How to determine if angular momentum is conserved?

[shawli]

Homework Statement

I'm not really sure what I'm supposed to be checking for in the following question:

A 60.0kg woman stands at the western rim of a horizontal turntable having a moment of inertia of 500 kg m^² and a radius of 2.00 m. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of 1.5m/s relative to the Earth.

Is the angular momentum of the system constant?

Homework Equations

This is mainly a conceptual question, but anywho:

L = I * omega

L initial = L final

The Attempt at a Solution

What do I do with the information that the woman is standing at the western rim? I'm a bit lost with the concepts of this unit that we're doing in class so any clarification/explanations would be much appreciated !

 

[shawli]

Oh, to add -- the back of the book says that L is *not* conserved since the bearing of the turn table exerts a force northward onto the turntable. I don't understand where that comes from at all, could someone please clarify?

 

[rude man]

Think of her and the table standing on the equator. Now think of what the spin axis of the table is doing as the Earth rotates. Is there a change in angular momentum about the spin axis over time? Remember that ω is a vector with magnitude and direction.

EDIT:EDIT: nope, there is no quantitative answer unless latitude λ is included in the answer as a parameter. But I notice none seems to be asked.

New hint: what is precession?

 

[Redbelly98]

Note: I have moved this thread to Intro Physics from Advanced Physics.

I assume "the system" refers to the turntable + woman.

Angular momentum is conserved when there is no net, external torque exerted on the system. Yes, a northward force is exerted on the turntable's axle. Question for the student: what is the torque that results from this force?

 

[Redbelly98]

Think of her and the table standing on the equator. Now think of what the spin axis of the table is doing as the Earth rotates. Is there a change in angular momentum about the spin axis over time? Remember that ω is a vector with magnitude and direction.

EDIT:EDIT: nope, there is no quantitative answer unless latitude λ is included in the answer as a parameter. But I notice none seems to be asked.

New hint: what is precession?

Surely the effects of the Earth's rotation are to be neglected here.

 

[rude man]

Surely the effects of the Earth's rotation are to be neglected here.

One might think so a priori, but not here, especially not in view of the back-of-the-book hint. If you think you can neglect Earth rotation then you would not be welcomed into the fraternity of gyroscope manufacturers where a good inertial navigation gyro is required to not only resolve but measure to better than 0.001 deg/hr! (Earth rotation being of course 15-odd deg/hr).

Besides: why else would there be a North force exerted on the gyro?

 

[rude man]

redbelly -I seem to have lost your last post.

It should be clear that Earth rate was meant to be included by the hint 'at the back of the book' which I think you noticed by now.

Given that, it seems like the table could not be located either at a pole or at the equator. At either pole there would be angular momentum magnitude |L| due to Earth + Sun rotation of 15.04 deg/hr but no change in the angular momentum vector direction, so no dL/dt, and so L would be conserved.

Similarly, at the Equator there would be no net angular momentum magnitude since the woman walking cancels out the table rotation, so even though the angular momentum vector would change due to direction, it doesn't because |L| = 0.

But at other latitudes there is (constant-magnitude) |B| and there is change in direction of L, both due to Earth rotation, so there is a finite dL/dt term and angular momentum is not conserved but changes over time as |dL/dt| = Iω*Ω = IΩsinλcosλ, Ω = 15.04 deg/hr changed to rad/s, λ = latitude.

 

[D H]

Besides: why else would there be a North force exerted on the gyro?

Because the woman starts walking clockwise when viewed from above from a position the western rim of the turntable. In which compass direction is her first step?

This problem has absolutely nothing to do with the Earth's rotation.

 

[rude man]

Because the woman starts walking clockwise when viewed from above from a position the western rim of the turntable. In which compass direction is her first step?

This problem has absolutely nothing to do with the Earth's rotation.

I think otherwise, as you have probably gathered by now. So apparently does her textbook author(s).

Let us amicably agree to disagree.

 

[D H]

I think otherwise, as you have probably gathered by now. So apparently does her textbook author(s).

This would happen on a non-rotating planet. This is homework, so I can't say much more than that.

 

[rude man]

This would happen on a non-rotating planet. This is homework, so I can't say much more than that.

