What determines that a system carries no angular momentum?

[KittyCat1534]

Homework Statement

So I had a question regarding a specific part of a question that I'm trying to understand regarding momentum? In the picture attached, I'm wondering why there is no angular momentum when the boy walks radially inward. Thanks!

Relevant Equations

P = M * Vector V

 

[PeroK]

To be precise, you are looking at angular momentum about the centre of the merry-go-round. Any object moving in a straight line towards the centre has no angular momentum about the centre.

If you throw something onto the merry-go round towards the centre, then it contributes no angular momentum to the system. If you throw something onto the merry-go-round in any other direction, then you have additional angular momentum (about the centre) in your system.

 

[hmmm27]

If the boy just leaps straight to the center, without otherwise stepping on the ride, then there is no angular momenta changes.

 

[haruspex]

If the boy just leaps straight to the center, without otherwise stepping on the ride, then there is no angular momenta changes.

That is not exactly the issue here. The text considers the angular momentum of the system consisting of the apparatus plus boy. Since that is a closed system, there is no angular momentum change. The next question is what the system's a.m. was before the boy jumped on. This is independent of whether he stepped onto the periphery or leapt to the centre.

Even if looking at just the a.m. of the apparatus, it wouldn't matter whether the boy leapt to the centre or made his way there more cautiously, on reaching the centre the a.m. would be restored.

@KittyCat1534 , in case you are wondering why a radial approach brings with it no angular momentum about the axis, it is because the angular momentum about an axis equals the linear momentum multiplied by the perpendicular distance from the approach path to the axis. For a radial approach, that distance is zero.

 

[Frodo]

If the boy just leaps straight to the center, without otherwise stepping on the ride, then there is no angular momenta changes.

He was not rotating before he jumped on so he had no angular momentum relative to the axis of rotation..

He is rotating when he is standing at the centre so he now has angular momentum.

Where did he get it from? The roundabout which must, therefore, slow down to give him his angular motion.

As an aside I was somewhat surprised to find that an object traveling in a straight line has angular momentum and obeys Kepler's Second Law of sweeping equal areas in equal time.

How much angular momentum does it have? Any value I want. All I do is locate the axis about which the angular momentum is measured appropriately to give me any value I want from zero to infinity.

 

[haruspex]

He is rotating when he is standing at the centre so he now has angular momentum.

True, but for the purposes of the question it is reasonable to treat the boy as a vertical pole.

 

[Frodo]

Homework Statement:: I'm wondering why there is no angular momentum when the boy walks radially inward.

1. Once the boy is on the roundabout there are no external forces being applied to the roundabout/boy system. You can consider it a closed system (ignoring friction, air resistance etc).

2. If there are no external forces being applied the angular momentum of the system cannot change.

So the angular momentum must remain constant.

But ...

... the angular velocity will increase as the boy walks inward because mass is moving closer to the axis. It is like a spinning skater: Her angular momentum is constant but when she draws in her arms, she moves some of her mass closer to the axis and so her angular velocity increases.

Walking towards the central pole will feel extremely peculiar to the boy because his legs will not go where he thinks they should be going.

If you get the opportunity to try a roundabout like this (one where you stand on a platform and have access to the central pole), face the pole, hold on tightly, and swing your leg back and forth in a radial direction. You will find your leg swings wildly to the side. It is a most strange feeling (and why you need to hold on tightly)!

Similarly, try to kick the central pole and you will find your leg swerves wildly to the side. You are thinking that you need to kick radially relative to the roundabout platform (along a radial line painted, say, on the platform). However, you are revolving so you actually need your foot to follow a curved path to compensate for your motion.

 

[hmmm27]

So... while he slows the spinning platform as he steps on, it will return to its original spin by the time he gets to the center... ?

 

[haruspex]

So... while he slows the spinning platform as he steps on, it will return to its original spin by the time he gets to the center... ?

In the idealised model, yes.

 

[Frodo]

So... while he slows the spinning platform as he steps on ...

Correct

... it will return to its original spin by the time he gets to the center... ?

Not quite correct. It will be rotating slightly slower than originally.

He will be rotating when he is at the centre so he will have some angular momentum. Hence the roundabout must have given him some, so the roundabout must be rotating slightly slower than it was originally rotating.

I say slightly because the roundabout is a lot heavier than the boy so has much more angular momentum than he does.

