How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[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.