How Does Angular Momentum Conservation Affect Asteroid Collision Dynamics?

[Hak]

Yes, I see now what you are asking. The general expression for the force at any point is $\vec {F}_c=-(dm)\omega^2 \vec {r}'$ where $\vec {r}'$ is the location of the point relative to the CM. The asteroid is at location $\vec {r}'=\vec d$ and has mass $dm = m$. So the force on the asteroid is $$\vec {F}_c=-m\omega^2 \vec {d}.$$

Is it correct?

 

[kuruman]

Ah, this is a big problem. I agreed with @erobz, but at this point, I can't really understand why my friend is wrong, mistaking $d$ for $R$. Why? Sorry to persist with this kind of soap opera....

It seems to me that acting as an information conduit between your friend and PF is confusing you because you don't seem to have the expertise to diagnose where and why your friend is wrong. Your friend needs to speak for him/herself and post here. Then we will be able to diagnose your friend's problem. Maybe your friend is stringing you along trying to see how many posts this thread will add up to before it is closed down.

 

[Hak]

It seems to me that acting as an information conduit between your friend and PF is confusing you because you don't seem to have the expertise to diagnose where and why your friend is wrong. Your friend needs to speak for him/herself and post here. Then we will be able to diagnose your friend's problem. Maybe your friend is stringing you along trying to see how many posts this thread will add up to before it is closed down.

Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...

 

[haruspex]

Isn't it simpler to note that any point on the composite rigid object rotates about the CM with constant angular velocity $\vec{\omega}$? Then that point has centripetal acceleration $\vec a_c=-\omega^2 \vec {r}'$ where $\vec {r}'$ is the position vector relative to the CM. Thus the net force on mass element $dm$ is $\vec {F}_c=-(dm)\omega^2 \vec {r}'.$

Yes, I was just providing confirmation from other evidence, since @Hak and @erobz seemed unsure.

 

[kuruman]

Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...

Please remind us, what issue is this?

 

[Hak]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

Moreover, you had said that the same result as $\omega$ would be arrived at without preliminary simplifications. By using $\omega = \frac{L}{I_{cm}}$, with $I_{cm}$ simplified moment of inertia, you arrive at the result $\omega = \frac{5mMv}{(2M + 5m)(M+m) R}$, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

The second paragraph. Thanks.

 

[Hak]

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.)

This is the post I am referring to.

 

[kuruman]

The second paragraph. Thanks.

Your expression in the second paragraph is dimensionally incorrect. The angular velocity has dimensions of [v/R]. Your expression has dimensions of [Mv/r].

 

[Hak]

Your expression in the second paragraph is dimensionally incorrect. The angular velocity has dimensions of [v/R]. Your expression has dimensions of [Mv/r].

I edited my message, correcting my result. I had mistakenly copied from the paper to the screen. My expression is $\omega = \frac{5mMv}{(2M + 5m)(M+m) R}$

 

[erobz]

but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass.

I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for $\omega.$

What was your objection to the statement; that you never said it, and/or that it is not true? Because in the end, we can take angular momentum about any axis ( as you showed previously in post #178), but it is ultimately conserved about the systems center of mass in either case. i.e. $\omega$ is to be applied about the systems center of mass, not about the arbitrary axis?

 

[kuruman]

What was your objection to the statement; that you never said it, and/or that it is not true? Because in the end, we can take angular momentum about any axis ( as you showed previously in post #178), but it is ultimately conserved about the systems center of mass in either case. i.e. $\omega$ is to be applied about the systems center of mass, not about the arbitrary axis?

My objection was to both. I never said it because in this case angular momentum is conserved about any point in space since there are no torques acting on the two-mass system about any point in space. I think I'll write something up to clarify these issues, because the idea is quite simple.

