How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

The theory we were laboring under there was updated 5 posts later. Have you properly updated the linear momentum equation?

Sorry, we had said that the linear momentum conservation equation was not necessary for the purpose of force calculation, so we did not update it. Our discussion on it ended there, so this is our final equation. How should it be modified?

 

[erobz]

Sorry, we had said that the linear momentum conservation equation was not necessary for the purpose of force calculation, so we did not update it. Our discussion on it ended there, so this is our final equation. How should it be modified?

What is the linear velocity of the imbedded asteroid in the ground frame just after impact?

 

[Hak]

What is the linear velocity of the imbedded asteroid in the ground frame just after impact?

We had said $V = v' + \omega d$, but it doesn't fit with what I said in post #149

 

[erobz]

We had said $V = v' + \omega d$, but it doesn't fit with what I said in post #149

What is the angular momentum of the little asteroid with respect to the center mass before impact?

 

[Hak]

What is the angular momentum of the little asteroid with respect to the center mass before impact?

Isn't it $L = mvd$?

 

[erobz]

Isn't it $L = mvd$?

Yeah

 

[Hak]

Yeah

Okay, but I can't hit the nail on the head. In post #149 I said that I arrived at the same angular momentum conservation equation by a different method, which involves a different linear momentum conservation equation, isn't that right?

 

[erobz]

Okay, but I can't hit the nail on the head. In post #149 I said that I arrived at the same angular momentum conservation equation by a different method, which involves a different linear momentum conservation equation, isn't that right?

We don't arrive at the same angular momentum by linear momentum arguments.

Angular momentum is not dependent on the linear momentum here, but the linear momentum is dependent on the angular momentum.

 

[Hak]

We don't arrive at the same angular momentum by linear momentum arguments.

Angular momentum is not dependent on the linear momentum here, but the linear momentum is dependent on the angular momentum.

So why is my expression in post #149 incorrect?

 

[erobz]

I believe the equation should be:

$$ mv = M v' + m(v' + \omega d)$$

where $\omega = \frac{mvd}{I_{com}}$

 

[Hak]

I believe the equation should be:

$$ mv = M v' + m(v' + \omega d)$$

where $\omega = \frac{mvd}{I_{com}}$

Okay, so far so good, but I'm talking about something else. I am saying that calculating angular momentum with respect to the center of the larger sphere, and not with respect to the center of mass of the system, leads to a different equation of conservation of linear momentum. What you expressed just now is a different procedure from mine in post #149, that's what we used yesterday.

 

[Hak]

@PeroK, @kuruman, @haruspex, @Chestermiller What do you think about it?

 

[erobz]

Okay, so far so good, but I'm talking about something else. I am saying that calculating angular momentum with respect to the center of the larger sphere, and not with respect to the center of mass of the system, leads to a different equation of conservation of linear momentum. What you expressed just now is a different procedure from mine in post #149, that's what we used yesterday.

Maybe now that I'm looking at it, the mass $M$ is rotating about the center of mass too after the collision albeit with smaller radius and in opposite direction.

$v'$ is the combined center of mass velocity...

So perhaps it should be:

$$ mv = M( v' - \omega ( R-d) ) + m( v' + \omega d ) $$

Or do we also need to adjust the referencing the center of mass velocity before the collision too like:

$$ \frac{m}{M+m}v = M( v' -\omega ( R-d) ) + m( v' + \omega d ) $$

 

[PeroK]

@PeroK, @kuruman, @haruspex, @Chestermiller What do you think about it?

Linear momentum and angular momentum are conserved separately. The point about which you choose to measure angular momentum does not affect the linear momentum calculation.

 

[Hak]

Maybe now that I'm looking at it, the mass $M$ is rotating about the center of mass too after the collision albeit with smaller radius.

$v'$ is the combined center of mass velocity?

So perhaps it should be:

$$ mv = M( v' + \omega ( R-d) ) + m( v' + \omega d ) $$

Or do we also need to adjust the referencing the center of mass velocity before the collision too like:

$$ \frac{m}{M+m}v = M( v' + \omega ( R-d) ) + m( v' + \omega d ) $$

I am struggling to understand this process. The velocity $v'$ is the velocity of the center of mass of the largest sphere, I think. Could you explain these steps better?

