How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

It seems to me there might be an algebraic error here. For $\frac{m}{M}<<\frac{2}{5}$, I get $$\omega^{'}=\frac{5}{2}\frac{m}{M}\frac{v}{R}$$

Why is $\frac{m}{M}<<\frac{2}{5}$? However you are right, I probably miscalculated $I_{cm}$, didn't I?

 

[erobz]

What expression did you get for $I_{cm}$? I couldn't find it, but maybe it is implied somewhere. If you don't have it, use the parallel axes theorem $I_O=I_{cm}+(M+m)(R-d)^2$ where $I_O=(\frac{2}{5}MR^2+mR^2)$ is the moment of inertia about the center of the planet and $(R-d)$ is the distance between the parallel axes.

I'm confused about this, shouldn't it be:

$$ I_{cm} = \frac{2}{5}MR^2 + M(R-d)^2 + md^2$$

?

 

[Hak]

I'm confused about this, shouldn't it be:

View attachment 332490

$$ I_{cm} = \frac{2}{5}MR^2 + M(R-d)^2 + md^2$$

?

With all these discordant calculations, I am understanding less and less...

 

[Chestermiller]

Why is $\frac{m}{M}<<\frac{2}{5}$?

What do the words "by another much smaller asteroid" mean to you? To me, the implication is m<<<M.

However you are right, I probably miscalculated $I_{cm}$, didn't I?

That part looks OK to me.

 

[erobz]

With all these discordant calculations, I am understanding less and less...

Well, how would you calculate the moment of inertia of the sphere ( mass $M$) and the point mass ( mass $m$) about their combined center of mass?

 

[erobz]

What do the words "by another much smaller asteroid" mean to you? To me, the implication is m<<<M.

I thought that too, but it doesn't explicitly state that. It does say $r \ll R$. I think @kuruman has it right and we should not be neglecting $m$ w.r.t. $M$, it say its two asteroids colliding, not an asteroid and a planet.

 

[Hak]

That part looks OK to me.

I don't understand. My calculation is different from @kuruman's, which in turn is different from @erobz's. Which one would be the correct one?

 

[Hak]

Well, how would you calculate the moment of inertia of the sphere ( mass $M$) and the point mass ( mass $m$) about their combined center of mass?

I had calculated it in my first message. Yours and @kuruman's expressions seem correct, but I cannot understand why they turn out as such. Could you please explain how you arrived at these values?

 

[erobz]

I had calculated it in my first message. Yours and @kuruman's expressions seem correct, but I cannot understand why they turn out as such. Could you please explain how you arrived at these values?

When you (we) did it the first time the fact that the center of mass of the system was shifted from the center of mass of the large asteroid (sphere) was not accounted for. We assumed the center of mass of the system was coincident with the center of mass of the sphere. Now accounting for that like @kuruman suggests I'm completely on board with.

But now that the center of mass is shifted, its like the diagram I posted in #62. But @kuruman doesn't agree with my computation of $I_{cm}$ as far as I can tell.

 

[Hak]

When you (we) did it the first time the fact that the center of mass of the system was shifted from the center of mass of the large asteroid (sphere) was not accounted for. We assumed the center of mass of the system was coincident with the center of mass of the sphere. Now accounting for that like @kuruman suggests I'm completely on board with.

But now that the center of mass is shifted, its like the diagram I posted in #62. But @kuruman doesn't agree with my computation of $I_{cm}$ as far as I can tell.

Let's wait for @kuruman's response so we can see why his solution is not the same as yours? What do you think?

 

[Chestermiller]

I thought that too, but it doesn't explicitly state that. It does say $r \ll R$. I think @kuruman has it right and we should not be neglecting $m$ w.r.t. $M$, it say its two asteroids colliding, not an asteroid and a planet.

Let me guess. You’re not an engineer, right?

 

[erobz]

What do you think?

All I know is the expressions for $I_{cm}$ in post #55 and #62 aren't equivalent. Do you concour that they aren't the same or am I missing something?

 

[Chestermiller]

Let's wait for @kuruman's response so we can see why his solution is not the same as yours? What do you think?

You need to learn how to make reasonable approximation.

 

[Hak]

You need to learn how to make reasonable approximation.

I don't understand what you mean.

 

[erobz]

Let me guess. You’re not an engineer, right?

No, I'm not a physicist! Believe it or not I am an engineer (It scares me sometimes too!)

 

[Chestermiller]

No, I'm not a physicist! Believe it or not I am an engineer (It scares me sometimes too!)

As an engineer, what approximations would you make in solving this problem?

 

[Hak]

All I know is the expressions for $I_{cm}$ in post #55 and #62 aren't equivalent. Do you concour that they aren't the same or am I missing something?

