Two Opposing Spinning Connected Flywheels

May 2014
10
0
Imagine two flywheels in space connected via a mechanism which ensures they spin in opposite directions.

An observer on a space platform at rest to the flywheels observes two incoming cannonballs from the right approaching the flywheels.

Each of the flywheels will capture one cannonball and lock it with a mechanism attached on the edge of the flywheels.

The flywheels weight 500kg and have a radius of 1 meter.

The cannonballs weight 100kg and have a radius of 20cm. They approach the flywheels at 10 m/s.

here is a picture to give a little better understanding of the above



http://imgur.com/E98x6ff


I have many questions about the outcome of this scnenario, but let me begin with a more general one.

How would the observer on the space platform, at rest to the flywheels which are not spinning before the cannonballs hit, see the flywheels behave once they captured the cannonballs?
 
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ChipB

PHF Helper
Jun 2010
2,369
294
Morristown, NJ USA
The two flywheels are initially not spinning, so have no kinetic energy nor angular momentum. The two canon balls carry both energy and momentum. After they are captured the flywheel/canon ball system will be spinning with momentum conserved. The fact that the two flywheels are tied together is immaterial. Suppose the flywheels are called 'A' and 'B': from conservation of momentum for flywheel A you have:

mrv_1 = I v_2/r + mrv_2

and for B:

-mrv_1 = -I v_2/r - mrv_2

where 'm' is the mass of the canon ball, v_1 is the initial velocity of the canon ball, I is the moment of inertia for the flywheel, and r is the radius of the flywheel. From this you can calculate the value for v_2, which is the final velocity of each canon ball (which also equals the velocity of the rim of the flywheel). Note that total angular momentum for the combined A-B system is conserved - it is initially 0 and is also 0 after the system starts turning. Also note that because this is not an elastic collision some of the initial kinetic eregy of the two canonballs is lost.

Why do you emphasize that the observer on the adjacent platform is at rest? Are you thinking of relativistic effects? At a mere 10 m/s classic Newtonian mechanics works just fine.
 
May 2014
10
0
Why do you emphasize that the observer on the adjacent platform is at rest?

You say the angular momentum is conserved, but what about the linear momentum of this system.
From an observer's perspective on the platform, the whole system initially has zero angular momentum and just linear momentum from the cannon balls.

After the cannon balls hit, it remains at zero angular momentum, but where is the linear momentum?

Does the flywheel stay at rest relative to the platform or does it have any linear momentum, constant/altering velocity to the left?


Are you thinking of relativistic effects? At a mere 10 m/s classic Newtonian mechanics works just fine.
Maybe, but the platform is required nevertheless if you want to describe the motion of the flywheels. Without a reference point, it is impossible.


edit: Adding to the above, we also add another object to this experiment. Let's say a hollow tube which weight just as much as the flywheels combined (1000kg). The hollow tube also captures two cannonballs.
This is added in order to compare the energies.
Will the total energy of the hollow tube capturing the two cannon balls inside (not spinning) be the same as that of the flywheels?
 
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ChipB

PHF Helper
Jun 2010
2,369
294
Morristown, NJ USA
After the cannon balls hit, it remains at zero angular momentum, but where is the linear momentum?

Does the flywheel stay at rest relative to the platform or does it have any linear momentum, constant/altering velocity to the left?
Yes, good point. The flywheel/canon ball system will have the same final linear momentum as the two canon balls start with. If M = mass of each flywheel and m = mass of each canon ball:

(2m+2M)V_2 = 2mV_1


edit: Adding to the above, we also add another object to this experiment. Let's say a hollow tube which weight just as much as the flywheels combined (1000kg). The hollow tube also captures two cannonballs.
This is added in order to compare the energies.
Will the total energy of the hollow tube capturing the two cannon balls inside (not spinning) be the same as that of the flywheels?
Remember that these are inelastic collisions, so energy is not conserved. If you run the calculation to find V_2 and v_2 you can then calculate the final kinetic energy. For the flywheels you get:

KE = (1/2)(2I omega^2 + (2m +2M) V_2^2)

and for the cylinder (whose mass is 2M):

KE = (1/2) (2m +2M)V_2^2

From this you can see that the flywheel system ends up with more final KE. Stated another way - it looses less energy to heat in the collision with the canon balls. By my calculations the system in both cases starts with total KE of 10,000 Joules, and after collision the spinning flywheel system has 4523 Joules while the cylinder system has only 1667 Joules.
 
