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The gravitational field of the asteroid is the same as a point mass only if it has a spherically symmetrical distribution of mass. If the asteroid has 3 different moments of inertia then it cannot be spherically symmetric. Therefore when it flips the gravitational forces on its satellite (or moon) will change. What effect this will have on the satellite ...


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Not a book recommendation, but a quick overview of the process in hopes to fill some holes in your understanding, or to spark the correct questions to ask. 1. How do I calculate the kinetic energy of a connected system of rigid bodies? Like you mention in your post, you need to know the the translational velocity of each body center of mass $\boldsymbol{v}...


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Q1: Without table top friction, the top won’t invert. This proves that frictional forces cause the inverting effect. And when the spindle hits the tabletop, the Tippe Top jumps up to inversion - presumably due to the increased distance between the spindle top and the axis of spin. This would increase any torque effects. I guess it’s similar to gyroscopic ...


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Unsurprisingly, there have been no answers so far, as it is a rather niche area. In the meantime I think I may have found the answer myself. It occurs to me the following. Let us define angular momentum in the following way (using Einstein index notation): $L_{i}=I_{ir}\omega^{r}$ Thus here I have written angular momentum is a contravariant tensor. We can ...


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Let us first recall how the Euler equations simplifies sufficiently close to a principal axis, cf. e.g. this Phys.SE post: The major axis $\vec{\Omega}\approx (\Omega_1,0,0)$ is stable because $$\ddot{\Omega}_i~\approx~-\omega^2_1 \Omega_i, \qquad i\in{2,3}, \tag{1a}$$ where $$\omega_1~:=~\Omega_1\sqrt{\frac{(I_1-I_2)(I_1-I_3)}{I_2I_3}}. \tag{1b}$$ The ...


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Interesting. However narrator interpretation is totally wrong. Car moves because of reaction force which tries to compensate net force produced by gyroscope + arm system movement. Car movement stops when gyroscope+arm movement kinetic energy is exhausted. To make things more simple, you can take into account a similar system - a ball wobbling on swings, ...


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Actually the problem with your equation for energy is that the term $\frac12 m V_{CM}^2$ is already included in the term $\frac12 I_1\omega^2$ for the energy of rotation about the point of contact with the large cylinder. To understand why this is true analyse the circular motion of a point mass about a fixed point with constant speed. For its kinetic ...


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This question might be closed because of the multiple questions inside, so I am expanding on my comment: There are multiple purposes for teaching physics at school. Most of them are for students who end up in engineering or other disciplines using physics in a background way. The courses are designed by physicists for future physists who may be doing ...


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