 

[kuruman]

Consider an irregular rigid object of mass $M$ simultaneously rotating and translating in space in the absence of external forces and torques. This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is $\vec V_{cm}$ and its rotational angular velocity is $\vec \omega.$

First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that $\vec V_{cm}= \rm{const.}$ If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point at position $\vec {r}'$ from the CM is $\vec {v}'=\vec{\omega}\times \vec {r}'.$

Now consider mass element $dm$ at an arbitrary point P the position vector of which relative to A is $\vec r$. Let $\vec {r}'$ be the position vector of $dm$ relative to the CM and $\vec R$ be the position of the CM relative to A. Finally, let $\vec v_P$ be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is $\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.$

The angular momentum contribution of $dm$ about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.

1. $~(dm) \vec R \times \vec V_{cm}$
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ This is often called the angular momentum of the center of mass.

2. $~(dm)\vec R\times (\vec{\omega}\times \vec {r}')$
Using the triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because $\vec R$ is perpendicular to $\vec {\omega}$. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, $\int (dm)\vec {r}'=0.$

3. $~(dm)\vec {r}'\times \vec V_{cm}$
This term also vanishes upon integration for the same reason when we take constant $\vec R$ out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. $~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')$
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$This is called the angular momentum about the center of mass. Thus, the total angular momentum about point A of the translating and rotating rigid body is $$\begin{align} & \vec L=\vec L_1+\vec L_4=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\end{align}$$ The first term in this expression, angular momentum of the center of mass, depends on the choice of reference point A. The second term, angular momentum about the center of mass, does not.

The bottom line is that one can always write the angular momentum of a translating and rotating rigid body in the form of Equation (1).

Note that a simplification occurs if point A is chosen anywhere on the straight line path of the CM. Then $\vec {R}$ and $\vec V_{cm}$ are parallel and the angular momentum of the CM vanishes. This is why in the asteroid problem and all problems like it, it is convenient to choose the CM as a point of reference for the angular momentum and immediately write the angular momentum conservation equation in the form $$m~v~d=I_{cm}~\omega$$ where $d$ is the vertical distance between the path of the asteroid and the CM of the combined object. This equation says that the initial orbital angular momentum about the CM of the asteroid before the collision is converted to spin angular momentum of the combined object after the collision. Furthermore, as we have seen, choosing point A away from the path of the CM adds equal amounts of (orbital) angular momentum of the CM on both sides of the momentum conservation equation which cancel out. (See post #178.)

 

[Hak]

Consider an irregular rigid object of mass $M$ simultaneously rotating and translating in space in the absence of external forces and torques. This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is $\vec V_{cm}$ and its rotational angular velocity is $\vec \omega.$

View attachment 332757

First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that $\vec V_{cm}= \rm{const.}$ If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point at position $\vec {r}'$ from the CM is $\vec {v}'=\vec{\omega}\times \vec {r}'.$

Now consider mass element $dm$ at an arbitrary point P the position vector of which relative to A is $\vec r$. Let $\vec {r}'$ be the position vector of $dm$ relative to the CM and $\vec R$ be the position of the CM relative to A. Finally, let $\vec v_P$ be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is $\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.$

The angular momentum contribution of $dm$ about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.

1. $~(dm) \vec R \times \vec V_{cm}$
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ This is often called the angular momentum of the center of mass.

2. $~(dm)\vec R\times (\vec{\omega}\times \vec {r}')$
Using the triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because $\vec R$ is perpendicular to $\vec {\omega}$. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, $\int (dm)\vec {r}'=0.$

3. $~(dm)\vec {r}'\times \vec V_{cm}$
This term also vanishes upon integration for the same reason when we take constant $\vec R$ out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. $~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')$
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$This is called the angular momentum about the center of mass. Thus, the total angular momentum about point A of the translating and rotating rigid body is $$\begin{align} & \vec L=\vec L_1+\vec L_4=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\end{align}$$ The first term in this expression, angular momentum of the center of mass, depends on the choice of reference point A. The second term, angular momentum about the center of mass, does not.

