How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

  • #1
Hak
709
56
Homework Statement
An asteroid has a spherical shape and uniform mass distribution. Its radius is ##R## and its mass ##M##. The asteroid is stationary in interstellar space when it is struck by another much smaller asteroid of mass ##m## and negligible radius relative to ##R##. The minor asteroid has velocity ##v## before the impact, and its direction is tangent to the surface of the major asteroid. The impact is completely inelastic, so that the minor asteroid remains embedded inside the major one, but it is not destroyed; it is just embedded inside the rock. The gravity is negligible.

After the collision has occurred, what is the value of the force acting on the minor asteroid?
Relevant Equations
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Before the collision, the total linear momentum of the system is ##p = mv##, where ##m## is the mass of the minor asteroid and ##v## is its velocity. The total angular momentum of the system is ##L = mRv##, where ##R## is the radius of the major asteroid. Since there are no external forces or torques acting on the system, both ##p## and ##L## are conserved during and after the collision.

After the collision, the minor asteroid becomes embedded inside the major one, so they move together as a single rigid body. The linear momentum of this body is still ##p = mv##, so its velocity is ##v' = \frac{p}{(M + m)}##, where ##M## is the mass of the major asteroid. The angular momentum of this body is also still ##L = mRv##, so its angular velocity is ##\omega' = \frac{L}{I}##, where ##I## is the moment of inertia of the body.

The moment of inertia of a sphere with uniform mass distribution and radius ##R## is ##I = (2/5)MR^2##. However, since there is a small mass ##m## embedded inside the sphere at a distance ##R## from its center, we need to use the parallel axis theorem to find the moment of inertia of the body. The parallel axis theorem states that ##I = I_{cm} + md^2##, where ##I_{cm}## is the moment of inertia about the center of mass and ##d## is the distance from the center of mass to the axis of rotation. In this case, ##I_cm = (2/5)MR^2## and ##d = R##, so we get ##I = (\frac{2}{5})MR^2 + mR^2 = (\frac{2}{5} + \frac{m}{M})MR^2##.

Now we can find the angular velocity ##\omega'## by plugging in ##L## and ##I## into ##\omega' = \frac{L}{I}##. We get ##\omega' = \frac{(mRv)}{[(\frac{2}{5} + \frac{m}{M})MR^2]} = \frac{(5v)}{[2R(\frac{5}{2} + \frac{m}{M})]}##. This means that after the collision, both asteroids rotate together with an angular velocity ##\omega'## around their common center of mass.

The force acting on the minor asteroid after the collision is due to its centripetal acceleration toward the center of rotation. The centripetal acceleration is given by ##a_c = \omega'^2 R##, where ##\omega'## is the angular velocity and ##R## is the distance from the center. The force is then given by ##F = ma_c##, where ##m## is the mass of the minor asteroid. Plugging in ##\omega'## and ##R##, we get $$F = m(\omega '^2 R) = m \left[\frac{(5v)}{\left[2R \left(\frac{5}{2} + \frac{m}{M}\right)\right]}\right]^2 R = \frac{(25mv^2)}{\left[4R \left(\frac{5}{2} + \frac{m}{M}\right)^2 \right]}$$.

Where am I going wrong?
 
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  • #2
I'm not sure about this, but it seems like your equation for linear momentum conservation is not quite correct. What is the linear velocity of the center of mass of the impacting asteroid ##m## immediately after impact?
 
  • #3
erobz said:
I'm not sure about this, but it seems like your equation for linear momentum conservation is not quite correct. What is the linear velocity of the center of mass of the impacting asteroid ##m## immediately after impact?
I don't think I understand. I would set the equation ##mv = (M+m) V_{cm}##. What do you think?
 
  • #4
Hak said:
I don't think I understand. I would set the equation ##mv = (M+m) V_{cm}##. What do you think?
The way I see it immediately after it is imbedded it's on the periphery of an asteroid that has some translational velocity ##v'## and angular velocity ##\omega##. What is the linear velocity of the little mass?
 