Do find a roundabout and do the tests I suggested - it is a weird feeling. If you want to learn more look up Coriolis force.

 

[hmmm27]

He will be rotating when he is at the centre so he will have some angular momentum. Hence the roundabout must have given him some, so the roundabout must be rotating slightly slower than it was originally rotating.

OK, but by the convention of physics homework problems, he's a massy point or line.

 

[hutchphd]

His name would be Delta Function, 2D then?
The physics is not changed by whether it is zero or just small. Ann don't forget the parallel axis theorem is at work here.

 

[Frodo]

OK, but by the convention of physics homework problems, he's a massy point or line.

And that assumption opens up a whole can of worms we had better not get into here - it deserves a thread of its own.

And a word of advice for KittyCat1534. If in doubt, give both answers and explain the assumptions leading to them. " If I assume he is a point then ...", and "If I assume he is not a point, then ...". Somewhat paradoxically, this problem (the boy at the centre) is simpler if you assume he has size.

 

[haruspex]

give both answers and explain the assumptions leading to them

Why? None of this arcane discussion about whether the boy is a point mass and what happens if he migrates to the centre of the merry-go-round has anything to do with the question the student was given.
The boy approaches the merry-go-round, running radially, and leaps on. The natural assumption is that he leaps onto the periphery of the merry-go-round and stays there. Nothing in the problem statement suggests otherwise.

Post #1 does not quote the book's answer, but it looks like it is heading for "the merry-go-round's angular velocity decreases". As one would hope.

It would be interesting to know what the book says about the girl.

 

[hmmm27]

It would be interesting to know what the book says about the girl.

Your wish, etc...

■ On the Playground A merry-go-round is rotating freely when a boy runs radially inward, straight toward the merry-go-round’s center, and leaps on. Later, a girl runs tangent to the merry-go-round’s edge, in the same direction the edge is moving, and also leaps on. Does the merry-go-round’s angular speed increase, decrease, or stay the same in each case? EVALUATE Because the merry-go-round is rotating freely, the only torques are those exerted by the children as they leap on. If we consider a system consisting of the merry-go-round and both children, then those torques are internal, and the system’s angular momentum is conserved. In Fig. 11.7 we’ve sketched the situation, before either child leaps onto the merry-go-round and after both are on board. The boy, running radially, carries no angular momentum (his linear! momentum and the radius vector are in the same direction, making L zero), so you might think he doesn’t change the merry-go-round’s angular speed. Yet he does, because he adds mass and therefore rotational inertia. At the same time he doesn’t change the angular momentum, so with I increased, v must therefore drop. Running in the same direction as the merry-go-round’s tangential velocity, the girl clearly adds angular momentum to the system—an addition that would tend to increase the angular speed. But she also adds mass, and thus increases the rotational inertia—which, as in the boy’s case, tends to decrease angular speed. So which wins out? That depends on her speed. Without knowing that, we can’t tell whether the merry-go-round speeds up or slows down. ASSESS The angular momentum the girl adds is the product of her linear momentum mv and the merry-go-round’s radius R, while she increases the rotational inertia by mR2. With small m and large v, she could add a lot of angular momentum without increasing the rotational inertia significantly. That would increase the merry-go-round’s rotation rate. But with a large m and small v—giving the same additional angular momentum—the increase in rotational inertia would more than offset the angular momentum added, and the merry-goround would slow down. We can’t answer the question about the merry-go-round’s angular speed without knowing the numbers.

 

[haruspex]

Your wish, etc...

Looks good.

Thanks

 

[KittyCat1534]

That is not exactly the issue here. The text considers the angular momentum of the system consisting of the apparatus plus boy. Since that is a closed system, there is no angular momentum change. The next question is what the system's a.m. was before the boy jumped on. This is independent of whether he stepped onto the periphery or leapt to the centre.

Even if looking at just the a.m. of the apparatus, it wouldn't matter whether the boy leapt to the centre or made his way there more cautiously, on reaching the centre the a.m. would be restored.

@KittyCat1534 , in case you are wondering why a radial approach brings with it no angular momentum about the axis, it is because the angular momentum about an axis equals the linear momentum multiplied by the perpendicular distance from the approach path to the axis. For a radial approach, that distance is zero.

Oh I see, now I understand it a little bit more, thanks for the help! @haruspex

 

[KittyCat1534]

@haruspex I can show you the the diagram of what happen as well if you want