How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

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...

 

[erobz]

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?

 

[Hak]

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?

 

[erobz]

 

[erobz]

Certainly not. Would you have any hints on how to proceed?

Carefully re-read post 4.

 

[Hak]

Carefully re-read post 4.

I carefully re-read post #4. Didn't we say, in this connection, that $V_{cm} = v' + \omega' R$?

 

[erobz]

I carefully reread post #4. Didn't we say, in this connection, that $V_{cm} = v' + \omega' R$?

Yeah, Apparently you did not understand which body has that as its center of mass velocity at that instant.

 

[Hak]

Yeah, Apparently you did not understand which body has that as its center of mass velocity at that instant.

I thought it was the common velocity of the two bodies, but as you well explained in post #32, it is not so in this case. I did not understand what really changes in the case where the body of mass $m$ is tangent to the body of mass $M$, according to what you depicted in post #34.

 

[erobz]

Imagine instead wheel of mass $M$, radius $R$, rolling with angular velocity $\omega$, it has center of mass velocity $v'$ in the ground frame. On the top of this wheel (at this instant) is a piece of mud (stuck to it) of mass $m$. What is the center of mass velocity of the mud in the ground frame?

 

[Hak]

Imagine instead wheel of mass $M$, radius $R$, rolling with angular velocity $\omega$, it has center of mass velocity $v'$ in the ground frame. On the top of this wheel (at this instant) is a piece of mud (stuck to it) of mass $m$. What is the center of mass velocity of the mud in the ground frame?

Isn'it $V_{top} = 2 v'$?

 

[erobz]

Isn'it $V_{top} = 2 v'$?

If its not slipping ( Which is a special case) .

What is it in general.; i.e. in terms of $v'$ and $R \omega$

 

[Hak]

If its not slipping. Which is a different problem.

What is it in terms of $v'$ and $R \omega$

Is it not the sum of center-of-mass velocity and instantaneous velocity with respect to the ground reference, i.e., $V_{top} = v' + \omega R$? If it is not this, I do not understand where I am wrong.

 

[erobz]

Is it not the sum of center-of-mass velocity and instantaneous velocity with respect to the ground reference, i.e., $V_{top} = v' + \omega R$? If it is not this, I do not understand where I am wrong.

It is exactly that! That is the center of mass velocity of that little piece of mud in the ground frame. What is the center of mass velocity of the wheel?

 

[Hak]

It is exactly that!

And so, where is the error in the previous messages?

 

[erobz]

We have: $$mv = (m+M) V_{cm}$$, where $V_{cm} = (v' + \omega ' R)$. So:What do you think?

This is not correct. The big asteroid is the "the wheel", the little asteroid is "the mud". Do they have the same center of mass velocities?

 

[Hak]

This is not correct. The big asteroid is the "the wheel", the little asteroid is "the mud". Do they have the same center of mass velocities?

Does the large asteroid have center-of-mass velocity $v'$, while the small asteroid has center-of-mass velocity $v$?

 

[erobz]

Does the large asteroid have center-of-mass velocity $v'$

This part is correct.

The rest is not. We are talking about "after the mud is stuck to the rotating wheel", what is its center of mass velocity just after its embedded (almost immediately after the collision)?

 

[Hak]

This part is correct.

we are talking about "after the mud is stuck to the rotating wheel"

So the small asteroid has center-of-mass velocity $v' + \omega R$ after getting stuck in the larger one, right?

 

[erobz]

So the small asteroid has center-of-mass velocity $v' + \omega R$ after getting stuck in the larger one, right?

Correct, do you see how to revise the equation?

 

[Hak]

Correct, do you see how to revise the equation?

Is $mv = M v' + m (v'+ \omega R)$ correct? All this is irrelevant for calculating the force $F$, right?

 

[erobz]

Is $mv = M v' + m (v'+ \omega R)$ correct?

Good.

All this is irrelevant for calculating the force $F$, right?

Correct. This was just to check conceptual understanding.

 

[Hak]

Good.

Correct. This was just to check conceptual understanding.

Thank you so much. Is the expression for $\omega$, on the other hand, correct?

 

[erobz]

Thank you so much. Is the expression for $\omega$, on the other hand, correct?

Yeah, I get what you got:

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

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

 

[Hak]

Yeah, I get what you got:

So my result for $F$ should be right. Let's wait for confirmation from other members.

 

[kuruman]

So my result for $F$ should be right. Let's wait for confirmation from other members.

I cannot confirm this answer. This is how to do it.

Angular momentum is conserved about the CM of the composite mass.
The distance from the center of mass to the embedded asteroid is $d=\dfrac{M}{M+m}R.$
The angular momentum about the CM before the collision is the orbital angular momentum of the asteroid only. We are told that the planet is not spinning.
$L_{\text{before}}=mvd=\dfrac{mMv}{M+m}R.$
The angular momentum about the CM after the collision is
$L_{\text{after}}=I_{cm}~\omega.$

What expression did you get for $I_{cm}$? I couldn't find it, but maybe it is implied somewhere. If you don't have it, use the parallel axes theorem $I_O=I_{cm}+(M+m)(R-d)^2$ where $I_O=(\frac{2}{5}MR^2+mR^2)$ is the moment of inertia about the center of the planet and $(R-d)$ is the distance between the parallel axes.

Then solve the momentum conservation equation for $\omega$.

 

[Hak]

I cannot confirm this answer. This is how to do it.

Angular momentum is conserved about the CM of the composite mass.
The distance from the center of mass to the embedded asteroid is $d=\dfrac{M}{M+m}R.$
The angular momentum about the CM before the collision is the orbital angular momentum of the asteroid only. We are told that the planet is not spinning.
$L_{\text{before}}=mvd=\dfrac{mMv}{M+m}R.$
The angular momentum about the CM after the collision is
$L_{\text{after}}=I_{cm}~\omega.$

What expression did you get for $I_{cm}$? I couldn't find it, but maybe it is implied somewhere. If you don't have it, use the parallel axes theorem $I_O=I_{cm}+(M+m)(R-d)^2$ where $I_O=(\frac{2}{5}MR^2+mR^2)$ is the moment of inertia about the center of the planet and $(R-d)$ is the distance between the parallel axes.

Then solve the momentum conservation equation for $\omega$.

Thank you. I knew something was wrong with my process. The reasoning and the unfolding are similar to yours, what changes is the algebra. I am sure your procedure is more correct than mine. After solving the momentum conservation equation for $/omega$, can I calculate the force $F$ as I did in post #1, that is, by $F = m \omega ^2 R$? How would this force be directed? I am waiting for confirmation, thanks again.

Postscript. Since I did not understand some steps, could you draw a graphic diagram of the physical situation that would help me understand it better? How would the conservation of linear momentum discussed with @erobz change, considering your corrections and observations? Thank you.

 

[erobz]

How would the conservation of linear momentum discussed with @erobz change, considering your corrections and observations?

My guess would be the term “ $\omega R$ “ gets replaced with “$\omega d$”. How about you?

 

[Hak]

My guess would be the term “ $\omega R$ “ gets replaced with “$\omega d$”. How about you?

I also think it may be so. Let's see if there is any other opinion disagreeing with this....

 

[erobz]

I also think it may be so. Let's see if there is any other opinion disagreeing with this....

That should go for the force $F$ too (replace $R$ with $d$)

 

[Chestermiller]

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: /

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

It seems to me there might be an algebraic error here. For $\frac{m}{M}<<\frac{2}{5}$, I get $$\omega^{'}=\frac{5}{2}\frac{m}{M}\frac{v}{R}$$