  • #5
erobz said:
The way I see it immediately after it is imbedded it's on the periphery of an asteroid that has some translational velocity ##v'## and angular velocity ##\omega##. What is the linear velocity of the little mass?

Is it ##V = \sqrt{v' ^ 2 + \omega ^2 R^2}##?
 
  • #6
Hak said:
Is it ##V = \sqrt{v' ^ 2 + \omega ^2 R^2}##?
I think we are talking about very short timescales for the imbedding here, meaning its velocity is pretty much parallel to ##v'##.
 
  • #7
erobz said:
I think we are talking about very short timescales for the imbedding here, meaning its velocity is pretty much parallel to ##v'##.
So, is it ##V = v' + \omega R##?
 
  • #8
Hak said:
So, is it ##V = v' + \omega R##?
Thats what I think. Does that seem plausible?
 
  • #9
erobz said:
Thats what I think. Does that seem plausible?
Yes, it seems plausible to me, but ... shall we wait for some other opinions along these lines? In any case, if this is the correct value of ##V##, how should we proceed to find ##F##?
 
  • #10
Hak said:
Yes, it seems plausible to me, but ... shall we wait for some other opinions along these lines?
Sure
Hak said:
In any case, if this is the correct value of ##V##, how should we proceed to find ##F##?
I would continue in the same manner as you were before, but I haven't solved it yet., maybe there is an issue.
 
  • #11
erobz said:
I would continue in the same manner as you were before, but I haven't solved it yet., maybe there is an issue.
Meaning? Substitute ##V## in place of ##v'## in my previous procedure?
 
  • #12
Hak said:
Meaning? Substitute ##V## in place of ##v'## in my previous procedure?
Did you utilize ##v'## in your procedure? Because it might not make a bit of difference as far as this problem is concerned.
 
  • #13
erobz said:
Did you utilize ##v'## in your procedure? Because it might not make a bit of difference as far as this problem is concerned.
No, you are right. It seems that my process is deficient. So, what do you recommend? Thank you for your answer.
 
  • #14
Hak said:
No, you are right. It seems that my process is deficient. So, what do you recommend? Thank you for your answer.
Do you know that you have the wrong result ( answer key)? Or do you just think it's not correct?
 
  • #15
erobz said:
Do you know that you have the wrong result ( answer key)? Or do you just think it's not correct?
I think my result is not correct. I do not know if I have the wrong result.
 
  • #16
Hak said:
I think my result is not correct.
Well, what bothers you about it?
 
  • #17
erobz said:
Well, what bothers you about it?
The fact that we should consider, as you said, ##V = v' + \omega R##, and not simply ##v'##.
 
  • #18
Hak said:
The fact that we should consider, as you said, ##V = v' + \omega R##, and not simply ##v'##.
While what you stated about linear momentum was not correct (and needed correcting), do we need to worry about linear momentum for this problem?
 
  • #19
Just so you are aware, in case you are not,
Hak said:
The asteroid is stationary in interstellar space
describes a nonsensical situation, since it implies a universal reference frame in which something can be "stationary".
 
  • #20
erobz said:
While what you stated about linear momentum was not correct (and needed correcting), do we need to worry about linear momentum for this problem?
I have absolutely no idea at this point in the discussion.
 
  • #21
phinds said:
Just so you are aware, in case you are not,

describes a nonsensical situation, since it implies a universal reference frame in which something can be "stationary".
Good point. So how would this assertion change the course of the problem? What would it entail?
 
  • #22
Hak said:
I have absolutely no idea at this point in the discussion.
There has to be something in the back of your mind that is telling you its important here.

Let me ask you this; if the asteroid were moving with velocity ##2v'## instead of ##v'## and its angular velocity was still ## \omega ## with the same radius, would it make a difference in the force the imbedded mass was feeling?
 
  • #23
erobz said:
There has to be something in the back of your mind that is telling you its important here.

Let me ask you this; if the asteroid were moving with velocity ##2v'## instead of ##v'## and its angular velocity was still ## \omega ## with the same radius, would it make a difference in the force the imbedded mass was feeling?