 

[Hak]

Linear momentum and angular momentum are conserved separately. The point about which you choose to measure angular momentum does not affect the linear momentum calculation.

Ok, but I can't find the error in post #149. Could you help me in that regard?

 

[erobz]

I am struggling to understand this process. The velocity $v'$ is the velocity of the center of mass of the largest sphere, I think. Could you explain these steps better?

I think $v'$ actually needs to be the velocity of the center of mass. Before collision are looking at $v_{cm} = \frac{m}{M+m}v$, and after the collision we are looking at $v_{cm} = v'$.

 

[Hak]

Before collision are looking at $v_{cm} = \frac{m}{M+m}v$, and after the collision we are looking at $v_{cm} = v'$.

Why?

 

[erobz]

Why?

The rotation of each mass ( $M$ and $m$) about the center of mass the instant after collision has me speculating.

 

[Hak]

The rotation of each mass ( $M$ and $m$) about the center of mass the instant after collision has me speculating.

Yes, but I cannot understand such assumptions. From where do you derive this velocity $v_{cm}$, and under what assumptions do you place it as the initial velocity? Could it not be, more simply, $mv = (M+m) v_{cm}$?

 

[erobz]

I don't know, I'm getting confused. You might be correct and we only need to look at the change in center of mass velocity before and after collision, not the velocities of each component mass...

 

[Hak]

I don't know, I'm probably confused. You might be correct and we only need to look at the change in center of mass velocity before and after collision, not the velocities of each component mass...so:

$$ \frac{m^2}{m+M}v = ( M+m)v'$$

?

Although the problem is solved, this point is interesting, isn't it? I can't understand the $(M+m)$ factor in the denominator, I'm more confused than you are. What do you say, let's wait for some input from the other members, like @kuruman, @haruspex, @PeroK, @Chestermiller, or maybe others who haven't participated in the thread?

 

[Chestermiller]

Although the problem is solved, this point is interesting, isn't it? I can't understand the $(M+m)$ factor in the denominator, I'm more confused than you are. What do you say, let's wait for some input from the other members, like @kuruman, @haruspex, @PeroK, @Chestermiller, or maybe others who haven't participated in the thread?

Considering the apparent source and level of this problem, I agree with @haruspex's determination.

 

[erobz]

Although the problem is solved, this point is interesting, isn't it? I can't understand the $(M+m)$ factor in the denominator, I'm more confused than you are. What do you say, let's wait for some input from the other members, like @kuruman, @haruspex, @PeroK, @Chestermiller, or maybe others who haven't participated in the thread?

I deleted that equation. It might be correct, but I'm not sure. So I didn't want to stir the pot with it ( but you are very quick on the draw).

As for that factor in the denominator it would have came from this idea:

to calculate the velocity of the system center of mass:

$$ v_{cm} = \frac{mv+ M\cdot 0}{m+M} = \frac{m}{M+m}v$$

 

[Hak]

Considering the apparent source and level of this problem, I agree with @haruspex's determination.

What is "@haruspex's determination"?

 

[Hak]

I deleted that equation. It might be correct, but I'm not sure. So I didn't want to stir the pot with it ( but you are very quick on the draw).

As for that factor in the denominator it would have came from this idea:

View attachment 332540

to calculate the velocity of the system center of mass:

$$ v_{cm} = \frac{mv+ M\cdot 0}{m+M} = \frac{m}{M+m}v$$

The problem is that all the assumptions made are worthy of attention and make sense, so as far as I can tell, they could all be correct. Unfortunately, I do not have much experience (I am only a student), so I cannot know for sure which of them is the correct answer. Anyway, I don't want to be presumptuous, I just try to learn by asking questions.

 

[erobz]

The problem is that all the assumptions made are worthy of attention and make sense, so as far as I can tell, they could all be correct. Unfortunately, I do not have much experience (I am only a student), so I cannot know for sure which of them is the correct answer. Anyway, I don't want to be presumptuous, I just try to learn by asking questions.

Its possible... the same way @kuruman and I had different approaches to the moment of inertia that appeared to be different on inspection, but were in fact the same.

$$ mv = M( v' - \omega ( R-d) ) + m( v' + \omega d ) \tag{1} $$

This equation (1) seems like its keeping track of all the individual mass center momentums

$$( M+m) \frac{m}{m+M}v = ( M+m )v' \tag{2} $$

This equation (2) seems like forget about that and just focus on the change in center of mass momentum.

