How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[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.

 

[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 results$$mvR=mv'R+m\omega R^2+(2/5)MR^2\omega$$ 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 results$$mvR=mv'R+m\omega R^2+(2/5)MR^2\omega$$ 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?