No, on a non-rotating planet, angular momentum of the table & woman would be conserved.

 

[D H]

No, it wouldn't. At what point is the sole external horizontal force to the turntable+woman system applied? Is this the same as the center of mass of the turntable+woman system?

 

[rude man]

No, it wouldn't. At what point is the sole external horizontal force to the turntable+woman system applied? Is this the same as the center of mass of the turntable+woman system?

There is no external torque applied to the table axis in the absence of Earth rate. Where do you dig up this mysterious 'external' force? The distance betw. the table/woman c.g. and the woman has no bearing on the problem.

 

[D H]

There is no external torque applied to the table axis in the absence of Earth rate. Where do you dig up this mysterious 'external' force?

The turntable might be rotating (and it will be), but as a whole it isn't moving. This means the net force on the turntable must be zero. However, the woman is exerting a force on the turntable. To keep the turntable as a whole from moving, there must be a constraint force that balances this force. This constraint force is applied at the bearing of the turntable; see post #2. Ignoring aerodynamic drag, this force at the bearing of the turntable is the sole external horizontal force acting on the turntable+woman system.

 

[rude man]

The turntable might be rotating (and it will be), but as a whole it isn't moving. This means the net force on the turntable must be zero. However, the woman is exerting a force on the turntable. To keep the turntable as a whole from moving, there must be a constraint force that balances this force. This constraint force is applied at the bearing of the turntable; see post #2. Ignoring aerodynamic drag, this force at the bearing of the turntable is the sole external horizontal force acting on the turntable+woman system.

Huh?? Quote from the problem, "The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. >>

This will be my last shot, and may the best man/woman eventually be vindicated:

1. assume non-rotating planet. Initially, woman and table are at rest. As she starts to walk up to speed, at every instant of time the total angular momentum of the table plus her, about the table axis, remains zero. The table moves counterclockwise in step with her walking clockwise, and the relative angular rates of the woman and the table are the ratio of their respective moments of inertia. No external forces possible; angular momentum is conserved.

2. Rotating planet: done that already above.

Cheers!

 

[D H]

1. assume non-rotating planet. Initially, woman and table are at rest. As she starts to walk up to speed, at every instant of time the total angular momentum of the table plus her, about the table axis, remains zero.

This is wrong.

Do the math, but do not post it here. This is a homework problem. Posting solutions to homework problems is not allowed at this site.

 

[tiny-tim]

hi shawli!

Oh, to add -- the back of the book says that L is *not* conserved since the bearing of the turn table exerts a force northward onto the turntable. I don't understand where that comes from at all, could someone please clarify?

never mind whether this is right or wrong … it seems to me to make no sense anyway …

a "force northward" doesn't necessarily create a torque (it depends whether it acts on or off the axis), so it doesn't necessarily affect L

also, since the axle is frictionless, the external force can only be perpendicular to the surface of the axle, ie along the radius and through the centre of the axle, and so exerts no torque about the axle

(two external forces along different lines of action can of course constitute a torque)

Is the angular momentum of the system constant?

since I'm disagreeing with the answer in the book, i'll take the unusual step of giving a full answer …

Euler's equations (for a frame of reference fixed in the body along three (perpendicular) principal axes, and therefore rotating with it) are:
$$\tau_1\ =\ I_1\,\frac{d\omega_1}{dt} + (I_3\ -\ I_2)\omega_2\omega_3$$

$$\tau_2\ =\ I_2\,\frac{d\omega_2}{dt} + (I_1\ -\ I_3)\omega_3\omega_1$$

$$\tau_3\ =\ I_3\,\frac{d\omega_3}{dt} + (I_2\ -\ I_1)\omega_1\omega_2$$

we're only interested in the component along the axis of the turntable (say, the third equation), and since presumably I₂ - I₁ = 0, that reduces to (for the turntable only):

$$\tau_3\ =\ I_3\,\frac{d\omega_3}{dt}$$

the component of the turntable's angular momentum parallel to the axis of the turntable caused by the Earth's rotation is constant, so it does not affect $\omega_3$, so the third equation is the same whether the Earth is rotating or not

so the component of external torque on the turntable parallel to its axis is the same as if the Earth was not rotating, and is equal and opposite to the component of the external torque on the woman, and the total angular momentum parallel to the axis is conserved

(there's probably a simpler way of doing this that doesn't involve Euler's equations … but it's just after midnight here, and i don't see it! :zzz:)

 

[D H]

OK. The cat's out of the bag.