The bottom line is that one can always write the angular momentum of a translating and rotating rigid body in the form of Equation (1).

Note that a simplification occurs if point A is chosen anywhere on the straight line path of the CM. Then $\vec {R}$ and $\vec V_{cm}$ are parallel and the angular momentum of the CM vanishes. This is why in the asteroid problem and all problems like it, it is convenient to choose the CM as a point of reference for the angular momentum and immediately write the angular momentum conservation equation in the form $$m~v~d=I_{cm}~\omega$$ where $d$ is the vertical distance between the path of the asteroid and the CM of the combined object. This equation says that the initial orbital angular momentum about the CM of the asteroid before the collision is converted to spin angular momentum of the combined object after the collision. Furthermore, as we have seen, choosing point A away from the path of the CM adds equal amounts of (orbital) angular momentum of the CM on both sides of the momentum conservation equation which cancel out. (See post #178.)

Thank you very much for your interesting contribution!

 

[Hak]

Consider an irregular rigid object of mass $M$ simultaneously rotating and translating in space in the absence of external forces and torques. This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is $\vec V_{cm}$ and its rotational angular velocity is $\vec \omega.$

View attachment 332757

First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that $\vec V_{cm}= \rm{const.}$ If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point at position $\vec {r}'$ from the CM is $\vec {v}'=\vec{\omega}\times \vec {r}'.$

Now consider mass element $dm$ at an arbitrary point P the position vector of which relative to A is $\vec r$. Let $\vec {r}'$ be the position vector of $dm$ relative to the CM and $\vec R$ be the position of the CM relative to A. Finally, let $\vec v_P$ be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is $\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.$

The angular momentum contribution of $dm$ about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.

1. $~(dm) \vec R \times \vec V_{cm}$
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ This is often called the angular momentum of the center of mass.

2. $~(dm)\vec R\times (\vec{\omega}\times \vec {r}')$
Using the triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because $\vec R$ is perpendicular to $\vec {\omega}$. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, $\int (dm)\vec {r}'=0.$

3. $~(dm)\vec {r}'\times \vec V_{cm}$
This term also vanishes upon integration for the same reason when we take constant $\vec R$ out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. $~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')$
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$This is called the angular momentum about the center of mass. Thus, the total angular momentum about point A of the translating and rotating rigid body is $$\begin{align} & \vec L=\vec L_1+\vec L_4=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\end{align}$$ The first term in this expression, angular momentum of the center of mass, depends on the choice of reference point A. The second term, angular momentum about the center of mass, does not.

The bottom line is that one can always write the angular momentum of a translating and rotating rigid body in the form of Equation (1).

Note that a simplification occurs if point A is chosen anywhere on the straight line path of the CM. Then $\vec {R}$ and $\vec V_{cm}$ are parallel and the angular momentum of the CM vanishes. This is why in the asteroid problem and all problems like it, it is convenient to choose the CM as a point of reference for the angular momentum and immediately write the angular momentum conservation equation in the form $$m~v~d=I_{cm}~\omega$$ where $d$ is the vertical distance between the path of the asteroid and the CM of the combined object. This equation says that the initial orbital angular momentum about the CM of the asteroid before the collision is converted to spin angular momentum of the combined object after the collision. Furthermore, as we have seen, choosing point A away from the path of the CM adds equal amounts of (orbital) angular momentum of the CM on both sides of the momentum conservation equation which cancel out. (See post #178.)

@kuruman I got it. My only doubt is with respect to the conservation of linear momentum. In post #178, you say that:

Note that after the collision, the CM is the only point that moves in a straight line with velocity $V_{cm}$ while all other points rotate about it with angular velocity $\omega.$ This means that the linear momentum of any mass element $dm$ away from the CM is not conserved.

