How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[PeroK]

Okay, so what we have is the full calculation giving:
$$F = \frac{25mv^2}{4R}\bigg [\frac 1 {1 +\mu} \bigg ] \bigg [\frac{\mu}{1 + \frac{7\mu}{2}} \bigg ]^2 \approx \frac{25mv^2}{4R}\big [\mu^2 -8\mu^3 \big ]$$Where $\mu = \frac m M$

Just to explain this calculation for the uninitiated. For $\mu \ll 1$ we can expand the expression using the binomial theorem (which is a special case of Taylor series), and neglect all but the lowest order term:
$$\frac 1 {1 +\mu} = (1 + \mu)^{-1} = 1 - \mu + \mu^2 + \dots$$$$(1 + \frac{7\mu}{2})^{-2} = 1 - 7\mu + \frac{147}{4}\mu^2 + \dots$$Hence:
$$\bigg [\frac 1 {1 +\mu} \bigg ] \bigg [\frac{\mu}{1 + \frac{7\mu}{2}} \bigg ]^2 = \mu^2(1 - \mu + \mu^2 + \dots)(1 - 7\mu + \frac{147}{4}\mu^2 + \dots) = \mu^2(1 -8\mu + \dots)$$And,for $\mu \ll 1$, this reduces to the approximation $\mu^2$.

 

[Hak]

Just to explain this calculation for the uninitiated. For $\mu \ll 1$ we can expand the expression using the binomial theorem (which is a special case of Taylor series), and neglect all but the lowest order term:
$$\frac 1 {1 +\mu} = (1 + \mu)^{-1} = 1 - \mu + \mu^2 + \dots$$$$(1 + \frac{7\mu}{2})^{-2} = 1 - 7\mu + \frac{147}{4}\mu^2 + \dots$$Hence:
$$\bigg [\frac 1 {1 +\mu} \bigg ] \bigg [\frac{\mu}{1 + \frac{7\mu}{2}} \bigg ]^2 = \mu^2(1 - \mu + \mu^2 + \dots)(1 - 7\mu + \frac{147}{4}\mu^2 + \dots) = \mu^2(1 -8\mu + \dots)$$And,for $\mu \ll 1$, this reduces to the approximation $\mu^2$.

Ok, thank you very much. I didn't understand why in post #130 you say:

With $\mu = 0.01$ there is still an 8% difference. That seems quite significant. You need $\mu = 0.001$ for the early simplifications to give only a 1% error.

whereas, in post #145, you say:

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 $\mu \equiv \frac m M \le 0.01$.

Especially the two sentences in bold, seem to me to contradict each other. Could you elucidate this for me?

 

[PeroK]

Ok, thank you very much. I didn't understand why in post #130 you say:

whereas, in post #, you say:

Especially the two sentences in bold, seem to me to contradict each other. Could you elucidate this for me?

It's just a typo in the second one. I meant $0.001$. This is a reason I think the full calculation was worth doing. Normally, we might consider $0.01 \ll 1$, because it's two orders of magnitude less than 1. But, in this case, when we do the full calculation, we find that even when $\mu = 0.01$, we have a significant error in the basic approximation. Whereas, if we want less than 1% error, we need $\mu = 0.001$ or less.

By doing the full calculation, we know the range of $\mu$ for which the approximation is valid. Without that, it would be easy to assume (wrongly) that $\mu = 0.01$ is sufficiently small.

In general, I'd be wary of making approximations too early in a calculation. Even though in this case @haruspex and @Chestermiller were correct.

 

[Hak]

It's just a typo in the second one. I meant $0.001$. This is a reason I think the full calculation was worth doing. Normally, we might consider $0.01 \ll 1$, because it's two orders of magnitude less than 1. But, in this case, when we do the full calculation, we find that even when $\mu = 0.01$, we have a significant error in the basic approximation. Whereas, if we want less than 1% error, we need $\mu = 0.001$ or less.

By doing the full calculation, we know the range of $\mu$ for which the approximation is valid. Without that, it would be easy to assume (wrongly) that $\mu = 0.01$ is sufficiently small.

In general, I'd be wary of making approximations too early in a calculation. Even though in this case @haruspex and @Chestermiller were correct.

It is all clear now. Thank you for everything!

 

[haruspex]

In general, I'd be wary of making approximations too early in a calculation.

