How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[PeroK]

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 $F = 0$.

PS I did make a mistake and authomatically used $R$ instead of $d$ when calculating $\omega$.

 

[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 $F = 0$.

I can't understand it. Why?

 

[PeroK]

I can't understand it. Why?

Because you didn't think about it for long enough?

 

[Hak]

Because you didn't think about it for long enough?

No, maybe I didn't explain myself well. I can't understand why you said that algebraic operations are not set up, nor where that result of $F$ for $M = 9 m$ came from.

 

[PeroK]

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.

 

[Chestermiller]

My understanding of the problem statement is that you are supposed to assume that $$\frac{m}{M}<<1$$and$$\frac{r}{R}<<1$$

 

[PeroK]

My understanding of the problem statement is that you are supposed to assume that $$\frac{m}{M}<<1$$

Then $F \approx 0$.

and$$\frac{r}{R}<<1$$

Yes.

 

[haruspex]

We get $\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}$.

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

 

[Hak]

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

Yes, your reasoning makes sense to me. The problem is that that value of $\omega$ would be wrong even under your (correct) assumptions, because the previous calculations regarding $I_{cm}$ and conservation of momentum are wrong.

 

[erobz]

Shouldn’t that be $= \frac{(5v)}{[2R(\frac{5}{2} + \frac{M}{m})]}$?
Reading through the thread, it seems some approaches are failing to make full use of $m<<M$. That allows us to gloss over the difference between the resulting CM and the larger planet's centre, and to simplify the above to $\frac{(5v)}{[2R(\frac{M}{m})]}=\frac{5mv}{2MR}$.

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 $\omega \approx 0$?

They say not clear things in the problem statement. "An asteroid hitting an asteroid" is stated and $m$ is much smaller than $M$. I'm thinking like its a large ( but not "planetary" large) less dense body, being impacted by a small dense body.

 

[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.