This is a conceptual reasoning problem. It did not ask for an exact answer. The numbers are a bit of a red herring; Euler's equations are even more of a red herring.

Start by looking at the turntable itself. The woman is exerting a force on the turntable. Since the turntable as a whole is not moving, the Earth must be exerting a constraint force that exactly counterbalances the force exerted by the woman. This constraint force is applied at the bearing of the turntable, which is presumably at the turntable's center of mass, so there is no torque on the turntable from this constraint force.

Since the force exerted by the woman is 2 meters away from the center of the turntable, the woman does exert a torque on the turntable. End result: While there is no net force on the turntable but there is a net torque on the turntable.Now look at the woman plus the turntable as a system. From this perspective, the force exerted by the woman on the turntable is an internal force. The force she exerts on the turntable is equal but opposite to the force the turntable exerts on her.This internal force has no effect on the linear momentum of the woman+turntable system. It similarly has no effect on the angular momentum of the woman+turntable system.

However, that external force exerted by the Earth at the turntable bearing does have an effect on both the linear and angular momentum of the woman+turntable system. The woman+turntable system gains a net linear momentum to the north. While the turntable as a whole doesn't move, the woman does.

What about angular momentum? The center of mass of the woman+turntable system is located somewhere between the center of the turntable and the woman. The sole external force on the woman+turntable system is directed northward and is applied at a point that is some distance from the center of mass of the the woman+turntable system. This force results in a non-zero torque on the woman+turntable system, and thus angular momentum is not conserved, either.

 

[Doc Al]

Oh, to add -- the back of the book says that L is *not* conserved since the bearing of the turn table exerts a force northward onto the turntable. I don't understand where that comes from at all, could someone please clarify?

I believe the back of the book (Serway) says that momentum is not conserved, not angular momentum.

 

[Redbelly98]

This has all the makings of an introductory physics problem. I'm still convinced that the Earth's rotation is negligible.

Oh, to add -- the back of the book says that L is *not* conserved

Can you double check that -- are they referring to the angular or the linear momentum?

since the bearing of the turn table exerts a force northward onto the turntable. I don't understand where that comes from at all, could someone please clarify?

The turntable exerts a force on the woman in the ___ direction, in order for her to start moving clockwise from the western edge of the turntable.

By Newton's 3rd law, the woman must exert an equal-but-opposite force on the turntable.

By Newton's 2nd law, the net force on the turntable must be zero since it's CM does not move. Question for you: if the woman exerts a force in the ___ direction, how do we get zero net force on the turntable?

Moreover ... after thinking through this problem, it seems to me there is an ambiguity here. The value of the angular momentum, as well as any torques, depend on where you choose the reference axis to be. Two obvious choices are either (1) the turntable axis, or (2) the system's center of mass. The problem statement does not specify which, at least as it was copied here, so strictly speaking the question is unanswerable. However the question is most easily answered if we use the turntable axis.

 

[rude man]

This has all the makings of an introductory physics problem. I'm still convinced that the Earth's rotation is negligible. .

I will stipulate that there is room for (mis)interpretation here. On condition that you don't think the Earth's rotation is negligible in explaining sunrises and sunsets ... :rofl:

But it would be nice if shawli (the OP) would let us all know what the outcome per his instructor was...

 

[Redbelly98]

I will stipulate that there is room for (mis)interpretation here. On condition that you don't think the Earth's rotation is negligible in explaining sunrises and sunsets ... :rofl:

Agreed Though you could easily start arguing whether air resistance is negligible in projectile motion problem too. It often isn't, yet we assume it is in order to have easily solvable problems that help teach the basic concepts to students.

But it would be nice if shawli (the OP) would let us all know what the outcome per his instructor was...

Yes, especially given that Doc Al appears to have the same book with the same problem but has different information in his back-of-the-book solution.