How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

My advice, let your professor explain it to them... or you can show them the thread in which it was explained to you?

I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts. It was just to get a complete overview of the possibilities of this problem.

 

[erobz]

I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts.

Why are they authoring problems they have doubts about?

It was just to get a complete overview of the possibilities of this problem.

Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?

 

[Hak]

Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?

You are right, but they don't feel like it. They prefer me to report their interpretations. Sorry if I am annoying, that is not my intention. I just try to help others and always learn more from those who know more than me.

 

[Hak]

Why are they authoring problems they have doubts about?

I have no idea. The author seems to me to be a very good professor, but he does not know all the possible solutions of his own problems. He works them out over time, making various observations and attempts.

 

[kuruman]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

$$\begin{align} & I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2 \nonumber \\
& I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right) \nonumber \\
\end{align}$$When ##m<<M~## you can approximate in the denominator ##M+m\approx M##. Hence $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right).$$

 

[kuruman]

##\dots## but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass.

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##

 

[Hak]

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##

Ah, this is a big problem. I agreed with @erobz, but at this point, I can't really understand why my friend is wrong, mistaking ##d## for ##R##. Why? Sorry to persist with this kind of soap opera....

 

[erobz]

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##

I don't mean to misquote you...I could have sworn you said that talking about angular momentum not being conserved for a differential mass ##dm## away from the center of mass axis. Maybe I'm imaging it.

EDIT: ( yeah, scrolling though the saga, I don't see it ...very sorry for the misquote)

 

[Hak]

$$\begin{align} & I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2 \nonumber \\
& I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right) \nonumber \\
\end{align}$$When ##m<<M~## you can approximate in the denominator ##M+m\approx M##. Hence $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right).$$

Ok, thank you, but...

Moreover, you had said that the same result as ##\omega## would be arrived at without preliminary simplifications. By using ##\omega = \frac{L}{I_{cm}}##, with ##I_{cm}## simplified moment of inertia, you arrive at the result ##\omega = \frac{5mMv}{(2M + 5m) R}##, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

What about it, @kuruman?

 

[erobz]

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##

But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ## This is the crux of the issue with the @Hak friends solution to the problem.

 

[Hak]

But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ##

This is exactly what I wanted to say... It's really a dilemma...

 

[haruspex]

But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ## Becuase this is the crux of the issue with the @Hak friends solution to the problem.

The algebra for finding ##\omega## in post #206 is quite correct and involves no approximation. But as you note, for finding the force, we must decide what radius to use.
Note that whatever force is exerted on m, an equal and opposite force is exerted on M. I think that makes it clear that the correct radius to use is the distance to the common mass centre.

 

[Hak]

The algebra for finding ##\omega## in post #206 is quite correct and involves no approximation. But as you note, for finding the force, we must decide what radius to use.
Note that whatever force is exerted on m, an equal and opposite force is exerted on M. I think that makes it clear that the correct radius to use is the distance to the common mass centre.

Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.

 

[erobz]

Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.

I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.

 

[Hak]

I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.

Yes, it is correct to me. However, I don't understand what this has to do with the Newton's Third Law argument...

 

[kuruman]

I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.

I view this as a single rigid body rotating in space with angular velocity ##\vec {\omega}.## It follows that mass element ##dm## at any point P on the body at position vector ##\vec {r}'_p## relative to the CM will have velocity (also relative to the CM) ##\vec v'_p=\vec {\omega}\times \vec {r}'_p##. Now if the CM is moving relative to a point in an inertial frame with velocity ##\vec V_{cm}##, the velocity of P relative to the inertial frame will be $$\vec v_p=\vec V_{cm}+\vec v'_p=\vec V_{cm}+\vec {\omega}\times \vec {r}'_p.$$The centripetal force on mass ##m## is ##\vec F_c=-(dm) \omega^2 \vec{r}'_p.##

 

[Hak]

I view this as a single rigid body rotating in space with angular velocity ##\vec {\omega}.## It follows that mass element ##dm## at any point P on the body at position vector ##\vec {r}'_p## relative to the CM will have velocity (also relative to the CM) ##\vec v'_p=\vec {\omega}\times \vec {r}'_p##. Now if the CM is moving relative to a point in an inertial frame with velocity ##\vec V_{cm}##, the velocity of P relative to the inertial frame will be $$\vec v_p=\vec V_{cm}+\vec v'_p=\vec V_{cm}+\vec {\omega}\times \vec {r}'_p.$$The centripetal force on mass ##m## is ##\vec F_c=-(dm) \omega^2 \vec{r}'_p.##

This would imply that since the minor asteroid is a homogeneous body, it has mass ##dm = m##? And also that ##r'_p = d##?

 

[kuruman]

This would imply that since the minor asteroid is a homogeneous body, it has mass ##dm = m##? And also that ##r'_p = d##?

No, ##dm## is a very small mass at position ##\vec {r}'## relative to the center of the planet. Imagine chopping up the composite body into many little pieces. ##dm## is the mass of the little piece that you will find at position ##\vec {r}'##; it could be small piece of the asteroid or a small piece of the planet. The only thing you can say is that $$\int dm=M+m.$$

 

[Hak]

No, ##dm## is a very small mass at position ##\vec {r}'## relative to the center of the planet. Imagine chopping up the composite body into many little pieces. ##dm## is the mass of the little piece that you will find at position ##\vec {r}'##; it could be small piece of the asteroid or a small piece of the planet. The only thing you can say is that $$\int dm=M+m.$$

OK, so yours is just a general overview of the physical situation. The result of ##F## you wrote does not lead to the form we found earlier, right?

