Trouble with conservation laws

[LKL]

Suppose we have a system, devoid of any external forces acting on it, consisting of a cylindrical flywheel (mass 10 kg = mf, diameter 1 m) and an object located on the rim (mass 1 kg = mo), and rotating about a central axis at 100 rad/s = w. Thus, the object would have an orbital velocity of 50 m/s = v. The moment of inertia for the flywheel would be 1.25 kgm^2 = Jf, and for the object 0.25 kgm^2 = Jo.

 

The total mechanical energy of the system consists of the rotational energies of both the flywheel and the object. Etot = 0.5*(Jf + Jo)*w^2 = 7500 J.

 

Now, once the link between the object and flywheel is discontinued, the object will continue moving along the tangent at 50 m/s. If we calculate the kinetic energy for the object, Eko = 0.5*mo*v^2 = 1250 J and the rotational energy of the flywheel Erf = 0.5*Jf*w^2 = 6250 J, we'll notice that together they add up to Etot, so energy is conserved.

 

The question is, what happens to both the angular and linear momentum of the system. If we calculate the situation on the basis of the conservation of energy, it would seem like the total angular momentum decreases once the object is let loose and the system somehow gains linear momentum at the same time.

 

If we calculate the situation on the basis of the conservation of momentum, it would seem like the system gains 2875 J from nowhere, which is clearly impossible!

 

I've never heard of such a thing as transition from angular to linear momentum, but at the moment, I just don't know how to calculate the situation taking into consideration all three conservation laws. Could someone with a little more knowledge in mechanics please help?

[swansont]

Suppose we have a system, devoid of any external forces acting on it, consisting of a cylindrical flywheel (mass 10 kg = mf, diameter 1 m) and an object located on the rim (mass 1 kg = mo), and rotating about a central axis at 100 rad/s = w

 

You have a system rotating about an axis that does not go through its center of mass. You don't get to claim that it is also devoid of forces.

[LKL]

Wow, missed that! The "devoid of external forces" was mainly for simplicity, but I guess if you added a counterweight on the opposite side of the flywheel, ditching one of the weights would now have some sort of a recoil effect on the flywheel+counterweight-system due to the shifting of the center of mass. That's a lot of help already! Any ideas how to calculate the resulting linear/angular momentum changes in the flywheel, while still making the energies match?

[Johnny5]

Wow, missed that! The "devoid of external forces" was mainly for simplicity, but I guess if you added a counterweight on the opposite side of the flywheel, ditching one of the weights would now have some sort of a recoil effect on the flywheel+counterweight-system due to the shifting of the center of mass. That's a lot of help already! Any ideas how to calculate the resulting linear/angular momentum changes in the flywheel, while still making the energies match?

 

Do you have a diagram of this somewhere? Maybe provide a link to a similiar problem? I think it's a good problem.

[swansont]

Wow, missed that! The "devoid of external forces" was mainly for simplicity, but I guess if you added a counterweight on the opposite side of the flywheel, ditching one of the weights would now have some sort of a recoil effect on the flywheel+counterweight-system due to the shifting of the center of mass. That's a lot of help already! Any ideas how to calculate the resulting linear/angular momentum changes in the flywheel, while still making the energies match?

 

Once you make the modification, you should be able to apply conservation of linear and angular momentum and describe the resulting behavior - the center of mass of the flywheel will move opposite the piece that flies off, and you will have to measure the angular momentum consistently (I think the parallel axis theorem will be used here), but it should all work out.

[LKL]

Yeah, I'll have to look into this more closely. First time I heard about the parallel axis theorem, dug it up and it looks like tailor-made for this situation. Thanks for help! Please let me know if you stumble across sites that deal with similar situations of "breaking from orbit".

[Johnny5]

(I think the parallel axis theorem will be used here), but it should all work out.

 

You both seem to think the parallel axis theorem can be used here.

 

Do you have an explanation of how that gets used?

 

I understand the focus of the post is using three conservation laws.

 

Conservation linear momentum

Conservation of angular momentum

Conservation of energy

 

The conservation laws give you the equations, you need as many equations as you have unknowns, and then you can solve for whatever it is you are looking for. But I don't see how the parallel axis theorem fits in.

[swansont]

You both seem to think the parallel axis theorem can be used here.

 

Do you have an explanation of how that gets used?

 

I understand the focus of the post is using three conservation laws.

 

Conservation linear momentum

Conservation of angular momentum

Conservation of energy

 

The conservation laws give you the equations' date=' you need as many equations as you have unknowns, and then you can solve for whatever it is you are looking for. But I don't see how the parallel axis theorem fits in.[/quote']

 

 

If you know the moment of inertia around one axis, you can use the theorem to find the moment about another axis - the moments won't be the same. Once the part breaks off the orbital axis will change. Say you wanted to calculate the angular momentum about this new axis. Rather than calculate the moment directly, which would be difficult because you don't have symmetry about the rotational axis, you can use the parallel axis theorem to find the new moment of inertia.

