How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[erobz]

Is $mv = M v' + m (v'+ \omega R)$ correct?

Good.

All this is irrelevant for calculating the force $F$, right?

Correct. This was just to check conceptual understanding.

 

[Hak]

Good.

Correct. This was just to check conceptual understanding.

Thank you so much. Is the expression for $\omega$, on the other hand, correct?

 

[erobz]

Thank you so much. Is the expression for $\omega$, on the other hand, correct?

Yeah, I get what you got:

By clearing the fractions in the denominator, the expression for $\omega'$ becomes:

$$\omega' = \frac{5mv}{(2M + 5 m)R}$$. Therefore:

 

[Hak]

Yeah, I get what you got:

So my result for $F$ should be right. Let's wait for confirmation from other members.

 

[kuruman]

So my result for $F$ should be right. Let's wait for confirmation from other members.

I cannot confirm this answer. This is how to do it.

Angular momentum is conserved about the CM of the composite mass.
The distance from the center of mass to the embedded asteroid is $d=\dfrac{M}{M+m}R.$
The angular momentum about the CM before the collision is the orbital angular momentum of the asteroid only. We are told that the planet is not spinning.
$L_{\text{before}}=mvd=\dfrac{mMv}{M+m}R.$
The angular momentum about the CM after the collision is
$L_{\text{after}}=I_{cm}~\omega.$

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.

Then solve the momentum conservation equation for $\omega$.

 

[Hak]

I cannot confirm this answer. This is how to do it.

Angular momentum is conserved about the CM of the composite mass.
The distance from the center of mass to the embedded asteroid is $d=\dfrac{M}{M+m}R.$
The angular momentum about the CM before the collision is the orbital angular momentum of the asteroid only. We are told that the planet is not spinning.
$L_{\text{before}}=mvd=\dfrac{mMv}{M+m}R.$
The angular momentum about the CM after the collision is
$L_{\text{after}}=I_{cm}~\omega.$

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.

Then solve the momentum conservation equation for $\omega$.

Thank you. I knew something was wrong with my process. The reasoning and the unfolding are similar to yours, what changes is the algebra. I am sure your procedure is more correct than mine. After solving the momentum conservation equation for $/omega$, can I calculate the force $F$ as I did in post #1, that is, by $F = m \omega ^2 R$? How would this force be directed? I am waiting for confirmation, thanks again.

Postscript. Since I did not understand some steps, could you draw a graphic diagram of the physical situation that would help me understand it better? How would the conservation of linear momentum discussed with @erobz change, considering your corrections and observations? Thank you.

 

[erobz]

How would the conservation of linear momentum discussed with @erobz change, considering your corrections and observations?

My guess would be the term “ $\omega R$ “ gets replaced with “$\omega d$”. How about you?

 

[Hak]

My guess would be the term “ $\omega R$ “ gets replaced with “$\omega d$”. How about you?

I also think it may be so. Let's see if there is any other opinion disagreeing with this....

 

[erobz]

I also think it may be so. Let's see if there is any other opinion disagreeing with this....

That should go for the force $F$ too (replace $R$ with $d$)

 

[Chestermiller]

Homework Statement: An asteroid has a spherical shape and uniform mass distribution. Its radius is $R$ and its mass $M$. The asteroid is stationary in interstellar space when it is struck by another much smaller asteroid of mass $m$ and negligible radius relative to $R$. The minor asteroid has velocity $v$ before the impact, and its direction is tangent to the surface of the major asteroid. The impact is completely inelastic, so that the minor asteroid remains embedded inside the major one, but it is not destroyed; it is just embedded inside the rock. The gravity is negligible.

After the collision has occurred, what is the value of the force acting on the minor asteroid?
Relevant Equations: /

Before the collision, the total linear momentum of the system is $p = mv$, where $m$ is the mass of the minor asteroid and $v$ is its velocity. The total angular momentum of the system is $L = mRv$, where $R$ is the radius of the major asteroid. Since there are no external forces or torques acting on the system, both $p$ and $L$ are conserved during and after the collision.

After the collision, the minor asteroid becomes embedded inside the major one, so they move together as a single rigid body. The linear momentum of this body is still $p = mv$, so its velocity is $v' = \frac{p}{(M + m)}$, where $M$ is the mass of the major asteroid. The angular momentum of this body is also still $L = mRv$, so its angular velocity is $\omega' = \frac{L}{I}$, where $I$ is the moment of inertia of the body.

The moment of inertia of a sphere with uniform mass distribution and radius $R$ is $I = (2/5)MR^2$. However, since there is a small mass $m$ embedded inside the sphere at a distance $R$ from its center, we need to use the parallel axis theorem to find the moment of inertia of the body. The parallel axis theorem states that $I = I_{cm} + md^2$, where $I_{cm}$ is the moment of inertia about the center of mass and $d$ is the distance from the center of mass to the axis of rotation. In this case, $I_cm = (2/5)MR^2$ and $d = R$, so we get $I = (\frac{2}{5})MR^2 + mR^2 = (\frac{2}{5} + \frac{m}{M})MR^2$.

Now we can find the angular velocity $\omega'$ by plugging in $L$ and $I$ into $\omega' = \frac{L}{I}$. We get $\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}$.

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

 

[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:

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

 

[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.