I agree.
The strictly correct procedure is to obtain the exact equation before making any approximations. That allows you to figure out the range of validity, and avoids the blunder of discarding important terms. E.g. if we have $y=f(x)+g(x)$, approximate $g(x)=c_0x+o(x)$ (I am using $o(), O ()$ 'order' notation here), so simplify to $y=f(x)+c_0x$, then find that $f(x)=-c_0x+o(x)$ and again discard the $o(x)$ term we end up with $y=0$. We have thrown the baby out with the bathwater.

 

[Hak]

Sorry to bother, and to clog the thread again.
The author of the problem has informed me that the solution we found is correct. However, I would like to clarify an observation that the same author made. He says:

Immediately after the collision, we can see instantaneous motion (only instantaneous motion!) as one of these two options:

1. $M$ translates and rotates around its center; $m$ translates. However, this description works only at the initial instant!
2. $M$ and $m$ rotate around the CoM, and CoM translates.

He says also that this can also be justified in a physical way...

In my opinion, option 2. is related to equation $mv = (M+m) V_{cm}$. Option 1. relates to which equation? I cannot understand it. Any help would be appreciated.

Also, my friend solves the problem in the following way:

If $v'$ is the translation velocity of the asteroid that of embedded m is v' + \omega R. Then for conservation of momentum it results mv= mv'+ m\omega R+Mv' while for conservation of angular momentum it resultsmvR=mv'R+m\omega R^2+(2/5)MR^2\omega The rotational momentum of the asteroid is zero. Doing the math if I have not made a mistake results in v'= \frac{m(v-\omega R)}{m+M} and \omega= \frac{mv}{(R/5)(7m+2M)} so F_c= mR[\frac{mv}{(R/5)(7m+2M)}]^2.

I cannot understand how it is possible that although the calculations are not convincing (perhaps they are wrong), the angular velocity is the same as we found. Is there a possibility that these calculations are correct instead?

 

[Hak]

Folks who are interested in expansions should note that $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{m}{M}\right).$$

Could you explain how this expansion can be obtained? Thank you very much.

 

[kuruman]

Could you explain how this expansion can be obtained? Thank you very much.

Actually, it should be $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{7m}{2M}\right).$$Is that what you got? I corrected post #128.

 

[Hak]

Actually, it should be $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{7m}{2M}\right).$$Is that what you got? I corrected post #128.

No, I have not understood by which process this is to be achieved. The previous result made me suspicious because you had said that one would obtain $\omega$ without approximations with this reduced value of $I_{cm}$, but using that value it was not so. Could you explain how it should be obtained?

 

[Hak]

Sorry to bother, and to clog the thread again.
The author of the problem has informed me that the solution we found is correct. However, I would like to clarify an observation that the same author made. He says:

Immediately after the collision, we can see instantaneous motion (only instantaneous motion!) as one of these two options:

1. $M$ translates and rotates around its center; $m$ translates. However, this description works only at the initial instant!
2. $M$ and $m$ rotate around the CoM, and CoM translates.

He says also that this can also be justified in a physical way...

In my opinion, option 2. is related to equation $mv = (M+m) V_{cm}$. Option 1. relates to which equation? I cannot understand it. Any help would be appreciated.

Also, my friend solves the problem in the following way:

If $v'$ is the translation velocity of the asteroid that of embedded m is v' + \omega R. Then for conservation of momentum it results mv= mv'+ m\omega R+Mv' while for conservation of angular momentum it resultsmvR=mv'R+m\omega R^2+(2/5)MR^2\omega The rotational momentum of the asteroid is zero. Doing the math if I have not made a mistake results in v'= \frac{m(v-\omega R)}{m+M} and \omega= \frac{mv}{(R/5)(7m+2M)} so F_c= mR[\frac{mv}{(R/5)(7m+2M)}]^2.

I cannot understand how it is possible that although the calculations are not convincing (perhaps they are wrong), the angular velocity is the same as we found. Is there a possibility that these calculations are correct instead?

By chance, do you have any ideas about this?

 

[erobz]

By chance, do you have any ideas about this?

If it's not a coincidental set of mistakes that fortuitously cancel (which does happen), then its claims must be justifiable withing the accepted framework. What parts of the argument can you justify given what we painstakingly took the time to discover earlier?

 

[Hak]

If it's not a coincidental set of mistakes that fortuitously cancel (which does happen), then its claims must be justifiable withing the accepted framework. What parts of the argument can you justify given what we painstakingly took the time to discover earlier?