I did not explain myself well. This question was not referring to the calculation of the center of mass, but to ask if you would agree with me to wait for @kuruman's answer.

 

[erobz]

As an engineer, what approximations would you make in solving this problem?

With all due respect, I don't think this is an engineering problem. The problem statement says the mass $m$ is much smaller than mass $M$, (I'll agree that is a confusing statement) but it doesn't say explicitly that mass $m$ is negligible w.r.t. mass $M$, like it does the radius.

 

[kuruman]

I'm confused about this, shouldn't it be:

View attachment 332490

$$ I_{cm} = \frac{2}{5}MR^2 + M(R-d)^2 + md^2$$

?

Yes you are confused. The parallel axis theorem says $$I_{\text{parallel axis}}=I_{\text{cm}}+Md^2$$ where $~M~$ is the mass of the object and $d$ is the separation between the parallel axes.
Here

• The mass of the object is $M+m$.
• The two parallel axes are the one going through the CM and the one going through the center of the planet O.
• The distance between them is $R-d$.

So in the parallel axis equation substitute
$I_{\text{parallel axis}}\rightarrow I_O=\frac{2}{5}MR^2+mR^2$
$Md^2\rightarrow (M+m)(R-d)^2.$
What do you get?

 

[Hak]

Yes you are confused. The parallel axis theorem says $$I_{\text{parallel axis}}=I_{\text{cm}}+Md^2$$ where $~M~$ is the mass of the object and $d$ is the separation between the parallel axes.
Here

• The mass of the object is $M+m$.
• The two parallel axes are the one going through the CM and the one going through the center of the planet O.
• The distance between them is $R-d$.

So in the parallel axis equation substitute
$I_{\text{parallel axis}}\rightarrow I_O=\frac{2}{5}MR^2+mr^2$
$Md^2\rightarrow (M+m)(R-d)^2.$
What do you get?

Thank you, @kuruman. I believe, however, that you are wrong in the expression of $I_O$, since $r$ is not among the known data and is negligible compared to $R$.

 

[kuruman]

Thank you, @kuruman. I believe, however, that you are wrong in the expression of $I_O$, since $r$ is not among the known data and is negligible compared to $R$.

Oops, I forgot to capitalize. I corrected the typo. The moment of inertia of the just the asteroid about the center of the planet is $mR^2.$ Thanks for pointing it out and sorry about the confusion.

 

[Hak]

Oops, I forgot to capitalize. I corrected the typo. The moment of inertia of the just the asteroid about the center of the planet is $mR^2.$ Thanks for pointing it out and sorry about the confusion.

No problem, don't worry. Therefore, the force $F$ is equal to $F= m \omega ^2 d$, right?

 

[kuruman]

No problem, don't worry. Therefore, the force $F$ is equal to $F= m \omega ^2 d$, right?

Right.

 

[Hak]

However, I found that one hint of the problem is as follows: asteroids are very far from other bodies and are not interacting with anything else, and they are not rotating in any way. This doesn't change anything in our reasoning, right?

 

[Hak]

Right.

Thank you.

 

[erobz]

@kuruman We are looking top down, the axis of rotation is into the page, through the combined systems center of mass I have labeled $com$ is that true?

 

[kuruman]

We are looking top down, the axis of rotation is into the page, through the combined systems center of mass I have labeled $com$ is that true?

View attachment 332492

Yes.

 

[kuruman]

However, I found that one hint of the problem is as follows: asteroids are very far from other bodies and are not interacting with anything else, and they are not rotating in any way. This doesn't change anything in our reasoning, right?

This says that you should not worry about spin angular momentum before the collision. The only angular momentum before the collision is $mvd.$

 

[Hak]

This says that you should not worry about spin angular momentum before the collision. The only angular momentum before the collision is $mvd.$

Okay, thank you very much. I have one more question: is it possible to make after-the-fact observations on this problem, some interesting notes or whatever? I have noticed that for other problems here on the Forum, very interesting observations have been made by you in addition to the final result.

 

[erobz]

Yes.

Then for just the sphere alone about the axis of rotation that is into the page labeled "com""

I get using parallel axis thereom:

$$ \overbrace{I_{com - axis }}^{\text{sphere}} = 2/5 MR^2 + M(R-d)^2 $$

You are saying that is incorrect for the sphere alone that is rotating about the combined center of mass axis labled "com"?

Because for the point mass we have that:

$$ \overbrace{I_{com - axis }}^{\text{point mass}} = \cancel{I_{cm}}^0 + md^2 $$

Do we not?

Then we combine them:

$$ \overbrace{I_{com - axis }}^{\text{combined}} = \overbrace{I_{com - axis }}^{\text{sphere}}+ \overbrace{I_{com - axis }}^{\text{point mass}} = 2/5 MR^2 + M(R-d)^2+ md^2 $$