# Not really homework still an intro physics question : moment of inertia/system

1. Dec 1, 2008

### fluidistic

1. The problem statement, all variables and given/known data
Hi, I don't know if it's the appropriate forum to post this question... Basically I've been talking with a professor at my university (a professor that will evaluate me in less than 15 days) about a problem we had on a test and I must admit I got very stubborn (I hate me for that) because or I didn't understand her or I was right. I told her the problem we had on a test was badly made.
I'll show you my argument : The problem was "a bar of mass m with 2 points of mass m at its extremity is rotating with an angular velocity of $$\vec \omega$$ around an axis passing by the center of mass of the system bar-points.
At a moment a point drop out the bar.
1) Calculate the angular velocity of the system bar-point. (I assumed it was rotating on an axis passing by the center of mass of the new system but I was wrong and I got a result of $$\vec \omega _f$$ greater than $$\vec \omega$$ . While according to her the angular velocity doesn't change. I trust her on this.)
Then the next question really surprised me :
2) Is the energy conserved? Calculate the energy of the system before and after the mass drop."
I did calculate it and I found it to be superior after the drop than before. (Which was against my intuition that says the energy is conserved. I got this false result because of my mistake in the first question). But, by calculating out $$\vec \omega$$, don't we suppose that the energy is conserved? That's what I've been telling her again and again. As they don't say "the falling point is dropped with a velocity of $$\vec v$$" which could have let us thought about an explosive drop or a drop that "consume" energy, I concluded we simply assume that the energy is conserved. She told me that no, we cannot say this because the axis of rotation will suffer a force $$\vec F$$ (I agree with her on that... the next question was to calculate it) that could cause some loss of energy and we don't know if the body would suffer deformations. I told her that if the body suffers deformations then it's not a rigid body. She didn't change her mind. I told her that as we aren't told about any friction in any part of the body (not even it's axis of rotation), then THERE'S NO dissipative force that could consume energy. According to her I was wrong and I have to calculate the energy before and after. It was asked so I simply did it. But what I meant by all this discussion is that we could have answered the second question very simply without any calculus...
Was I right? I feel a bit sad to have been so stubborn but it seems I'm like that and despite she will judge me for the final exam I stood my ground.

2. Dec 1, 2008

### D H

Staff Emeritus
Short answer: Linear momentum, angular momentum, and kinetic energy are conserved. (You are correct in that there are no dissipative forces.) Somewhat surprisingly, angular velocity doesn't change. In fact, that it doesn't is a consequence of the conservation laws.

I'll try to show some math later.

3. Dec 1, 2008

### flatmaster

If your quote of the orogional question is verbatum, I see the ambiguity you're talking about. She wanted you to assume the bar and one remaining mass continued to rotate around the same fixed point, while you calculated how the bar and mass would rotate in free space. This ambiguity is in the question.

I wouldn't worry about being stuborn on this point. It seems you had the "let's figure the question out" attitude rather than "give me my points" attitude. It may have felt like you were arguing, but that's usually what working out the kinks in a problem can look like from the outside. I've worked problems with fellow students where we're nearly yelling at each other, but we're actually just working things out and actually having a good time of it!!

4. Dec 1, 2008

### fluidistic

Thanks a lot D H. Yeah I suppose it's because even if the distribution of the mass is different after the drop, the rotating system has lost mass so it is possible that it conserves its angular velocity as said my professor.
I'll do the problem tomorrow (1 a.m. now) but I'd appreciate if you could show some math as you said, just in case I need it.

So I was right?! According to me we suppose the dropped mass leaves the bar at $$v=\omega r$$ with $$r$$ being the distance from the center of mass of the bar to the mass. By supposing this we already suppose that the energy is conserved so asking how the energy is worth before and after the drop is redundant. We could just calculate it once. But according to my professor I cannot say the energy will be conserved only by this fact.

Thanks for the input. In fact I don't care anything about my score on the test. I already succeed in 2 test and succeeding in the third is just a matter of taste but don't affect my average. So yes I wanted her to understand how my intuition thinks the problem has something wrong. It's like you assume the energy is conserved (because no information is given to say the contrary) and soon after they ask you if it is conserved.
EDIT: No, I know I was wrong by thinking the bar would rotates about its center of mass. It was clearly told that it rotates around a "fixed" axis. I just made a mistake I agree. I realized it when I calculated the energy before and after the drop. But I didn't realize where I made my error up till today.

Last edited: Dec 1, 2008
5. Dec 1, 2008

### D H

Staff Emeritus
That's it. If it wasn't specified whether the bar was rotating on a fixed pivot or in free space there is indeed an ambiguity. I assumed free space, but I have a bias; all my work is in free space. In the free space problem everything is conserved. With a fixed pivot point the pivot provides the dissipative force that makes energy and linear momentum non-conserved quantities.

6. Dec 1, 2008

### flatmaster

This problem reminds me of one type of pumpkin chunker I saw once. Your mass that flys off is obviously the pumpkin. As you can expect, once the pumpkin is thrown, the device wobbles wildly as it is no longer spinning around it's desired center of mass.