Both of them make sense to me, and for some reason I decided to mix them...which I think was wrong.

The second is easy to solve. The first one is obviously more work, but I ( we) should probably try it to see if they yield to each other.

EDIT: fixed equation 2 to make it correct.

 

[kuruman]

Linear momentum conservation throughout a collision between two masses means one thing and one thing only: The momentum of the center of mass of the two-mass system is the same before the collision as it is after the collision. Since linear momentum is a vector, this means that both its magnitude and direction do not change. We can find the velocity of the center of mass using the definition $$V_{cm}=\frac{mv_{\text{before}}+MV_{\text{before}}}{m+M}.$$ Before the collision, $v_{\text{before}}=v$ and $V_{\text{before}}=0$ so that $$V_{cm}=\frac{mv}{m+M}$$ and that's all there is. Note that after the collision, the CM is the only point that moves in a straight line with velocity $V_{cm}$ while all other points rotate about it with angular velocity $\omega.$ This means that the linear momentum of any mass element $dm$ away from the CM is not conserved.

As for conserving angular momentum about point O, the center of the planet, one needs to also consider the orbital angular moment about O after the collision.
$L_{\text{before}}=mvR$
$L_{\text{after}}=L_{\text{about CM}}+L_{\text{of cm}}$
Now
$L_{\text{about cm}}=I_{cm}~\omega$ which is the same as before.
$L_{\text{of cm}}=(M+m)V_{cm}(R-d)$ because the cm is at distance $(R-d)$ from point O.
Angular momentum conservation about point O demands that $$\begin{align}
& L_{\text{before}} = L_{\text{after}} \nonumber \\
&mvR=I_{cm}~\omega+(M+m)V_{cm}(R-d) \nonumber \\
& mvR=I_{cm}~\omega+\cancel{(M+m)}\frac{mv}{\cancel{M+m}}(R-d) \nonumber \\
& \cancel{mvR}=I_{cm}~\omega+mv(\cancel{R}-d) \nonumber \\
& 0=I_{cm}~\omega -mvd \nonumber \\
& mvd=I_{cm}~\omega. \nonumber
\end{align} $$ Surprise, surprise!

 

[Hak]

Linear momentum conservation throughout a collision between two masses means one thing and one thing only: The momentum of the center of mass of the two-mass system is the same before the collision as it is after the collision. Since linear momentum is a vector, this means that both its magnitude and direction do not change. We can find the velocity of the center of mass using the definition $$V_{cm}=\frac{mv_{\text{before}}+MV_{\text{before}}}{m+M}.$$ Before the collision, $v_{\text{before}}=v$ and $V_{\text{before}}=0$ so that $$V_{cm}=\frac{mv}{m+M}$$ and that's all there is. Note that after the collision, the CM is the only point that moves in a straight line with velocity $V_{cm}$, while all other points rotate about it with angular velocity $\omega.$ This means that the linear momentum of any mass element $dm$ away from the CM is not conserved.

As for conserving angular momentum about point O, the center of the planet, one needs to also consider the orbital angular moment about O after the collision.
$L_{\text{before}}=mvR$
$L_{\text{after}}=$L_{\text{about CM}}+$L_{\text{of cm}}$
Now
$L_{\text{about cm}}=I_{cm}~\omega$ which is the same as before.
$L_{\text{of cm}}=(M+m)V_{cm}(R-d)$ because the cm is at distance $(R-d)$ from point O.
Angular momentum conservation about point O demands that $$\begin{align}
& L_{\text{before}} = L_{\text{after}} \nonumber \\
&mvR=I_{cm}~\omega+(M+m)V_{cm}(R-d) \nonumber \\
& mvR=I_{cm}~\omega+\cancel{(M+m)}\frac{mv}{\cancel{M+m}}(R-d) \nonumber \\
& \cancel{mvR}=I_{cm}~\omega+mv(\cancel{R}-d) \nonumber \\
& 0=I_{cm}~\omega -mvd \nonumber \\
& mvd=I_{cm}~\omega. \nonumber
\end{align} $$ Surprise, surprise!