[Johnny5]

If you know the moment of inertia around one axis, you can use the theorem to find the moment about another axis - the moments won't be the same. Once the part breaks off the orbital axis will change. Say you wanted to calculate the angular momentum about this new axis. Rather than calculate the moment directly, which would be difficult because you don't have symmetry about the rotational axis, you can use the parallel axis theorem to find the new moment of inertia.

 

Here is hyperphysics on parallel axis theorem

[LKL]

Think I got it almost solved now, only one conservation law to go! The shifting of the center of mass changes the rotational inertia for the flywheel+counterweight-system only minutely, the new rotational (axis being the center of mass) is 1.47727 kgm^2 as opposed to 1.5 kgm^2 (axis being the center of the flywheel). In the beginning the system's (flywheel + two 1kg weights) total mechanical energy was 8750 J. Once the weight is released, its kinetic energy amounts to 1250 J. Applying the conservation of linear momentum law, the flywheel will gain an opposite velocity of about 4.55 m/s, kinetic energy 113.64 J. Its rotational energy is 7386.35 J, if we assume it's angular velocity doesn't change. Adding up, we get 8750 J. The problem is, since the rotational inertia of the flywheel system decreases and since its angular velocity isn't increased in the process, its angular momentum is not conserved. Could the released object be thought of as having angular momentum even if it's not actually orbiting anything? Any ideas?

[swansont]

Yes. You measure angular momentum about a particular axis. L = r x p So if an object is not moving radially away from the axis, it will have some value of rotational angular momentum.

[LKL]

Three out of three! Thanks for help.

[LKL]

"Breaking from orbit" is still causing me trouble...

 

Suppose we have a uniform rod of any given length rotating about its center of mass. For simplicity, let's say that the rod's axle is supported in such a way that all forces acting on it cause only diminutive changes in the velocity of the system.

 

Now, let's say that one end of the rod hits a baseball. Suppose the baseball clings to the rod. Before impact the baseball's velocity relative to the axle was 0, after impact its velocity is Wr (angular velocity*length of rod/2) in the line of the tangent. Let's call the impulse that accelerates the baseball from rest to orbital velocity I1. An equal and opposite impulse will act on the axle, let's call this I2. The baseball will cling to the end of the rod for one full revolution. Since the rod is now rotating about an axis that is not its center of mass, there will be a net force acting on the axle in the direction of the baseball. After one revolution the impulse caused by this net force, I3, will be 0. Once the rod has rotated for one rev, the baseball is let go. According to the previous example in this thread, there should now be a recoil impulse, I4, acting on the axle. But since I1 + I2 + I3 = 0, this would lead to linear momentum not being conserved. So does I4 exist in this situation? Where's the flaw?

 

Help appreciated.

[swansont]

there will be a net force acting on the axle

 

As soon as that is true, linear momentum will not be conserved.

[LKL]

The axle will exert a (centripetal?) force on the baseball to keep it in orbit. The baseball will exert an equal and opposite force on the axle. The axle mount will exert a countering force on the axle (since the axle is supported rigidly). So no net force on the axle. Anyway, if we consider the impulse of the force exerted on the axle by the orbiting baseball, it will be 0 after one revolution. You can't get a net impulse by rotation alone, so therefore I3 is still 0. Right? What bothers me is that once the baseball is caught, there will be an impulse to the axle, I2, and once it is released, there should also be an impulse on the axle, I4, according to the previous example. Could you consider the baseball receiving an impulse I5 = -I4 upon breaking from orbit?

[swansont]

Lienar momentum/impulse isn't the best way to attack the problem. Linear momentum is not conserved, and you are looking at impulses that happen at different times.

 

You can look at the baseball right before and after the release and use conservation of linear momentum of the ball (i.e. assume no force on the ball). Other than that it's the wrong tool for the job.

[LKL]

Thanks, swansont. I'd still have some questions though. I've been doing analysis on variable orbits (cycloids) lately. I've especially studied the force exerted on the axis in those situations. The impulse method did seem to work fine if you knew the force acting on the orbiting body/the axis at all times during the cycle. Then it was just simple integration to find out the impulses. They always added up to 0 after a full cycle except in a situation like this.

 

So when the baseball is released, its linear momentum doesn't change, so there should be no recoil impulse on the axel, correct?

 

And when you say that "linear momentum is not conserved", do you really mean this: (linear momentum of baseball in the beginning) + (linear momentum of rod in the beginning) DOES NOT EQUAL (linear momentum of baseball in the the end) + (linear momentum of rod in the end)?

 

And what would be a better tool for this kind of a problem?