Conservation of angular momentum

[ruggerodk]

http://www.lsa.umich.edu/physics/demola ... spx?id=376

Mount a spring weight-meter to the handle, and read the pull/push (inward/outward) force.

It's like an old brake governor mech with springs...

Ruggero ;-)

[Trevor Lyn Whatford]

Hi all,

So what happens when there is just one weight at the outer rim, with a wheel speed about 10 RPM, what happens when it is pulled half way in? what is the input force to move the weight in? and how much acceleration would there be? and what would the wheel speed be when the weight is returned to the outer rim?

Regards Trevor

[eccentrically1]

The same thing would happen; the wheel would accelerate to 40 rpm. The force required depends on the mass of the weight. You would need some method to pull it in like the string on the batteries. The wheel speed would decel to 10 rpm when the weight is let go.

[smith66]

@ruggerodk,
That's something I thought of tryinb once but didn't tbink people would find interesting.
What might be missed is witb a 1 meter radius that honza suggested, momentum and inertia have the same potential.
Yet inertia is the force necessary to hold a moving mass in it's orbit. In a wheel it would be considered stress.

@Trevor,
The 10 rpm would increase to 20 rpm if 1/2 the radius. If the weight started at top center and had a line attached to it like the previously mentioned tether ball, then it's radius might be able to be reduced. It would still have it's same momentum but would have a shorter distance to travel.

@eccintrically1,
There is a difference between a weight gaining momentum because gravity is accelerating and having a motor power a shaft which is accelerating it.
With gravity, torque comes from the weight. This is why the velocity of the weight wouldn't change.

edited to add; there might be a simple way to test conservation of momentum wbile attempting perpetual motion.
It would use 2 weights and something similar to the tether ball example to conserve energy.

[Wubbly]

To answer Honza's original question, the work required to pull the weights in does result in an increase in linear momentum, no increase in angular momentum, and no increase in overall system energy.

There is NO increase in overall energy because the work performed (energy used) pulling the mass in, is equal to the energy gain (kinetic energy increase) of the spinning system.


(This assumes zero friction and point masses, so you would actually get a little less kinetic energy out of it than the work you put into it.)

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Details of the attached spreadsheet

You can use a spreadsheet to calculate the work required to pull the spinning mass inward (as in the example of the batteries shown in the previous youtube link ( http://www.youtube.com/watch?v=9MGQJar8dNg ) but this spreadsheet example uses a single mass.

-Start with a 1 kg mass, traveling at 1 m/s at a radius of 2 meters.
-Decrement the radius by a small amount.
-Calculate the new tangential velocity to satisfy conservation of angular momentum. v2=r1*v1/r2
-Calculate the new centripital force on the mass.
-Calculate the work performed to move the mass inward based on the centripital force and the distance moved.
-Keep a running total of the work performed.
-Calculate angular momentum to verify it remains constant
-Calculate tangential linear momentum and see what it does.
-Calculate the new moment of inertia at the new radius and calculate the new angular velocity.
-Recalculate the kinetic energy based on the moment of inertia and the angular velocity, and verify it matches the kinetic energy based on the tangential linear velocity.
-Decrement the radius and recalculate all fields.
-When the radius becomes half its original value, compare the total work performed (moving the mass inward) to the total increase in kinetic energy.
-The total work required to pull the mass inward was 1.5 Joules. The rotational velocity increased to have a kinetic energy increase of 1.5 Joules.
-Conclusion: The work required to pull the mass inward is equal to the increase in kinetic energy of the rotating system.

[eccentrically1]

Nice chart, Wubbly! I didn't know the values for angular velocity are supposed to match the KE values.
This is a good reference for everyone.

[smith66]

I guess what you 2 are saying is that gravity can not be converted into mechanical energy. OK :-)

[Wubbly]

ecc. It just happened to work out that way for the numbers selected. Angular velocity and KE have different units, so a matching number between the columns is meaningless.

If you download the spreadsheet, there's a tab where the radius changes from 1 to 0.5 meters and it has a different initial velocity. The numbers in that tab between the angular velocity and KE columns do not match.

The point is to notice that the work performed is equal to the KE increase. While linear momentum increases, you still don't get any energy increase in the overall system.