In post #214, @haruspex says:

It can be written as equations in more than one way, but what the author is saying is that the direction of motion of m is initially unchanged, and M's motion can be thought of as the sum of a rotation about its centre and linear motion of its centre (which is always true).
If M's rotation rate is $\omega$ and initial post-collision mass centre velocity is v' then m's velocity is $v'+R\omega$ and we have $Mv'+m(v'+R\omega)=mv$.

As can be seen, in @haruspex's message the linear momentum is not conserved with respect to the CoM. Instead, you say that linear momentum is only ever conserved with respect to the CoM. How is this discrepancy possible. Where am I wrong? What wrong assumptions am I making? Thank you very much.

 

[Hak]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

Moreover, you had said that the same result as $\omega$ would be arrived at without preliminary simplifications. By using $\omega = \frac{L}{I_{cm}}$, with $I_{cm}$ simplified moment of inertia, you arrive at the result $\omega = \frac{5mMv}{(2M + 5m)(M+m)R}$, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

However, I have been trying since yesterday, with the simplified moment of inertia calculated by you (even with subsequent corrections) I always arrive at the same result, the one quoted above. This result does not correspond to the original one, whereas you said we would get the same value of $\omega$ with this moment of inertia. I can't see where the error lies and trying still doesn't make sense. Could you tell me which way is correct for what you are saying to start making sense? Thank you very much. @kuruman

 

[haruspex]

in @haruspex's message the linear momentum is not conserved with respect to the CoM.

I don’t know how you deduce that from what I wrote, and I don't know what you mean by "with respect to the CoM" in the context of linear momentum. Conservation of linear momentum is not in respect of any particular point.

 

[Hak]

I don’t know how you deduce that from what I wrote.

I deduced it from the fact that that equation relates to the conservation of linear momentum with respect to the centre of the major asteroid, not with respect to the centre of mass, as @erobz and I have said many times...

 

[haruspex]

I deduced it from the fact that that equation relates to the conservation of linear momentum with respect to the centre of the major asteroid, not with respect to the centre of mass, as @erobz and I have said many times...

Please see the text I added to post #286.

 

[Hak]

the linear momentum of any mass element $dm$ away from the CM is not conserved.

So what does this assertion mean? How can it be justified by that equation, which considers $v'$ as the velocity of the centre of the major asteroid, and not the centre of mass? Thank you.

 

[haruspex]

So what does this assertion mean? How can it be justified by that equation, which considers $v'$ as the velocity of the centre of the major asteroid, and not the centre of mass? Thank you.

@kuruman refers to the momentum conservation of an individual element mass dm. The equations I posted are for momentum conservation of the whole system.
One mass element may gain momentum in a given direction while another loses it, but the total sum is conserved.

 

[Hak]

@kuruman refers to the momentum conservation of an individual element mass dm. The equations I posted are for momentum conservation of the whole system.
One mass element may gain momentum in a given direction while another loses it, but the total sum is conserved.

OK, everything is now clear. Thank you very much.

 

[kuruman]

This result does not correspond to the original one, whereas you said we would get the same value of ω with this moment of inertia.

Are you saying that when you use the approximate expression $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right)$$ instead of the exact expression $$I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right)$$ in $\omega = \dfrac{mvd}{I_{cm}}$ you get expressions that look different? If so, that is is to be expected because the expressions for $I_{cm}$ are different.

The real question is whether that difference affects $\omega$ significantly. The quickest way to find out is to plug in numbers and see for yourself how close the approximate expression is to the exact expression. Make sure that $m<<M$, say $m=10^{-6}M$.

On edit: Corrected typo. See #295.

 

[Hak]

Are you saying that when you use the approximate expression $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right)$$ instead of the exact expression $$I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right)$$ in $\omega = \dfrac{mv}{I_{cm}}$ you get expressions that look different? If so, that is is to be expected because the expressions for $I_{cm}$ are different.

The real question is whether that difference affects $\omega$ significantly. The quickest way to find out is to plug in numbers and see for yourself how close the approximate expression is to the exact expression. Make sure that $m<<M$, say $m=10^{-6}M$.

