How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[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 $F = 0$.

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

$$F = \frac{25 m^3 M v^2}{(2M + 7m)^2 (M+m) R}$$.

Therefore, for $M = 9m$, I get:

$$F = \frac{9mv^2}{250 R}$$.

Where do I go wrong?

 

[PeroK]

Substituting in $F = M \omega ^2 d$,
Where do I go wrong?

That should be $F = m \omega ^2 d$.

 

[Hak]

That should be $F = m \omega ^2 d$.

I edited my message. Sorry.

 

[Hak]

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 $\omega \approx 0$?

Actually no, because that value of $\omega$ is wrong.

 

[haruspex]

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 $\omega \approx 0$?

m<<M means we are looking for the smallest order nonzero approximation for small m/M. If a function of $x$ is $\Sigma_{n=0}c_nx^n$ that is $c_ix^i$, where $c_i\neq 0$ and $c_j=0$ for $j<i$.
So no, we cannot end up with F=0.

 

[PeroK]

Note that I used $R$ instead of $d$ in calculating $\omega$, so I'll have to redo my calculation.

 

[haruspex]

the previous calculations regarding Icm and conservation of momentum are wrong.

Not wrong enough to matter. $I_{cm}\approx \frac 25MR^2$, $I_{cm}\omega\approx mvR$.

 

[Hak]

Not wrong enough to matter. $I_{cm}\approx \frac 25MR^2$, $I_{cm}\omega\approx mvR$.

Yes, if $m \ll M$.

 

[Hak]

Note that I used $R$ instead of $d$ in calculating $\omega$, so I'll have to redo my calculation.

Okay, I am waiting for a response. I didn't understand what result there is in Kleppner and Kolenkow....

 

[Chestermiller]

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 $\omega \approx 0$?

m/M is supposed to be small compared to unity, not approach zero absolutely. It can't be eliminated when it stands alone.

 

[Chestermiller]

Yes, if $m \ll M$.

So what's your problem?

 

[Hak]

So what's your problem?

None, I had not noticed this aspect before @haruspex's message.

 

[erobz]

When I don't use $m \ll M$ and plug in $M= 9m$ I get:

$$ I_{com} = \frac{9}{2}m R^2 $$

$$ \omega = \frac{v}{5R} $$

$$F = \frac{9mv^2}{250 R}$$

 

[PeroK]

When I don't use $m \ll M$ and plug in $M= 9m$ I get:

$$ I_{com} = \frac{9}{2}m R^2 $$

$$ \omega = \frac{v}{5R} $$

$$F = \frac{9mv^2}{250 R}$$

That's what I get now.

 

[Hak]

That's what I get now.

Yes, that should be right. So, is my generic result in post #101 correct?

 

[PeroK]

m/M is supposed to be small compared to unity, not approach zero absolutely. It can't be eliminated when it stands alone.

I'd like to see the maths. There is a term in $\frac {m^2v}{M}$ in the final calculation for $F$. I don't know how to handle that if we have $m \ll M$.

 

[Hak]

I'd like to see the maths. There is a term in $\frac {m^2v}{M}$ in the final calculation for $F$. I don't know how to handle that if we have $m \ll M$.

Where is this term? I can't see it.

 

[PeroK]

PS it's not like the calculations are particularly complicated without the simplification. And, ultimately, the force per unit mass on $m$ is proportional to $\frac m M$.

 

[erobz]

I am missing the utility in the approximation $m \ll M$ that is being talked about, given that a person like me can solve it without.

 

[Hak]

PS it's not like the calculations are particularly complicated without the simplification. And, ultimately, the force per unit mass on $m$ is proportional to $\frac m M$.

Excuse me, but I cannot see what you are saying. Could you check my result in post #101 and tell me where I am wrong? Thank you very much.

 

[PeroK]

Excuse me, but I cannot see what you are saying. Could you check my result in post #101 and tell me where I am wrong? Thank you very much.

Yes, that looks right to me.

 

[Hak]

Yes, that looks right to me.

Okay, but how to make my equation for force per unit mass visibly proportional to $\frac{m}{M}$? In the case where $m \ll M$, how would such an equation become?

