I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts. It was just to get a complete overview of the possibilities of this problem.erobz said:My advice, let your professor explain it to them... or you can show them the thread in which it was explained to you?
Why are they authoring problems they have doubts about?Hak said:I can't understand what you mean. However, even the author of the problem (he is not really my professor) seems to have doubts.
Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?Hak said:It was just to get a complete overview of the possibilities of this problem.
You are right, but they don't feel like it. They prefer me to report their interpretations. Sorry if I am annoying, that is not my intention. I just try to help others and always learn more from those who know more than me.erobz said:Why bring up all your friends interpretations in every problem then? You can just tell them about PF forums, and they can see for themselves?
I have no idea. The author seems to me to be a very good professor, but he does not know all the possible solutions of his own problems. He works them out over time, making various observations and attempts.erobz said:Why are they authoring problems they have doubts about?
$$\begin{align} & I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2 \nonumber \\Hak said:Thank you, that's very comforting to me. Could I know what method you used to arrive at this result? How did you get there through the expansions?
I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##erobz said:##\dots## but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass.
Ah, this is a big problem. I agreed with @erobz, but at this point, I can't really understand why my friend is wrong, mistaking ##d## for ##R##. Why? Sorry to persist with this kind of soap opera....kuruman said:I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##
I don't mean to misquote you...I could have sworn you said that talking about angular momentum not being conserved for a differential mass ##dm## away from the center of mass axis. Maybe I'm imaging it.kuruman said:I never stated or implied that. In post #178 I conserved angular momentum about the center of the planet, not about the CM, and showed that you end up with the same equation for ##\omega.##
Ok, thank you, but...kuruman said:$$\begin{align} & I_{cm}=\frac{2}{5}MR^2+\frac{Mm}{M+m}R^2 \nonumber \\
& I_{cm}=\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M+m}\right) \nonumber \\
\end{align}$$When ##m<<M~## you can approximate in the denominator ##M+m\approx M##. Hence $$I_{cm}\approx\frac{2}{5}MR^2\left(1+\frac{5}{2}\frac{m}{M}\right).$$
What about it, @kuruman?Hak said: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) 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.
But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ## This is the crux of the issue with the @Hak friends solution to the problem.kuruman said: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.##
This is exactly what I wanted to say... It's really a dilemma...erobz said:But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ##
The algebra for finding ##\omega## in post #206 is quite correct and involves no approximation. But as you note, for finding the force, we must decide what radius to use.erobz said:But where then is that to be applied? If that ##\omega ## was to be applied about ##O## then ##F = m \omega^2 R ## Becuase this is the crux of the issue with the @Hak friends solution to the problem.
Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.haruspex said:The algebra for finding ##\omega## in post #206 is quite correct and involves no approximation. But as you note, for finding the force, we must decide what radius to use.
Note that whatever force is exerted on m, an equal and opposite force is exerted on M. I think that makes it clear that the correct radius to use is the distance to the common mass centre.
I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.Hak said:Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.
Yes, it is correct to me. However, I don't understand what this has to do with the Newton's Third Law argument...erobz said:I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.
I view this as a single rigid body rotating in space with angular velocity ##\vec {\omega}.## It follows that mass element ##dm## at any point P on the body at position vector ##\vec {r}'_p## relative to the CM will have velocity (also relative to the CM) ##\vec v'_p=\vec {\omega}\times \vec {r}'_p##. Now if the CM is moving relative to a point in an inertial frame with velocity ##\vec V_{cm}##, the velocity of P relative to the inertial frame will be $$\vec v_p=\vec V_{cm}+\vec v'_p=\vec V_{cm}+\vec {\omega}\times \vec {r}'_p.$$The centripetal force on mass ##m## is ##\vec F_c=-(dm) \omega^2 \vec{r}'_p.##erobz said:I don't know the details of Newtons Third Law argument, but for me it is clear that masses are orbiting their common center of mass, just like we expect of any other composite body freely rotating.