I don't know if you noticed, but it is roughly the same procedure as mine in post #149. The equation I arrived at is the same.

 

[Hak]

Its possible... the same way @kuruman and I had different approaches to the moment of inertia that appeared to be different on inspection, but were in fact the same.
This equation (1) seems like its keeping track of all the individual mass center momentums
This equation (2) seems like forget about that and just focus on the change in center of mass momentum.

Both of them make sense to me, and for some reason I decided to mix them...which I think was wrong.

The second is easy to solve. The first one is obviously more work, but I ( we) should probably try it to see if they yield to each other.

Equation (2) gives:

$$v' = \left(\frac{m}{m+M} \right)^2 v$$.

This equation has a power of 2 to the exponent over the expression of $v_{cm}$ that I had calculated and that @kuruman also calculated.

Equation (1) gives:

$$v' = \left(\frac{m}{m+M}\right) v + \frac{10 m^2 M}{(M+m)^2 (2M + 7m)}v$$.

This expression, in my opinion, has an awkward term on the right hand side. Most interestingly, if you change a minus sign to a plus sign in only one of the two round brackets referring to $m$ and $M$, you get the expression:

$$v' = \left(\frac{m}{m+M} \right) v$$, that is, the same as before.

How can this be explained?

 

[erobz]

Linear momentum conservation throughout a collision between two masses means one thing and one thing only: The momentum of the center of mass of the two-mass system is the same before the collision as it is after the collision.

Ahh. No external forces are acting on the center of mass, i.e no change in $V_{cm}$

 

[kuruman]

Ahh. No external forces are acting on the center of mass, i.e no change in $V_cm$

Exactly. That follows from Newton's 3rd law.

 

[Hak]

Equation (2) gives:

$$v' = \left(\frac{m}{m+M} \right)^2 v$$.

This equation has a power of 2 to the exponent over the expression of $v_{cm}$ that I had calculated and that @kuruman also calculated.

Equation (1) gives:

$$v' = \left(\frac{m}{m+M}\right) v + \frac{10 m^2 M}{(M+m)^2 (2M + 7m)}v$$.

This expression, in my opinion, has an awkward term on the right hand side. Most interestingly, if you change a minus sign to a plus sign in only one of the two round brackets referring to $m$ and $M$, you get the expression:

$$v' = \left(\frac{m}{m+M} \right) v$$, that is, the same as before.

How can this be explained?

Can this have relevance?

 

[kuruman]

Can this have relevance?

Can what have relevance?

 

[Hak]

Can what have relevance?

What I wrote in post #180. It seems to me that by modifying @erobz's initial equation (1), we can arrive at the same result for $V_{cm}$. Maybe I'm wrong.

 

[erobz]

Equation (2) gives:

$$v' = \left(\frac{m}{m+M} \right)^2 v$$.

This equation has a power of 2 to the exponent over the expression of $v_{cm}$ that I had calculated and that @kuruman also calculated.

Equation (1) gives:

$$v' = \left(\frac{m}{m+M}\right) v + \frac{10 m^2 M}{(M+m)^2 (2M + 7m)}v$$.

This expression, in my opinion, has an awkward term on the right hand side. Most interestingly, if you change a minus sign to a plus sign in only one of the two round brackets referring to $m$ and $M$, you get the expression:

$$v' = \left(\frac{m}{m+M} \right) v$$, that is, the same as before.

How can this be explained?

I my EQ 2 is definitely incorrect. It should be

$$ ( M+m) \frac{m}{M+m}v = ( M+m)v' $$

That yieds:

$$v' = \frac{m}{M+m}v$$

I took the initial center of mass momentum as $m v_{cm}$, thats not right ( its momentum includes both masses $( M+m)$)

Equation (1) I'm on the fence about, because I still feel like we are keeping track of all the individual center of mass velocities w.r.t. the ground frame and that should be fine.

 

[Hak]

Equation (1) I'm on the fence about, because I still feel like we are keeping track of all the individual center of mass velocities w.r.t. the ground frame.

The problem, in my opinion, lies in the sign. As I said, by changing only one sign, the result is the correct one. We have to figure out where the wrong sign is....

 

[erobz]

The problem, in my opinion, lies in the sign. As I said, by changing only one sign, the result is the correct one. We have to figure out where the wrong sign is....