 

[PeroK]

Okay, but how to make my equation for force per unit mass visibly proportional to $\frac{m}{M}$? In the case where $m \ll M$, how would such an equation become?

That's with the simplifications using $\frac m M \ll 1$. There's a discussion about whether that's a valid approximation to make in this case.

 

[Hak]

That's with the simplifications using $\frac m M \ll 1$. There's a discussion about whether that's a valid approximation to make in this case.

Sorry to bother you, but I didn't understand very well, I understood only partially.

 

[PeroK]

Sorry to bother you, but I didn't understand very well, I understood only partially.

Don't worry about it. The thread has 125 posts already!

 

[Hak]

Don't worry about it. The thread has 125 posts already!

So?

 

[haruspex]

Okay, but how to make my equation for force per unit mass visibly proportional to $\frac{m}{M}$? In the case where $m \ll M$, how would such an equation become?

Not sure what you are asking. You have established $\omega\approx\frac{5mv}{2MR}$ and $F\approx mR\omega^2$. Put those together.

 

[kuruman]

Just for the record.

My expression in post#55 for finding the moment of inertia of the composite mass is $$\frac{2}{5}MR^2+md^2=I_{cm}+(M+m)(R-d)^2.$$ If one replaces $d=\dfrac{MR}{M+m}$ and solves for $I_{cm}$, one gets $$I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2.$$ @erobz 's expression for $I_{cm}$ in post#62 is $$I_{cm} = \frac{2}{5}MR^2 + M(R-d)^2 + md^2.$$
If one replaces $d=\dfrac{MR}{M+m}$ and solves for $I_{cm}$, one gets $$I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2.$$Just because the expressions look different doesn't mean that they are different. Folks who are interested in expansions should note that $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{\cancel{7}5m}{2M}\right).$$

Edited to fix error in the approximate expression. See posts #208 and #231

 

[Hak]

Not sure what you are asking. You have established $\omega\approx\frac{5mv}{2MR}$ and $F\approx mR\omega^2$. Put those together.

Thank you, I understand now. So, the force $F$ is not approximate to $0$ when $m \ll m$ as was said previously, right?

 

[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 $\mu = \frac m M$

Whereas, if we make the simplifications earlier in the calculation, we get:
$$F \approx \frac{25mv^2}{4R}\big [\mu^2 \big ]$$These are the same up to the second order in $\mu$. But, there is a factor of $8$ in the third order term.

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.

I'm not sure what conclusion to draw.

 

[PeroK]

Actually, my conclusion is that the question should be explicit about giving an answer to the first significant order of $\frac m M$. Also, it seems pointless to calculate the centripetal force on $m$. Calculating $\omega$ would make much more sense.

 

[kuruman]

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.

It's right unless I made the same mistake as you.

 

[PeroK]

It's right unless I I made the same mistake as you.

But, not really a mistake. I unwittingly made the simplification $d \approx R$. Which is significant if we take $\mu = 0.1$, which is too large for the simplifications to be valid.

 

[Hak]

Could you please explain the simplifications you have made? How did you come to such conclusions in retrospect?

 

[PeroK]

Could you please explain the simplifications you have made? How did you come to such conclusions in retrospect?

Just do what @haruspex suggested.

 

[Hak]

Just do what @haruspex suggested.

I did, but I didn't understand the percentages you set up.

 

[PeroK]

I did, but I didn't understand the percentages you set up.

Just plug in the numbers!

 

[kuruman]

But, not really a mistake. I unwittingly made the simplification $d \approx R$. Which is significant if we take $\mu = 0.1$, which is too large for the simplifications to be valid.

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.)

 

[Hak]

Funny thing, I did too. My corrected exact value for $\omega$ when $M=9m$ is $$\omega =\frac{1}{25}\frac{v}{R}.$$It turns out that no approximation is needed if one uses the simplified moment of inertia in post#128.

Sorry to bother you again, could you explain this assumption better? Thank you for your patience, sorry again.

 

[kuruman]

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.

 

[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?