No, that's right, you're right. I think I understand that. I got hung up on something that was not relevant to the exercise. So, in your opinion, is my result for ##F## correct? I still have my doubts, I don't really trust my solutions.... Anyway, thank you.
 
  • #24
Hak said:
No, that's right, you're right. I think I understand that. I got hung up on something that was not relevant to the exercise. So, in your opinion, is my result for ##F## correct? I still have my doubts, I don't really trust my solutions.... Anyway, thank you.
I haven't checked the algebra, but the methods seem correct to me. If I'm wrong, we both will be corrected shortly...that I know!
 
  • #25
erobz said:
I haven't checked the algebra, but the methods seem correct to me. If I'm wrong, we both will be corrected shortly...that I know!

Alright, thank you very much. Let me know whether the algebra is correct, so whether the result is okay in your opinion. In any case, it would be very nice if other members also give their opinions.
 
  • #26
Hak said:
Alright, thank you very much. Let me know whether the algebra is correct, so whether the result is okay in your opinion. In any case, it would be very nice if other members also give their opinions.
While we wait you could calculate ##v'##?

Or you could show more steps on your math. Just clear fractions in the denominator of ##\omega'## why are you leaving fractions in there?
 
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  • #27
Hak said:
Good point. So how would this assertion change the course of the problem? What would it entail?
Since you have specified that the "moving" one is at velocity "v", it doesn't change the math at all, you just need to leave out the "stationary" and say that the two have a relative velocity of "v", not that one of them is "stationary".
 
  • #28
erobz said:
While we wait you could calculate ##v'##?

Or you could show more steps on your math. Just clear fractions in the denominator of ##\omega'## why are you leaving fractions in there?
Ok.

We have: $$mv = (m+M) V_{cm}$$, where ##V_{cm} = (v' + \omega ' R)##. So:
$$mv = (m+M) (v' + \omega ' R) \Rightarrow \ mv = (m+M)v' + (m+M) \omega' R \Rightarrow \ (m+M) v' = mv - (m+M) \omega' R$$

By clearing the fractions in the denominator, the expression for ##\omega'## becomes:

$$\omega' = \frac{5mv}{(2M + 5 m)R}$$. Therefore:

$$(m+M) v' = mv - (m+M) \omega' R \Rightarrow \ (m+M) v' = mv - (m+M) \frac{5mv}{(2M + 5 m)} \Rightarrow \ (m+M) v' = mv \left[1- \frac{5 (M+m)}{2M + 5m}\right] \Rightarrow \ v' = \frac{m}{m+M} \left[1- \frac{5 (M+m)}{2M + 5m}\right] v$$.

What do you think?
 
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  • #29
phinds said:
Since you have specified that the "moving" one is at velocity "v", it doesn't change the math at all, you just need to leave out the "stationary" and say that the two have a relative velocity of "v", not that one of them is "stationary".
You are right, but I did not make up the text. I merely reported it. Thank you very much anyway, you were very precise and accurate.
 
  • #30
Hak said:
We have: $$mv = (m+M) V_{cm}, where V_{cm} = (v' + \omega ' R)$$
Think of them as two separate masses still. Do they both have center of mass velocity = ##v' + R \omega ##?
 
  • #31
erobz said:
Think of them as two separate masses still. Do they both have center of mass velocity = ##v' + R \omega ##?
How do you mean? Imposing conservation of momentum in an inelastic collision, isn't the velocity of the center of mass the same for both bodies? I guess I don't understand...
 
  • #32
Hak said:
How do you mean? Imposing conservation of momentum in an inelastic collision, isn't the velocity of the center of mass the same for both bodies? I guess I don't understand...
That is not necessarily the case on extended bodies. If the little mass came in with no angular momentum relative to the large mass, sure...but is that the case here?
 
  • #33
erobz said:
That is not necessarily the case on extended bodies. If the little mass came in with no angular momentum relative to the large mass, sure...but is that the case here?

Certainly not. Would you have any hints on how to proceed?
 
  • #34
1695518154091.png
 
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  • #35
Hak said:
Certainly not. Would you have any hints on how to proceed?
Carefully re-read post 4.
 

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