[smith66]

@All,
I've tried logging out on my phone 3x and it hasn't worked.
Am prepping for surger in the morning. It will be about a month or so before I'll be able to demonstrate one way angular momentum might be conservec in a closed system.

edited to add; the reason why I have a different perspective is because I would be considering 2 values.
And when the over balanced weight moves towards center, no energy will be required to move it. This should allow for either balanced (conservation) or over balanced states of energy.

[Trevor Lyn Whatford]

Hi all,

So what happens when there is just one weight at the outer rim, with a wheel speed about 10 RPM, what happens when it is pulled half way in? what is the input force to move the weight in? and how much acceleration would there be? and what would the wheel speed be when the weight is returned to the outer rim?

Regards Trevor

Hi all,

Could anyone give me a link to one weight only set of experiments, as I would like very much to see them, to know if they are the exact same effects ( chair experiments ) as the two weights experiments? I would expect a degree of oscillation and a quicker wind down of motive force in our gravity for one weight experiments, than that of the two weight experiments, In both the horizontal and vertical axis? Is there a maths variation in formulas, or is it just 50% less for the one weight (edit,) Horizontal axis experiments?

With thanks, Trevor

Edit, change, vertical axis to horizontal axis!

[Wubbly]

Here is a link for some of what you requested: http://www.besslerwheel.com/forum/viewt ... 755#109755

[eccentrically1]

ecc. It just happened to work out that way for the numbers selected. Angular velocity and KE have different units, so a matching number between the columns is meaningless.

If you download the spreadsheet, there's a tab where the radius changes from 1 to 0.5 meters and it has a different initial velocity. The numbers in that tab between the angular velocity and KE columns do not match.

The point is to notice that the work performed is equal to the KE increase. While linear momentum increases, you still don't get any energy increase in the overall system.

What a coincidence, huh?
But I don't see the radius at .5 m? It changes from 2 to 1 m by .1 increments.
Why do you have a column for linear momentum? That matches the linear tangential velocity, since they have different units? Isn't the momentum angular only, nothing is going in a straight line?

The other point is momentum is unchanged, not counting friction.

[Wubbly]

Why do you have a column for linear momentum?

The point mass has a tangential velocity. If the mass were to suddenly release from the rotating system, it would have a velocity and a linear momentum. The linear momentum column was put there to compare to the angular momentum. It is interesting to notice that both linear and angular momentum are not both conserved at the same time. If one is conserved, the other is not. And even though linear momentum was increasing, the overall system energy was not increasing. It might be confusing to refer to the linear momentum of a rotating system. Some people believe linear momentum is the only true momentum and they draw incorrect conclusions about rotating systems. I threw it in there just for fun.

But I don't see the radius at .5 m? It changes from 2 to 1 m by .1 increments.

The post has two attachments. One was a picture of the spreadsheet, the other attachment was the actual spreadsheet. If you want to see the other tabs on the spreadsheet, you have to download the actual spreadsheet, and have access to Microsoft Excel. The other tabs broke it down into .05 meter increments, then .025 meter increments, then fiddled with radius start and end points, and fiddled with the initial velocity, to see if the one case was a fluke. All cases produced the same results of the work performed equaled the kinetic energy increase.

The other point is momentum is unchanged, not counting friction.

That's the way the spreadsheet was designed. Angular momentum was forced to be conserved to calculate the new velocity at the next radius increment. The numbers I was interested in were the work and energy relationship if angular momentum was forced to be conserved. If a discrepancy was found, a loophole in the physics could be exploited, but no such discrepancy was discovered.

[eccentrically1]

The other tabs broke it down into .05 meter increments, then .025 meter increments, then fiddled with radius start and end points, and fiddled with the initial velocity, to see if the one case was a fluke.

Oh I see the tabs at the bottom now, thanks.

No loopholes that I see.

It might be confusing to refer to the linear momentum of a rotating system.

I have to admit I'm not getting it. I thought the system has to be rotating and moving in a straight line at the same time to have both. Like a baseball curving toward the plate for instance. Rather than spinning around a stationary axis.

[ruggerodk]

There is NO increase in overall energy because the work performed (energy used) pulling the mass in, is equal to the energy gain (kinetic energy increase) of the spinning system.

Let me introduce another perspective to this discussion:

Two discs, same weight but with different mass distribution rolling downhill on a slope (gravity) - They both reach the same uphill end position.

But what if...the mass distribution change while rolling downhill? From center to perimeter?

Will it reach the same end position?

I believe it will!

regards ruggero ;-)