How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[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:

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

 

[PeroK]

In Kleppner and Kolenkow, instead of algebriac answers, they would say (in this case), if $M = 9m$, then
$$F = \frac{4mv^2}{81R}$$That's what I get.

If we take $m \ll M$, then $F = 0$.

PS I did make a mistake and authomatically used $R$ instead of $d$ when calculating $\omega$.

 

[Hak]

In Kleppner and Kolenkow, instead of algebriac answers, they would say (in this case), if $M = 9m$, then
$$F = \frac{4mv^2}{81R}$$That's what I get.

If we take $m \ll M$, then $F = 0$.

I can't understand it. Why?

 

[PeroK]

I can't understand it. Why?

Because you didn't think about it for long enough?

 

[Hak]

Because you didn't think about it for long enough?

No, maybe I didn't explain myself well. I can't understand why you said that algebraic operations are not set up, nor where that result of $F$ for $M = 9 m$ came from.

 

[PeroK]

The idea is that I have got a formula for $F$ in terms of $m, M, v, R$. If I post that formula, then I've given you the answer. But, if I plug $M = 9m$ into that formula and simplify, then that gives you something to check. It also gives the others something to check and they can confirm whether I've got it right or not. I think it's right, but we all make mistakes.

 

[Chestermiller]

My understanding of the problem statement is that you are supposed to assume that $$\frac{m}{M}<<1$$and$$\frac{r}{R}<<1$$

 

[PeroK]

My understanding of the problem statement is that you are supposed to assume that $$\frac{m}{M}<<1$$

Then $F \approx 0$.

and$$\frac{r}{R}<<1$$

Yes.

 

[haruspex]

We get $\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}$.

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

 

[Hak]

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

Yes, your reasoning makes sense to me. The problem is that that value of $\omega$ would be wrong even under your (correct) assumptions, because the previous calculations regarding $I_{cm}$ and conservation of momentum are wrong.

 

[erobz]

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

If it was meant to be a small asteroid hitting a still planetary sized body where $m \ll M$ then we can tell without doing any calculation though that $F \approx 0$. by your own result as $\frac{m}{M} \approx 0$ , giving $\omega \approx 0$?

They say not clear things in the problem statement. "An asteroid hitting an asteroid" is stated and $m$ is much smaller than $M$. I'm thinking like its a large ( but not "planetary" large) less dense body, being impacted by a small dense body.

 

[Hak]

In Kleppner and Kolenkow, instead of algebriac answers, they would say (in this case), if $M = 9m$, then
$$F = \frac{4mv^2}{81R}$$That's what I get.

If we take $m \ll M$, then $F = 0$.

Following the procedure described by @kuruman in post #55, I find that $\omega = \frac{5mv}{(2M + 7m) R}$. Substituting in $F = m \omega ^2 d$, we have:

$$F = \frac{25 m^3 M v^2}{(2M + 7m)^2 (M+m) R}$$.

Therefore, for $M = 9m$, I get:

$$F = \frac{9mv^2}{250 R}$$.

Where do I go wrong?

 

[PeroK]

Substituting in $F = M \omega ^2 d$,
Where do I go wrong?

That should be $F = m \omega ^2 d$.

 

[Hak]

That should be $F = m \omega ^2 d$.

I edited my message. Sorry.

 

[Hak]

If it was meant to be a small asteroid hitting a still planetary sized body where $m \ll M$ then we can tell without doing any calculation though that $F \approx 0$. by your own result as $\frac{m}{M} \approx 0$ , giving $\omega \approx 0$?

Actually no, because that value of $\omega$ is wrong.

 

[haruspex]

If it was meant to be a small asteroid hitting a still planetary sized body where $m \ll M$ then we can tell without doing any calculation though that $F \approx 0$. by your own result as $\frac{m}{M} \approx 0$ , giving $\omega \approx 0$?

m<<M means we are looking for the smallest order nonzero approximation for small m/M. If a function of $x$ is $\Sigma_{n=0}c_nx^n$ that is $c_ix^i$, where $c_i\neq 0$ and $c_j=0$ for $j<i$.
So no, we cannot end up with F=0.