7. Dec 1, 2008

### fluidistic

ahahahah!! I was wrong then! There is a pivot, the bar cannot rotate around its center of mass after the little mass has been dropped... It wasn't ambiguous in the question. I made it ambiguous not on purpose. Sorry for that.
So the pivot will create a dissipative force? If yes then I feel the problem is VERY interesting to me, counter intuitive. By the way it means that my professor made a little mistake : while it was true that I was wrong on the conservation of energy, it is false to say that the mechanical energy is conserved (she said so).

8. Dec 1, 2008

### D H

Staff Emeritus
I was too quick. Mechanical energy is conserved. The only thing that is not conserved is linear momentum.

An easy way to look at it: After release, the released point mass continues moving with whatever linear velocity it had at the point of release. The bar doesn't translate because of the pivot. Before release the total linear momentum is zero. After release, it is non-zero.

The only force acting on the system after release is at the pivot point. This force produces no torque because it has a moment arm of zero. No torques => angular momentum is conserved. The angular momentum before the release is Ibarw-+mlbarw-. The angular momentum after the release is Ibarw++1/2mlbarw++1/2mlbarw-, and thus w+=w-.

Kinetic energy is also conserved.

9. Dec 1, 2008

### flatmaster

This also assumes the pivot point is fixed in space absolutely, rather than to a very massive object m<<<<M such as the earth. That's why conservation of momentum broke down. The lost momentum would go into the swinging-arm-massive-object system.

10. Dec 3, 2008

### fluidistic

I don't know why but I don't reach this answer.
I consider the origin of the system as being the pivot point. Angular momentum is conserved, so $$\vec L_i= \vec L_f$$.
$$\vec L_i= I \vec \omega_i$$.
$$I$$ of the bar is worth $$\frac{ML^2}{12}$$ and $$I$$ of each mass is worth $$\frac{ML^2}{4}$$. Thus $$\vec L_i= \frac{ML^2}{12}+2\cdot \frac{ML^2}{4}=\frac{7ML^2}{12} \omega_i$$.
$$\vec L_f=I\omega_f$$. But I need to calculate this $$I$$ which differs from the anterior. Furthermore I must consider Steiner's theorem to calculate it.
I have that $$I=\frac{ML^2}{12}+\frac{ML^2}{4}+M \left( \frac{L}{4} \right)^2=\frac{19ML^2}{48}$$. Thus $$L_f=\frac{19ML^2}{48} \omega_f$$.
Clearly $$\omega_i \neq \omega_f$$. I get that $$\omega_f=\frac{28}{19}\omega_i$$.
I'd appreciate if you could help me to find where are my errors.

Last edited: Dec 3, 2008
11. Dec 3, 2008

### D H

Staff Emeritus
You are omitted the angular momentum of the released point mass with respect to the pivot point from your calculations.

12. Dec 3, 2008

### fluidistic

Oh you're right. I still reach a non sense answer. I feel I really don't know enough the course.
Look what I've done.
The angular momentum of the released mass is an orbital one (not a spin one nor a combination of spin/orbital one) so $$L_o=\vec r \wedge m \vec v_M =\frac{L}{2}i \wedge M\omega _i r j=\frac{L^2M}{4}\omega _ik$$.
From it I wrote down $$\frac{19ML^2 \omega _f}{48}=\frac{7ML^2 \omega _i}{12}-\frac{L^2M\omega _i}{4}$$ and so $$\omega _i \neq \omega _f$$...

13. Dec 3, 2008

### D H

Staff Emeritus
The pivot point hasn't moved. It is still at the center of the bar.

14. Dec 3, 2008

### fluidistic

Right...
I really can't get the result! Correcting all I could, I get that $$L_i=\frac{7ML^2}{12}\omega _i$$.
Now $$L_f= \frac{ML^2}{8}\omega _f +\frac{L^2M}{4}\omega _i \Leftrightarrow \omega _f=\frac{8 \omega _i}{3}$$...
Where $$\frac{ML^2}{8}\omega _f$$ is the angular momentum of the bar after the release and $$\frac{L^2M}{4}\omega _i$$ is the angular momentum of the released mass. I don't know what's wrong this time.

15. Dec 3, 2008

### D H

Staff Emeritus
Correct.

How did you get that result?

16. Dec 3, 2008

### fluidistic

I considered the bar and the remaining mass as its center of mass. That is a point situated at $$\frac{L}{4}$$ from the pivot point. It has a mass of 2M. So its moment of inertia with respect to the pivot point is $$2M\left( \frac{L}{4} \right) ^2=\frac{ML^2}{8}$$.

17. Dec 3, 2008

### D H

Staff Emeritus
Forget about the bar+remaining mass center of mass because the bar+remaining mass is rotating around the pivot point, not it's center of mass.

18. Dec 3, 2008

### fluidistic

Yes I know, that's why I calculated the distance between the pivot point and the center of mass of the "bar+remaining mass". Why can't I calculate its moment of inertia with respect to the pivot??
EDIT: Oh... you're right. I'm just very confused.
So how would I find the final angular momentum of the system?

Last edited: Dec 3, 2008
19. Dec 3, 2008

### D H

Staff Emeritus
The moment of inertia with respect to the pivot is
bar: ml2/12
mass: ml2/4
total: ml2/3

20. Dec 3, 2008

### fluidistic

Can you believe it? I finally get the desired result.
Now calculating the energy before and after should be an easy task.
A big thank you D H!

EDIT : I confirm : The energy is conserved.

Last edited: Dec 3, 2008