How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[PeroK]

It's not an assumption. I am following @PeroK's lead in post#95. I derived an expression for $\omega$ that I think is correct, I plugged in $M=9m$ and simplified. This allows for other users to check their answers against mine without anyone giving the answer away which is against our rules.

The OP got the answer in post #101. The problem is solved. But, the thread goes on!

 

[Hak]

It's not an assumption. I am following @PeroK's lead in post#95. I derived an expression for $\omega$ that I think is correct, I plugged in $M=9m$ and simplified. This allows for other users to check their answers against mine without anyone giving the answer away which is against our rules.

I did not understand the final part of the message. I simply meant to say that I am not clear about the distinction between generic values and approximate values in your expressions of $\omega$, $I_{cm}$ and $F$. If you feel like explaining it to me, I thank you very much.

 

[Hak]

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.

I would like to know something about this request. Do you have any additional ideas about possible applications of this problem?

 

[Hak]

The OP got the answer in post #101. The problem is solved. But, the thread goes on!

My goal is not just to solve the problem, but to understand what is behind that problem, how many different ways it can be solved, and what and how many observations can be made in retrospect. So I try to learn as much information as I can from those who know more about the subject than I do. Sorry if I come across as annoying in asking all the time, it is just for good purposes.

 

[PeroK]

We have a series of calculations, all of which can be approximated at each stage:
$$d = \frac{MR}{m+M} = \frac{R}{\mu+1} \approx R$$$$I = \frac{MR^2}{m+M}\bigg [\frac{7m+2M}{5} \bigg] = \frac 2 5 MR^2\bigg [\frac{\frac 7 2 \mu + 1}{\mu + 1} \bigg ] \approx \frac 2 5 MR^2$$$$L = \frac{mvR}{\mu + 1} \approx mvR$$$$w = \frac L I = \frac{5mv}{2MR}\bigg [\frac 1 {\frac 7 2 \mu + 1} \bigg ] \approx \frac{5mv}{2MR}$$$$F = m\omega^2 d =\frac{25mv^2}{4R}\bigg [\frac 1 {1 +\mu} \bigg ] \bigg [\frac{\mu}{1 + \frac{7\mu}{2}} \bigg ]^2 \approx \frac{25mv^2}{4R}\big [\mu^2 \big ]$$The approximations are only really valid where $\mu \equiv \frac m M \le 0.01$.

Edit: The approximations are only really valid where $\mu \equiv \frac m M \le 0.001$.

 

[kuruman]

I did not understand the final part of the message. I simply meant to say that I am not clear about the distinction between generic values and approximate values in your expressions of $\omega$, $I_{cm}$ and $F$. If you feel like explaining it to me, I thank you very much.

You found an expression for $\omega$ in post #101.

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

What do you get when you substitute $M=9m$?

 

[Hak]

You found an expression for $\omega$ in post #101.

What do you get when you substitute $M=9m$?

Don't worry, I understand now. My problem was with the approximations, but @PeroK's post was maximally elucidating. Thank you all!

 

[Hak]

I would like to know something about this request. Do you have any additional ideas about possible applications of this problem?

Do you have any tips?

 

[Hak]

Is $mv = M v' + m (v'+ \omega R)$ correct? All this is irrelevant for calculating the force $F$, right?

I have a doubt. In post #50, I calculated the linear momentum equation, which @erobz found to be correct in post #51.
Nevertheless, however, if we calculate the angular momentum with respect to the center of the asteroid of mass $M$, we have that:

$$L_{before} = mvR$$
$$L_{after} = L_{cm} + y_{cm} V_{cm} (m+M)$$, where:

$$L_{cm} = I_{cm} \ \omega$$,
$$y_{cm} = \frac{m}{M+m}R$$ , obtained by placing an axis of reference $Oxy$ centered in the center of the large sphere,
$$V_{cm} = \frac{m}{M+m}v$$, obtained from equation $(1) \ mv = (m+M) V_{cm}$.

Equalizing $L_{before}$ and $L_{after}$, we obtain $mvd = I_{cm} \omega$, the same equation calculated by @kuruman in post #55.

So, why is there discordance between the equation in post #50 and the equation (1) I reported in this post?

 

[erobz]

I have a doubt. In post #50, I calculated the linear momentum equation, which @erobz found to be correct in post #51.
Nevertheless, however, if we calculate the angular momentum with respect to the center of the asteroid of mass $M$, we have that:

$$L_{before} = mvR$$
$$L_{after} = L_{cm} + y_{cm} V_{cm} (m+M)$$, where:

$$L_{cm} = I_{cm} \ \omega$$,
$$y_{cm} = \frac{m}{M+m}R$$ , obtained by placing an axis of reference $Oxy$ centered in the center of the large sphere,
$$V_{cm} = \frac{m}{M+m}v$$, obtained from equation $(1) \ mv = (m+M) V_{cm}$.

Equalizing $L_{before}$ and $L_{after}$, we obtain $mvd = I_{cm} \omega$, the same equation calculated by @kuruman in post #55.

So, why is there discordance between the equation in post #50 and the equation (1) I reported in this post?

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

 

[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"?