are not 0 A must be non-zero. So, here we have A=0 and A=something other than 0. If you use Ft=1/2g and then solve as before you get alpha=1/2*g/r and alpha=g/r. Its at this point we say “What da fuck is going on here”.

Well, what is going on here is that the assumption we made for the 3rd equation no longer applies. We know its the 3rd equation that is causing problems becuase if it were one of the first two, well, that would upset, give or take, 350 years of accepted Newtonian physics. But fear not, now that we know Ft we only have two unknown variables and two equations. We can do a force balance and find that A is equal to (mg-Ft)/m. We can sum moments around the center of mass and find that alpha is equal to (2Ft)/(rm). For the case of Ft=mg, A=0 and alpha=2*g/r

Now, what assumptions did we make concerning the 3rd equation? Well we assumed was that there was a point of zero velocity to an observer with the same velocity as the center of mass of the rotating object at the contact between the string and disk. Basically the assumption we make is that the velocity from rotation at that point exactly cancels out the forward linear velocity. From our results in the Ft=mg case we have no linear acceleration or velocity while we do have rotational acceleration and velocity. Obvious since our linear velocity is 0 it can not cancel out our rotational velocity. Thus our 3rd assumption is invalid and unsuprisingly leads to an invalid formula.

Lest you think these two equations fully describes the system we still would want to know how fast we have to pull our rope. To do this we have to use the energy and work relationships. We know that work=Forcedistance and rotational energy=1/2Iomega[sup]2[/sup]. Taking derivitives we get Power=Forcevelocity and Power=1/2Ialpha[sup]2[/sup]. Thus we have four equations now to describe the system:

Power=FtVrope
Power=1/2Ialpha[sup]2[/sup]
MdiskAdisk=Ft-mg
AlphaIdisk=Ftr

[treis]

Doh, I am going to have to do the same thing as David Simmons did and ask for a moment to reconsider my work.

[treis]

@treis

Now, what assumptions did we make concerning the 3rd equation? Well we assumed was that there was a point of zero velocity to an observer with the same velocity as the center of mass of the rotating object at the contact between the string and disk. Basically the assumption we make is that the velocity from rotation at that point exactly cancels out the forward linear velocity. From our results in the Ft=mg case we have no linear acceleration or velocity while we do have rotational acceleration and velocity. Obvious since our linear velocity is 0 it can not cancel out our rotational velocity. Thus our 3rd assumption is invalid and unsuprisingly leads to an invalid formula.

The bolded section should be “to an observer with the same velocity as the rope”. Now, this assumption remains valid for all cases so long as we are talking about the reference frame of the rope. But we are interested in the velocity of the disk relative to an observer on the earth. That means the only time this assumption is valid for an obsererver on the earth is when the velocity of the rope relative to the earth is 0. Now, for values other than Ft=1/3mg the velocity of the rope relative to the earth has to be something other than 0.

Why? Well think of the case of Ft=mg. Using the moment equation we know that the disk is accelerating rotational and thus the energy of the disk is increasing. The disk is not moving up or down so no energy is being changed from potential into rotational. The only other place for energy to come from in this case is applying a force through a distance. The only force that is being applied is from the rope so it must be in motion.

@treis

Lest you think these two equations fully describes the system we still would want to know how fast we have to pull our rope. To do this we have to use the energy and work relationships. We know that work=Forcedistance and rotational energy=1/2Iomega2. Taking derivitives we get Power=Forcevelocity and Power=1/2Ialpha2. Thus we have four equations now to describe the system:

Power=FtVrope
Power=1/2Ialpha2
MdiskAdisk=Ft-mg
AlphaIdisk=Ftr

The idea is right in this case but my execution is poor. The power equations I came up with are wrong but at the moment I don’t know how to make them right. I will come back to this when it isn’t 4 in the morning and figure it out though.

[treis]

To make it four posts in a row:

If you just care about the motion of the disk for a given Ft then you don’t need to worry about the energy stuff. The up and down acceleration of the disk is Ft-mg=mA and the rotational acceleration is Ftr=I*alpha.

[nd_n8]

Here’s a crack at it:

@brossa

However, what I would like to know is what the behavior of the disk is as different forces are applied to the wire. For example, my gut tells me that if a force of mg were applied to the wire the disk would not hang motionless; but I can’t seem to derive the relationship between the force on the wire and the motion of the disk’s center of mass. So, what happens when the force exceeds mg? Does the disk ever not fall, or does its linear acceleration only, say, approach zero as the force on the wire becomes infinite?

If you drop a yo-yo, keeping the speed of the string at 0, it accelerates toward the ground at the speed of gravity (normal gravity, frictionless string and frictionless environment to eliminate any terminal velocity, most yo-yos are too short for terminal velocity anyway) until it reaches the end. If you pull upward with a force equal to the weight of the yo-yo and a constant speed the yo-yo continues to fall relitive to the earth at an accelaration rate of speed of gravity minus speed of string. If the string accelarates at the speed of gravity the yo-yo remains fixed in relation to the earth. The force on the string increases in a linear fashion equal to the amount of downward force applied by gravity, the longer the yo-yo remains in “stationary free fall” relative to the earth, the greater the force required to accelerate the string to overcome gravity. As to the rotation of the disk (yo-yo) as the disk falls it rotates faster and makes the required acceleration of the string possible. I believe this relationship is also linear.