Specifically, what sign are you talking about changing in the equation(1) that makes it work out?

 

[Hak]

Specifically, what sign are you talking about changing in the equation(1) that makes it work out?

One of the two minus signs inside the round brackets, specifically those that follow $v'$ and precede $\omega$.

Edit. I realized that I made a mistake in copying the expression. The signs are all correct and the equation works out.

 

[erobz]

One of the two minus signs inside the round brackets, specifically those that follow $v'$ and precede $\omega$.

They should not both be minus, or plus. I have one as minus and one as plus in equation (1).

 

[Hak]

They should not both be minus, or plus. I have one as minus and one as plus in equation (1).

Yes, you are right. In fact, I edited the message, saying that the equation is correct. I apologize for the mistake.

 

[erobz]

Yes, you are right. In fact, I edited the message, saying that the equation is correct. I apologize for the mistake.

Thats good news! Do you see the idea behind eq 1then ( do the terms make sense to you )?

 

[Hak]

Thats good news! Do you see the idea behind eq 1then ( do the terms make sense to you )?

Certainly, in fact, as I predicted, for all I knew they could all have been correct. In fact, your equation turned out to be the same as mine and @kuruman's.

 

[erobz]

Certainly, in fact, as I predicted, for all I knew they could all have been correct. In fact, your equation turned out to be the same as mine and @kuruman's.

Thats happened a few times in this thread!

 

[Hak]

Anyway, thank you all very much, it was a very nice discussion! I would love to know some more ideas and observations afterwards about this physics problem, but if you don't have time, never mind. Thanks again.

 

[erobz]

Anyway, thank you all very much, it was a very nice discussion!

I would love to know some more ideas and observations afterwards about this physics problem, but if you don't have time, never mind. Thanks again.

It was a bit of a rollercoaster (I'll except some blame on that), but it's probably best if you leave it be. Threads that aimlessly walk around with variations aren't exactly looked upon as useful here. They want you to start a new thread for a new problem, and anything else would be a new problem.

 

[Hak]

It was a bit of a rollercoaster (I'll except some blame on that), but it's probably best if you leave it be. Threads that aimlessly walk around with variations aren't exactly looked upon as useful here. They want you to start a new thread for a new problem, and anything else would be a new problem.

I don't know how you might feel about it, but I really like discussions that turn out to be roller coasters in the end, they reveal a very good approach of problematizing and questioning that I think is fundamental. If everyone had the correct answer to everything, there would be no point in advancing a discussion in these Forums. You don't have to take responsibility for anything or blame yourself for anything, in my opinion.
Okay, if anything, I will open a new thread, thanks for the advice.

 

[erobz]

I don't know how you might feel about it, but I really like discussions that turn out to be roller coasters in the end, they reveal a very good approach of problematizing and questioning that I think is fundamental. If everyone had the correct answer to everything, there would be no point in advancing a discussion in these Forums. You don't have to take responsibility for anything or blame yourself for anything, in my opinion.
Okay, if anything, I will open a new thread, thanks for the advice.

Personally, I don't mind them. But with multiple contributors one can lose track of what is being said about what specific problem.

The site has yet other agendas. They are probably thinking (at least a bit) about search results. As far as I can tell they don't prefer others that search for the problem title to read through 200 posts of us debating about whether "the egg or the chicken comes first" in the problem so to speak. Sometimes it happens. It's not a perfect system, but its at least a good thing that you, the asker, got some clarity of thought through the ride. I did too.

 

[Hak]

Personally, I don't mind them. But with multiple contributors one can lose track of what is being said about what specific problem.

The site has yet other agendas. They are probably thinking (at least a bit) about search results. As far as I can tell they don't prefer others that search for the problem title to read through 200 posts of us debating about whether "the egg or the chicken comes first" in the problem so to speak. Sometimes it happens. It's not a perfect system, but its at least a good thing that you, the asker, got some clarity of thought through the ride. I did too.

You are absolutely right. I had not thought about this aspect. I am very sorry if I was intrusive, in future threads I will try to be less pretentious, so as to make the discussion on the thread clear and effective for anyone who visits it. I apologise again for this. Thanks for everything, everyone! See you soon!

 

[Chestermiller]

What is "@haruspex's determination"?

Post #127.