 

[kuruman]

OK, so yours is just a general overview of the physical situation. The result of ##F## you wrote does not lead to the form we found earlier, right?

At this point, with 264 posts and counting, I have no clue what form you are referring to. Please post it here so that I can see what you mean.

 

[Hak]

At this point, with 264 posts and counting, I have no clue what form you are referring to. Please post it here so that I can see what you mean.

##F = m \omega^2 d##

 

[haruspex]

Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.

There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.

 

[Hak]

There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.

OK, thank you very much. I cannot understand, however, how it can be deduced from this symmetric equation that ##d## is the radius to be chosen for the rotation. Sorry if I don't fully understand...

 

[kuruman]

There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.

Isn't it simpler to note that any point on the composite rigid object rotates about the CM with constant angular velocity ##\vec{\omega}##? Then that point has centripetal acceleration ##\vec a_c=-\omega^2 \vec {r}'## where ##\vec {r}'## is the position vector relative to the CM. Thus the net force on mass element ##dm## is ##\vec {F}_c=-(dm)\omega^2 \vec {r}'.##

 

[kuruman]

##F = m \omega^2 d##

Yes, I see now what you are asking. The general expression for the force at any point is ##\vec {F}_c=-(dm)\omega^2 \vec {r}'## where ##\vec {r}'## is the location of the point relative to the CM. The asteroid is at location ##\vec {r}'=\vec d## and has mass ##dm = m##. So the force on the asteroid is $$\vec {F}_c=-m\omega^2 \vec {d}.$$

 

[Hak]

Yes, I see now what you are asking. The general expression for the force at any point is ##\vec {F}_c=-(dm)\omega^2 \vec {r}'## where ##\vec {r}'## is the location of the point relative to the CM. The asteroid is at location ##\vec {r}'=\vec d## and has mass ##dm = m##. So the force on the asteroid is $$\vec {F}_c=-m\omega^2 \vec {d}.$$

Is it correct?

 

[kuruman]

Ah, this is a big problem. I agreed with @erobz, but at this point, I can't really understand why my friend is wrong, mistaking ##d## for ##R##. Why? Sorry to persist with this kind of soap opera....

It seems to me that acting as an information conduit between your friend and PF is confusing you because you don't seem to have the expertise to diagnose where and why your friend is wrong. Your friend needs to speak for him/herself and post here. Then we will be able to diagnose your friend's problem. Maybe your friend is stringing you along trying to see how many posts this thread will add up to before it is closed down.

 

[Hak]

It seems to me that acting as an information conduit between your friend and PF is confusing you because you don't seem to have the expertise to diagnose where and why your friend is wrong. Your friend needs to speak for him/herself and post here. Then we will be able to diagnose your friend's problem. Maybe your friend is stringing you along trying to see how many posts this thread will add up to before it is closed down.

Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...

 

[haruspex]

Isn't it simpler to note that any point on the composite rigid object rotates about the CM with constant angular velocity ##\vec{\omega}##? Then that point has centripetal acceleration ##\vec a_c=-\omega^2 \vec {r}'## where ##\vec {r}'## is the position vector relative to the CM. Thus the net force on mass element ##dm## is ##\vec {F}_c=-(dm)\omega^2 \vec {r}'.##

Yes, I was just providing confirmation from other evidence, since @Hak and @erobz seemed unsure.

 

[kuruman]

Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...

Please remind us, what issue is this?

 

[Hak]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

Moreover, you had said that the same result as ##\omega## would be arrived at without preliminary simplifications. By using ##\omega = \frac{L}{I_{cm}}##, with ##I_{cm}## simplified moment of inertia, you arrive at the result ##\omega = \frac{5mMv}{(2M + 5m)(M+m) R}##, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

The second paragraph. Thanks.

 

[Hak]

Funny thing, I did too. My corrected exact value for ##\omega## when ##M=9m## is $$\omega =\frac{1}{5}\frac{v}{R}.$$It turns out that no approximation is needed if one uses the simplified moment of inertia in post#128.

(Original expression edited to add missing factor of 5 in the numerator.)

This is the post I am referring to.

 

[kuruman]

The second paragraph. Thanks.

Your expression in the second paragraph is dimensionally incorrect. The angular velocity has dimensions of [v/R]. Your expression has dimensions of [Mv/r].

 

[Hak]

Your expression in the second paragraph is dimensionally incorrect. The angular velocity has dimensions of [v/R]. Your expression has dimensions of [Mv/r].

I edited my message, correcting my result. I had mistakenly copied from the paper to the screen. My expression is ##\omega = \frac{5mMv}{(2M + 5m)(M+m) R}##

 

[erobz]

but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass.

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##

What was your objection to the statement; that you never said it, and/or that it is not true? Because in the end, we can take angular momentum about any axis ( as you showed previously in post #178), but it is ultimately conserved about the systems center of mass in either case. i.e. ##\omega ## is to be applied about the systems center of mass, not about the arbitrary axis?