[nd_n8]

@nd_n8

If you pull upward while the yo yo is falling downward with a force equal to the weight of the yo-yo and a constant speed the yo-yo continues to fall relitive to the earth at an accelaration rate of speed of gravity minus speed of string. If the string accelarates at the speed of gravity the yo-yo remains fixed in relation to the earth. The force on the string increases in a linear fashion equal to the amount of downward force applied by gravity, the longer the yo-yo remains in “stationary free fall” relative to the earth, the greater the force required to accelerate the string to overcome gravity. As to the rotation of the disk (yo-yo) as the disk falls it rotates faster and makes the required acceleration of the string possible. I believe this relationship is also linear.

Forgot to add the part in bold, for the sake of this argument the assumption is that the yo-yo is always going down. If that assumption is removed then it is fairly easy to keep the yo-yo in place while moving the hand up and down. Kids like it and it comes in handy when trying to understand the acceleration of gravity.

[brossa]

David Simmons, treis, and nd_n8, thanks for your answers. I particularly liked nd_n8’s point that the wire must accelerate upwards at g in order to hold the spool in place.

I initially had a follow-up question: if the wire was not fixed to the ceiling, but instead passed over a massless frictionless pulley, what mass M would have to be added to the end of the wire to keep the mass m stationary (with respect to the ground; it would be spinning up continuously). It would appear to me now that this would require an infinite mass M, because only when M >>>>m can we have the state where M is essentially in free-fall, and thus accelerating downward at g.

Does that seem correct?

[nd_n8]

@brossa

David Simmons, treis, and nd_n8, thanks for your answers. I particularly liked nd_n8’s point that the wire must accelerate upwards at g in order to hold the spool in place.

I initially had a follow-up question: if the wire was not fixed to the ceiling, but instead passed over a massless frictionless pulley, what mass M would have to be added to the end of the wire to keep the mass m stationary (with respect to the ground; it would be spinning up continuously). It would appear to me now that this would require an infinite mass M, because only when M >>>>m can we have the state where M is essentially in free-fall, and thus accelerating downward at g.

Does that seem correct?

Ummm, no? I think that mass M would need to equal mass m exactly. I’m guessing of course but it seems that as mass M pulls the wire down accelerating at speed g, the unspooling wire from mass m would be flying up, also accelerating at speed g and thus hold mass m spinning stationary relative to the earth. This is because mass M is attached to the wire and mass m is not (ok, technicaly it is but only inside of the windings. Until the length runs out the only force the wire will exert on mass m will be the acceleration of g). In real life, mass M would need to be heavier, initially, by the factor of pully friction (and other frictions of course), but as the process continues, would need to get lighter due to the mass of the extra wire being pulled down with it to maintain equalibrium. In the original frictionless environment where the wire has no mass, both the original and counter masses need to be the same.

[brossa]

Originally posted by nd_n8:
Ummm, no? I think that mass M would need to equal mass m exactly. I’m guessing of course but it seems that as mass M pulls the wire down accelerating at speed g, the unspooling wire from mass m would be flying up, also accelerating at speed g and thus hold mass m spinning stationary relative to the earth.

But I don’t think that mass M falls at acceleration g, because there is tension on the wire that tends to ‘hold up’ M. Take the initial case of the wire being fixed to the ceiling - as the spool unravels, there is still tension = 1/3mg on the wire. So if the wire was looped over a massless frictionless pulley and a weight of M =1/3m were hooked onto it, it seems to me that M would hover in place while the spool unravelled and fell. A mass M greater than 1/3m would fall slowly, with the acceleration of the fall approaching g as M goes to infinity. At least I *think *so.

[treis]

@nd_n8

Here’s a crack at it:

If you drop a yo-yo, keeping the speed of the string at 0, it accelerates toward the ground at the speed of gravity (normal gravity, frictionless string and frictionless environment to eliminate any terminal velocity, most yo-yos are too short for terminal velocity anyway) until it reaches the end.

This is wrong, if the yo-yo is spinning then it cannot be accelerating downwards at the rate of g or you would be creating energy.

@brossa

So if the wire was looped over a massless frictionless pulley and a weight of M =1/3m were hooked onto it, it seems to me that M would hover in place while the spool unravelled and fell. A mass M greater than 1/3m would fall slowly, with the acceleration of the fall approaching g as M goes to infinity. At least I think so.