This would imply that since the minor asteroid is a homogeneous body, it has mass ##dm = m##? And also that ##r'_p = d##?kuruman said:I view this as a single rigid body rotating in space with angular velocity ##\vec {\omega}.## It follows that mass element ##dm## at any point P on the body at position vector ##\vec {r}'_p## relative to the CM will have velocity (also relative to the CM) ##\vec v'_p=\vec {\omega}\times \vec {r}'_p##. Now if the CM is moving relative to a point in an inertial frame with velocity ##\vec V_{cm}##, the velocity of P relative to the inertial frame will be $$\vec v_p=\vec V_{cm}+\vec v'_p=\vec V_{cm}+\vec {\omega}\times \vec {r}'_p.$$The centripetal force on mass ##m## is ##\vec F_c=-(dm) \omega^2 \vec{r}'_p.##
No, ##dm## is a very small mass at position ##\vec {r}'## relative to the center of the planet. Imagine chopping up the composite body into many little pieces. ##dm## is the mass of the little piece that you will find at position ##\vec {r}'##; it could be small piece of the asteroid or a small piece of the planet. The only thing you can say is that $$\int dm=M+m.$$Hak said:This would imply that since the minor asteroid is a homogeneous body, it has mass ##dm = m##? And also that ##r'_p = d##?
OK, so yours is just a general overview of the physical situation. The result of ##F## you wrote does not lead to the form we found earlier, right?kuruman said:No, ##dm## is a very small mass at position ##\vec {r}'## relative to the center of the planet. Imagine chopping up the composite body into many little pieces. ##dm## is the mass of the little piece that you will find at position ##\vec {r}'##; it could be small piece of the asteroid or a small piece of the planet. The only thing you can say is that $$\int dm=M+m.$$
At this point, with 264 posts and counting, I have no clue what form you are referring to. Please post it here so that I can see what you mean.Hak said:OK, so yours is just a general overview of the physical situation. The result of ##F## you wrote does not lead to the form we found earlier, right?
##F = m \omega^2 d##kuruman said:At this point, with 264 posts and counting, I have no clue what form you are referring to. Please post it here so that I can see what you mean.
There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.Hak said:Thank you, but what is the connection between Newton's Third Law and the choice of the most appropriate radius to use? Thank you if you would like to answer.
OK, thank you very much. I cannot understand, however, how it can be deduced from this symmetric equation that ##d## is the radius to be chosen for the rotation. Sorry if I don't fully understand...haruspex said:There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.
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}'.##haruspex said:There's a symmetry here. Each body is pulling the other towards the common mass centre. If that is displaced d from M's centre then ##m(R-d)=Md##, hence##m(R-d)\omega^2=Md\omega^2##.
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}.$$Hak said:##F = m \omega^2 d##
Is it correct?kuruman said:Yes, I see now what you are asking. The general expression for the force at any point is ##\vec {F}_c=-(dm)\omega^2 \vec {r}'## where ##\vec {r}'## is the location of the point relative to the CM. The asteroid is at location ##\vec {r}'=\vec d## and has mass ##dm = m##. So the force on the asteroid is $$\vec {F}_c=-m\omega^2 \vec {d}.$$
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 said: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....
Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...kuruman said: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, I was just providing confirmation from other evidence, since @Hak and @erobz seemed unsure.kuruman said: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}'.##
Please remind us, what issue is this?Hak said:Yes, you are completely right. Forget my friend's reasoning. One issue, however, remained unresolved: that of the simplified moment of inertia...
The second paragraph. Thanks.Hak said: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.
This is the post I am referring to.kuruman said: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.)
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 said:The second paragraph. Thanks.
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}##kuruman said:Your expression in the second paragraph is dimensionally incorrect. The angular velocity has dimensions of [v/R]. Your expression has dimensions of [Mv/r].
but I believe it was also stated by @kuruman that angular momentum is only conserved about the systems center of mass.
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 said: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.##