That's what I think (and continue to think) too. I cannot, however, justify his analysis. It seems to me totally different from the considerations we have been dealing with. I would be grateful if someone could help me out.

 

[Hak]

Actually, it should be $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{7m}{2M}\right).$$Is that what you got? I corrected post #128.

@kuruman I experience two annoying issues.
1) I cannot figure out where this result comes from or what process needs to be used to get it.

In post #138, you say:

Funny thing, I did too. My corrected exact value for $\omega$ when $M=9m$ is $$\omega =\frac{1}{5}\frac{v}{R}.$$It turns out that no approximation is needed if one uses the simplified moment of inertia in post#128.

(Original expression edited to add missing factor of 5 in the numerator.)

2) I cannot understand what you are referring to when you say there is no need for approximations. To achieve what? At first, I thought it was the angular velocity $\omega$. Using the expression $\omega = \frac{L}{I_{cm}}$, and employing the simplified moment of inertia above, I get none of the values of $\omega$ found earlier. What is wrong with this? Could @kuruman, or anyone else, tell me how this assertion should be understood, what it is meant to mean? Thank you very much.

 

[haruspex]

Option 1. relates to which equation?

It can be written as equations in more than one way, but what the author is saying is that the direction of motion of m is initially unchanged, and M's motion can be thought of as the sum of a rotation about its centre and linear motion of its centre (which is always true).
If M's rotation rate is $\omega$ and initial post-collision mass centre velocity is v' then m's velocity is $v'+R\omega$ and we have $Mv'+m(v'+R\omega)=mv$.

 

[Hak]

It can be written as equations in more than one way, but what the author is saying is that the direction of motion of m is initially unchanged, and M's motion can be thought of as the sum of a rotation about its centre and linear motion of its centre (which is always true).
If M's rotation rate is $\omega$ and initial post-collision mass centre velocity is v' then m's velocity is $v'+R\omega$ and we have $Mv'+m(v'+R\omega)=mv$.

Okay, thank you very much. So, does this mean that my friend's calculations are correct, since the first equation is basically the same?

 

[haruspex]

Okay, thank you very much. So, does this mean that my friend's calculations are correct, since the first equation is basically the same?

Which post?

Btw, in my experience, finding the common mass centre of a composite body is rarely helpful. Usually easier to calculate momenta, moments of inertia etc. separately and then add.

 

[Hak]

Which post?

Btw, in my experience, finding the common mass centre of a composite body is rarely helpful. Usually easier to calculate momenta, moments of inertia etc. separately and then add.

Thanks for the advice, but where was the common mass centre of a composite body found? What I am asking is in the last part of post #206. If you can, also look at post #213, there may be an issue there.

 

[erobz]

lets see if we can make sense of it.

Also, my friend solves the problem in the following way:
$\vdots$
while for conservation of angular momentum it resultsmvR=mv'R+m\omega R^2+(2/5)MR^2\omega The rotational momentum of the asteroid is zero.

Would you say that this is the diagram that goes along with your friends approach?

 

[Hak]

lets see if we can make sense of it.

Would you say that this is the diagram that goes along with your friends approach?

View attachment 332738

I think so. Is that correct?
P. S. Could I ask you how you made the FBD? It's very nice!

 

[erobz]

I think so. Is that correct?

Thats what I gather from the equation that is proposed by them for conservation of angular momentum:

$$mvR = m( v' + R \omega_s ) R + \frac{2}{5}MR^2\omega_s $$

How do you solve that for $\omega_s$? please leave nothing to the imagination.

P. S. Could I ask you how you made the FBD? It's very nice!

Thank You. its just PowerPoint.

 

[kuruman]

P. S. Could I ask you how you made the FBD? It's very nice!

It's not a free body diagram because it shows no forces. It's just a regular diagram showing a "before" and "after" picture in one.

 

[Hak]

Thats what I gather from the equation that is proposed by them for conservation of angular momentum:

$$mvR = m( v' + R \omega_s ) R + \frac{2}{5}MR^2\omega_s $$

How do you solve that for $\omega_s$? please leave nothing to the imagination.

I did not understand the request. Could you rephrase?

 

[Hak]

It's not a free body diagram because it shows no forces. It's just a regular diagram showing a "before" and "after" picture in one.

You are right, sorry. I have a bad habit of calling any kind of diagram FBD...

 

[erobz]

I'm asking you to solve that for $\omega_s$ in terms of $v$, like was done in the other approaches.