This is not possible for two reasons. The first is that if your string is unwinding the amount of string between the disk and the mass is increasin. That means that the distance between the mass and the disk has to be increasing since your rope isn’t bunching up or disappearing. Furthermore since your disk is accelerating rotationally the string is unwinding faster and faster. Becuae of that the amount of rope between is increasing faster and faster. With no bunching that means there must be relative acceleration between the disk and the mass. Since we want a Ft on the disk that means the mass must be accelerating downward. Therefore the mass must have a weight greater than Ft.

The second is that energy is going into the disk. Energy can not be created so in this system in must come from the change in height of the masses. Becuase the change in energy is not constant we know that the mass has to be falling accelerating downwards.

I know that in order to solve for the velocity you have to use energy equations but at the moment I can’t figure out how to do so.

[David_Simmons]

I haven’t given up yet. I’ve been a little occupied elsewhere. In addition, I got an answer that I don’t understand. Just need to clear a little more rust. Is it possible that rust is all there is left?

One thing though, brossa, the moment of inertia around the axis of the spool is the only one needed. The spool is rotating around its axis and not around a point on the circumferance.

[brossa]

David Simmons said:
The spool is rotating around its axis and not around a point on the circumferance.

Well, the reason that I put that in there was that at one point I was trying to think of the disk/wire system from a different frame of reference. I was trying to work out the problem as though the disk was trying to rotate around a point on its circumference. Just as much of a dead end as the other ways I was using to try to work it out.

treis said:
This is not possible for two reasons. The first is that if your string is unwinding the amount of string between the disk and the mass is increasin. That means that the distance between the mass and the disk has to be increasing since your rope isn’t bunching up or disappearing. Furthermore since your disk is accelerating rotationally the string is unwinding faster and faster. Becuae of that the amount of rope between is increasing faster and faster. With no bunching that means there must be relative acceleration between the disk and the mass. Since we want a Ft on the disk that means the mass must be accelerating downward. Therefore the mass must have a weight greater than Ft.

I’m not sure where our disagreement lies. I also think that one or both of the masses will fall, accelerating as it does so, with the length of wire between them increasing over time. It seems to me that the behavior of the system depends on the ratio of the masses. In case I didn’t make it clear, the second mass M is not a spool; just a sack of sugar or something. I’ll try to lay out what I *think *will happen for various mass ratios, but bear in mind that I can’t establish any of this rigorously (which is sort of what the O.P. was about):

Case 0: wire fixed to ceiling
The spool’s COM falls at 2/3g and there is tension 1/3mg on the wire.

Case 1: wire goes from spool, over pulley, to mass M where M < 1/3 m
The mass M rises slowly, while the spool falls faster than 2/3g. I suppose that it spins up at a slower rate than in case 0. The energy to lift M and spin/linearly accelerate m comes from the fall of m.

Case 2: same as 1 but M=1/3m
In this one I think the tension in the wire balances the weight of the second mass, so M remains stationary and m falls/spins as though the wire were fixed to the ceiling.

Case 3: M>1/3m
The tension on the wire is now greater than 1/3mg, and M falls at some acceleration less than g while the spool’s COM falls slower than 2/3g. I suppose that it spins up faster than in case 2? The spool probably falls faster than M for smaller M’s

Case 4: some ratio of M:m - not sure what, exactly.
M and m fall with the same linear acceleration; but m also has angular acceleration. The angular acceleration of m is also probably higher than in case 3. The linear accel of both masses is less than 2/3g.

Case 5: ratio M:m approaches infinity
M falls with acceleration approaching g. The wire is being ripped off the spool at acceleration approaching g. The spool hovers without falling, but gets its highest angular acceleration; greater than in case 4. Energy to spin up m comes from the fall of M. Tension on the wire approaches mg.

Now, the above comes from my (probably faulty) logic alone, not math, so may all be totally wrong. I wonder what the right ratio of masses is for case 5?

[treis]

Ok, I believe I have solved the problem. There are four unknowns: Acceleration of the mass=A, Acceleration of the disk=Ad, Rotational acceleration of the disk=alpha and the Force in the rope=Ft. The first three equations come from force and moment balances on the bodies. Other values are Ms=mass of the disk, M=mass of the mass, r=radius of disk, g=gravity and Is=moment of inertia of the disk. The last comes from the boundary conditions that result from the connection of the rope. I get:

MsAs=Ft-Msg
MA=Ft-mg
Isalpha=Ftr
-A=alphar+As

Solving those and inputting for I results in:

A=g*(Ms-3m)/(Ms+3m)
As=-g(m+ms)/(3m+ms)

For the cases you list your are right except for case 4 and case 5. In case 4 the ratio of masses is 1:1. Intuitavely this should make sense becuase you have identical weights being subject to identical Fts thus their As are the same.

For an infinately large M the acceleration of the mass is just g. Plugging in* infinity for m in the second equation results in an acceleration of 1/3g.

*I know plugging in isn’t the correct terminology but I don’t want to add any needless confusion.

[treis]

@treis

Acceleration of the disk=Ad

Oops, I used Acceleration of the spool=As instead of Ad.