@kuruman Why $\omega = \frac {mv}{I_{cm}}$? Shouldn't the expression in the numerator be the angular momentum before the collision, thus equal to $mvd$? I don't understand.
Besides, it was you who said we would get the same expression for $\omega$ with the simplified moment of inertia without approximations, so I didn't understand (and the maths don't add up). What did you mean? I am very uncertain and I don't seem to have understood what you meant (and mean).

 

[Hak]

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.)

This is the post I am referring to. What did you mean?

 

[kuruman]

This is the post I am referring to. What did you mean?

I meant that when I first posted my answer I had $\omega =\dfrac{1}{25}\dfrac{v}{R}$ which I then corrected. It's entirely possible that the correction is wrong but his point is totally irrelevant to the discussion.

@kuruman Why $\omega = \frac {mv}{I_{cm}}$? Shouldn't the expression in the numerator be the angular momentum before the collision, thus equal to $mvd$? I don't understand.

Yes, I meant to say $\omega = \dfrac {mvd}{I_{cm}}.$ It was a typo. Thanks for pointing it out.

 

[Hak]

Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?

Moreover, you had said that the same result as $\omega$ would be arrived at without preliminary simplifications. By using $\omega = \frac{L}{I_{cm}}$, with $I_{cm}$ simplified moment of inertia, you arrive at the result $\omega = \frac{5mMv}{(2M + 5m)(M+m)R}$, which does not correspond to the correct one. Try redoing the calculations, most likely I am wrong, however, I would like to understand it more. Thank you very much.

Thanks. I have noticed, however, that by placing $m \ll M$, one obtains: $\omega = \frac{5mv}{2MR}$, which corresponds to the value found previously for $m \ll M$. So, can we say that the simplified moment of inertia is an alternative way to find the simplified angular velocity?

 

[Hak]

Consider an irregular rigid object of mass $M$ simultaneously rotating and translating in space in the absence of external forces and torques. This means that both its linear and angular momentum are conserved. We will calculate its angular momentum about an arbitrary point A (see diagram below) given that the velocity of the center of mass is $\vec V_{cm}$ and its rotational angular velocity is $\vec \omega.$

View attachment 332757

First we point out that the axis of rotation must pass through the CM. That's because linear momentum is conserved which means that $\vec V_{cm}= \rm{const.}$ If the axis of rotation did not pass through the CM, the CM would rotate about that hypothetical axis and its velocity would change direction which means that it cannot be constant. Thus, the linear velocity relative to the CM of an arbitrary point at position $\vec {r}'$ from the CM is $\vec {v}'=\vec{\omega}\times \vec {r}'.$

Now consider mass element $dm$ at an arbitrary point P the position vector of which relative to A is $\vec r$. Let $\vec {r}'$ be the position vector of $dm$ relative to the CM and $\vec R$ be the position of the CM relative to A. Finally, let $\vec v_P$ be the instantaneous velocity of point P relative to point A. Addition of velocities requires that the velocity of point P relative to A is $\vec v_P=\vec V_{cm}+\vec{\omega}\times \vec {r}'.$

The angular momentum contribution of $dm$ about point A is $$d\vec L=(dm)\vec r\times \vec v_P=(dm)(\vec R+\vec {r}')\times(\vec V_{cm}+\vec{\omega}\times \vec {r}')$$ We multiply out the cross product on the RHS and consider each of the four resulting terms separately.

1. $~(dm) \vec R \times \vec V_{cm}$
The contribution from the first term to the total angular momentum is $$\vec L_1=\int (dm) \vec R \times \vec V_{cm}=\left( \int dm \right) \vec R \times \vec V_{cm}=M\vec R \times \vec V_{cm}.$$ This is often called the angular momentum of the center of mass.