 

[Hak]

@kuruman I experience two annoying issues.
1) I cannot figure out where this result comes from or what process needs to be used to get it.

In post #138, you say:

2) I cannot understand what you are referring to when you say there is no need for approximations. To achieve what? At first, I thought it was the angular velocity $\omega$. Using the expression $\omega = \frac{L}{I_{cm}}$, and employing the simplified moment of inertia above, I get none of the values of $\omega$ found earlier. What is wrong with this? Could @kuruman, or anyone else, tell me how this assertion should be understood, what it is meant to mean? Thank you very much.

Perhaps this request of mine escaped. Could I understand a little more? Thank you very much, I would be grateful.

 

[Hak]

I'm asking you to solve that for $\omega_s$ in terms of $v$, like was done in the other approaches.

My result is: $\omega_s = \frac{5mv}{R (2M+7m)}$, if we consider $Mv = Mv' + m(v'+ \omega_s R)$ as conservation of linear momentum.

 

[kuruman]

@kuruman I experience two annoying issues.
1) I cannot figure out where this result comes from or what process needs to be used to get it.

The exact expression (start point) is
$$I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2.$$ The approximate expression (end point) is
$$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{\cancel{7}5m}{2M}\right).$$ What did you try when you say you cannot figure out where this result comes from.

Do you understand approximations and how to make them? If you were to go from the start point to the end point when $m<<M$, what would you do? What is the logical first step to take given what the expressions look like?

 

[Hak]

The exact expression (start point) is
$$I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2.$$ The approximate expression (end point) is
$$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{7m}{2M}\right).$$ What did you try when you say you cannot figure out where this result comes from.

Do you understand approximations and how to make them? If you were to go from the start point to the end point when $m<<M$, what would you do? What is the logical first step to take given what the expressions look like?

I don't understand approximations very well; it's a subject I never dealt with in school. My process was as follows:

To go from the start point to the end point when $m<<M$, we need to use the approximation $\frac{Mm}{M+m}\approx\frac{m}{1+\frac{m}{M}}\approx\frac{m}{1}\approx m$. This is because when $m<<M$, the fraction $\frac{m}{M}$ is very small and can be neglected in the denominator. The logical first step is to apply this approximation to the start point expression and simplify it. This will give us:

$$I_{cm}\approx\frac{2}{5}MR^2+mR^2=\frac{2}{5}MR^2\left(1+\frac{5m}{2M}\right)$$.

Could you help me understand where I'm going wrong? I know for a fact that this procedure is wrong because I did not carry out an expansion.

 

[erobz]

My result is: $\omega_s = \frac{5mv}{R (2M+7m)}$, if we consider $Mv = Mv' + m(v'+ \omega_s R)$ as conservation of linear momentum.

Ok, that is also what I conclude. So what is the final expression for the force?

 

[Hak]

Ok, that is also what I conclude. So what is the final expression for the force?

My friend says $F = m \omega_s R$, but it's wrong. The correct expression is $F = m \omega_s d$, where $d = \frac{M}{M+m}R$, as we have seen.

 

[kuruman]

Could you help me understand where I'm going wrong?

You did nothing wrong. It's my mistake for putting a 7 in place of a 5. Sorry about the confusion.

 

[erobz]

My friend says $F = m \omega_s R$, but it's wrong. The correct expression is $F = m \omega_s d$, where $d = \frac{M}{M+m}R$, as we have seen.

So, using the equations your friend derives we find that "$\omega_s$" the angular velocity of the sphere about its own center of mass was equal to "$\omega$" which is the angular velocity of both masses about their common center of mass calculated by other methods previously discussed. Which leads to a clear discrepancy in the calculated force that $m$ feels. A force that your professor ( I presume ) has stated was correctly calculated with $F = m \omega^2 d$? Does that sum it up?

 

[kuruman]

So, using the equations your friend derives we find that $\omega_s$ the angular velocity of the sphere about its own center of mass was equal to ## \omega ## which the angular velocity of both masses about their common center of mass, Which leads to a clear discrepancy in the calculated forces. A force that your professor ( I presume ) has stated was correctly calculated with $F = m \omega d^2$

$m\omega d^2$ does not have dimensions of force. $m\omega^2 d$ does.

 

[Hak]

You did nothing wrong. It's my mistake for putting a 7 in place of a 5. Sorry about the confusion.