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May 2014
10
0
Yes, good point. The flywheel/canon ball system will have the same final linear momentum as the two canon balls start with. If M = mass of each flywheel and m = mass of each canon ball:

(2m+2M)V_2 = 2mV_1




Remember that these are inelastic collisions, so energy is not conserved. If you run the calculation to find V_2 and v_2 you can then calculate the final kinetic energy. For the flywheels you get:

KE = (1/2)(2I omega^2 + (2m +2M) V_2^2)

and for the cylinder (whose mass is 2M):

KE = (1/2) (2m +2M)V_2^2

From this you can see that the flywheel system ends up with more final KE. Stated another way - it losses less energy to heat in the collision with the canon balls. By my calculations the system in both cases starts with total KE of 10,000 Joules, and after collision the spinning flywheel system has 4523 Joules while the cylinder system has only 1667 Joules.
Unfortunately, it seems like i need a new experiment to get answers to the question i am trying to ask. I actually planned to ignore the losses via friction/heat, but there is maybe a better way to do it.
 
May 2014
10
0
What i am getting at is...

_Neglecting_ energy escaping via heat/friction from the cannonball/flywheels system...

How can we explain the linear momentum to be conserved in this system, when the cannonballs will pass some angular momentum to each of the two wheels.

While the total angular momentum remains zero at all times, the two flywheels certainly require energy to spin.

This energy the two flywheels take to spin, will be missing for the kinetic energy in the linear motion to the left. Hence the cannonballs/flywheel system cannot possibly have the same linear momentum to the left as the formerly incoming 2 cannonballs.

What am i missing here?
 

ChipB

PHF Helper
Jun 2010
2,369
294
Morristown, NJ USA
Don't confuse the principles of conservation of momentum with conservation of energy. Linear momentum is conserved and angular momentun is conserved, but kinetic energy is not conserved. Some of the initial linear KE of the canon balls is used to spin the wheels, some is used to give linear KE to the wheels, and some is lost as heat or other energy loss in the inelastic collision (i.e. to force the canon balls to "stick" to the flywheels and not bounce backwards).
 

BvU

Jun 2014
2
0
The two flywheels are initially not spinning, so have no kinetic energy nor angular momentum. The two canon balls carry both energy and momentum. After they are captured the flywheel/canon ball system will be spinning with momentum conserved. The fact that the two flywheels are tied together is immaterial. Suppose the flywheels are called 'A' and 'B': from conservation of momentum for flywheel A you have:

mrv_1 = I v_2/r + mrv_2

and for B:

-mrv_1 = -I v_2/r - mrv_2

where 'm' is the mass of the canon ball, v_1 is the initial velocity of the canon ball, I is the moment of inertia for the flywheel, and r is the radius of the flywheel. From this you can calculate the value for v_2, which is the final velocity of each canon ball (which also equals the velocity of the rim of the flywheel). Note that total angular momentum for the combined A-B system is conserved - it is initially 0 and is also 0 after the system starts turning. Also note that because this is not an elastic collision some of the initial kinetic eregy of the two canonballs is lost.

Why do you emphasize that the observer on the adjacent platform is at rest? Are you thinking of relativistic effects? At a mere 10 m/s classic Newtonian mechanics works just fine.
I have a lot of questions about this. Not only about the conclusions, but also about the assumptions. Your "conservation of momentum" looks like conservation of angular momentum to me. In a frame that gets a big kick and then has a uniformly, linearly moving center of mass that for the observer oscillates around the axis of rotation...

Later on:
KE = (1/2)(2I omega^2 + (2m +2M) V_2^2)
What is I here? The rotational energy of the disk only ? Don't the balls rotate around the disks axis (and also around their own....)

##and how does one present an equation to one's readers in this forum? As on a typewriter ? ## (Speechless)
 

ChipB

PHF Helper
Jun 2010
2,369
294
Morristown, NJ USA
Your "conservation of momentum" looks like conservation of angular momentum to me.
Yes, that's correct.

Later on:
What is I here?
I=Moment of inertia of a flywheel.


##and how does one present an equation to one's readers in this forum? As on a typewriter ? ##
Yes. Or you can use a 3rd-party site such as texify.com to convert a LaTeX expression into a link to a graphic. Take a look through some of the posts in the LaTeX Help forum: http://physicshelpforum.com/forumdisplay.php?f=51
 

BvU

Jun 2014
2
0
I=Moment of inertia of a flywheel.
aren't the cannonballs rotating arouond the same axis as the wheels ? Their moment of inertia doesn't count ?

I take it we've lost Geronimo by now, but I am intirigued by this problem, so I should invest some time to think it over; in the mean time any contributions wrt how to deal with it are definitely appreciated.