2. $~(dm)\vec R\times (\vec{\omega}\times \vec {r}')$
Using the triple-product rule,
$$\vec R\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec R\cdot \vec {r}') -\vec {r}'(\vec R \cdot \vec{\omega}).$$The second term on the RHS is zero because $\vec R$ is perpendicular to $\vec {\omega}$. To find the contribution of the first term, we integrate $$\vec L_2=\int (dm)\vec{\omega}(\vec R\cdot \vec {r}')=\vec{\omega}\left(\vec R\cdot \int (dm)\vec {r}'\right)=0$$ Note that by definition of the CM coordinates, $\int (dm)\vec {r}'=0.$

3. $~(dm)\vec {r}'\times \vec V_{cm}$
This term also vanishes upon integration for the same reason when we take constant $\vec R$ out of the integral. $$\vec L_3=\int (dm)\vec {r}'\times \vec V_{cm}=\left(\int (dm)\vec {r}'\right)\times \vec V_{cm}=0.$$ 4. $~(dm)\vec {r}'\times (\vec{\omega}\times \vec {r}')$
Triple-product rule once more
$$\vec {r}'\times (\vec{\omega}\times \vec {r}')=\vec{\omega}(\vec {r}'\cdot \vec {r}')-\vec {r}'(\vec{\omega}\cdot \vec {r}').$$The second term on the RHS vanishes because the vectors are orthogonal. Integrating the first term, $$\vec L_4=\int \vec{\omega}(\vec {r}'\cdot \vec {r}')=\vec{\omega}\int (dm){r'}^2=I_{cm}~\vec{\omega}.$$This is called the angular momentum about the center of mass. Thus, the total angular momentum about point A of the translating and rotating rigid body is $$\begin{align} & \vec L=\vec L_1+\vec L_4=M\vec R \times \vec V_{cm}+I_{cm}~\vec{\omega}.\end{align}$$ The first term in this expression, angular momentum of the center of mass, depends on the choice of reference point A. The second term, angular momentum about the center of mass, does not.

The bottom line is that one can always write the angular momentum of a translating and rotating rigid body in the form of Equation (1).

Note that a simplification occurs if point A is chosen anywhere on the straight line path of the CM. Then $\vec {R}$ and $\vec V_{cm}$ are parallel and the angular momentum of the CM vanishes. This is why in the asteroid problem and all problems like it, it is convenient to choose the CM as a point of reference for the angular momentum and immediately write the angular momentum conservation equation in the form $$m~v~d=I_{cm}~\omega$$ where $d$ is the vertical distance between the path of the asteroid and the CM of the combined object. This equation says that the initial orbital angular momentum about the CM of the asteroid before the collision is converted to spin angular momentum of the combined object after the collision. Furthermore, as we have seen, choosing point A away from the path of the CM adds equal amounts of (orbital) angular momentum of the CM on both sides of the momentum conservation equation which cancel out. (See post #178.)

@kuruman Also, I would like to ask you how you drew the diagram above, and how you obtained the depiction of the irregular rigid body. Thank you very much, if you feel like telling me.

 

[kuruman]

@kuruman Also, I would like to ask you how you drew the diagram above, and how you obtained the depiction of the irregular rigid body. Thank you very much, if you feel like telling me.

Like @erobz I use Powerpoint to make figures. There is a "Freeform" feature to draw closed 2d shapes freehand. In other versions of Powerpoint the icon shown below might look different.

 

[erobz]

Like @erobz I use Powerpoint to make figures. There is a "Freeform" feature to draw closed 2d shapes freehand. In other versions of Powerpoint the icon shown below might look different.

View attachment 332784

You can also use the curve function ( which I believe is just to the left ). It will make smooth/continuous curve, and closes automatically in the vicinity of the start point.

 

[kuruman]

You can also use the curve function ( which I believe is just to the left ). It will make smooth/continuous curve, and closes automatically in the vicinity of the start point.

Actually both functions on either side of the one with the arrow draw closed surfaces.