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

Moreover, you had said that the same result as $\omega$ would be arrived at without preliminary simplifications. By using $\omega = \frac{L}{I_{cm}}$, with $I_{cm}$ simplified moment of inertia, you arrive at the result $\omega = \frac{5mMv}{(2M + 5m) R}$, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

 

[Hak]

So, using the equations your friend derives we find that "$\omega_s$" the angular velocity of the sphere about its own center of mass was equal to "$\omega$" which is the angular velocity of both masses about their common center of mass calculated by other methods previously discussed. Which leads to a clear discrepancy in the calculated force that $m$ feels. A force that your professor ( I presume ) has stated was correctly calculated with $F = m \omega^2 d$? Does that sum it up?

Yes. However, I would like to understand why my friend considered $R$ instead of $d$. How should he introduce it into the analysis, since he has not mentioned it yet? Thanks.

 

[erobz]

Yes. However, I would like to understand why my friend considered $R$ instead of $d$. How should he introduce it into the analysis, since he has not mentioned it yet? Thanks.

The interpretation that your friend makes is not correct. @kuruman showed earlier how to conserve angular momentum about the mass center of $M$ using $R$, but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass. As far as I can tell your friend gets to the same angular velocity (symbolically), but it is doomed because it is about the wrong center of mass.

 

[Hak]

The interpretation that your friend makes is not correct. @kuruman showed earlier how to conserve angular momentum about the mass center of $M$ using $R$, but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass. As far as I can tell your friend gets to the same angular velocity (symbolically), but it is doomed because it is about the wrong center of mass.

That seems reasonable. However, I would like to ask for confirmation from others as well. Just to make sure.

 

[Hak]

@kuruman showed earlier how to conserve angular momentum about the mass center of $M$ using $R$.

Where? I cannot find it.

 

[erobz]

Where? I cannot find it.

You didn't look. Read though the 238 post prior this from the beginning. They are in there.

 

[Hak]

... for conservation of angular momentum it resultsmvR=mv&#039;R+m\omega R^2+(2/5)MR^2\omega The rotational momentum of the asteroid is zero.

So, is this equation wrong? Do you confirm? Thank you.

 

[erobz]

So, is this equation wrong? Do you confirm? Thank you.

Does it lead to the correct result? Anyhow it doesn't matter what I say, because you are just going to say, "ok, but lets wait for someone else to confirm".

 

[Hak]

You didn't look. Read though the 238 post prior this from the beginning. They are in there.

Is it in post #178?

 

[Hak]

Does it lead to the correct result?

Yes, but as you said, it is very likely that the angular momentum is conserved only with respect to the center of mass of the system. So how is it possible that the expression of $\omega$ is correct? I cannot explain how this is possible, while then my friend considers $R$ instead of $d$. Thank you.

 

[Hak]

Does it lead to the correct result? Anyhow it doesn't matter what I say, because you are just going to say, "ok, but lets wait for someone else to confirm".

No, I did not explain myself. I don't mean that I would like to wait for someone else to confirm, I simply mean that I am waiting for confirmation from others as well. If many people will say that this option is correct, then it will be reasonable to think that it is. I agree with you, however. Only, I cannot explain these discordances. Sorry if you were offended, that was not my intention.

 

[erobz]

Yes, but as you said, it is very likely that the angular momentum is conserved only with respect to the center of mass of the system. So how is it possible that the expression of $\omega$ is correct? I cannot explain how this is possible, while then my friend considers $R$ instead of $d$. Thank you.

My advice... let your professor explain it to them... or you can show them the thread in which it was explained to you?

 

[Hak]

My advice, let your professor explain it to them... or you can show them the thread in which it was explained to you?

I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts. It was just to get a complete overview of the possibilities of this problem.

 

[erobz]

I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts.

Why are they authoring problems they have doubts about?

It was just to get a complete overview of the possibilities of this problem.

Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?

 

[Hak]

Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?

You are right, but they don't feel like it. They prefer me to report their interpretations. Sorry if I am annoying, that is not my intention. I just try to help others and always learn more from those who know more than me.

 

[Hak]

Why are they authoring problems they have doubts about?

I have no idea. The author seems to me to be a very good professor, but he does not know all the possible solutions of his own problems. He works them out over time, making various observations and attempts.

 

[kuruman]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

$$\begin{align} & I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2 \nonumber \\
& I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right) \nonumber \\
\end{align}$$When $m<<M~$ you can approximate in the denominator $M+m\approx M$. Hence $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right).$$