Recent content by anuttarasammyak

I am sorry for my poor English. Tension force is zero. The wire outside the disk is straight and becomes short with pace ##r\Omega## during winding up operation around the disk bobbin. Wire tension T is the only force acting on mass m so we need to know how it takes place. Does your...

I try to make the Lagrangian. Say coordinate of m is (X,Y) $$X=r \sec \phi \ \cos (\theta - \phi) $$ $$Y=r \sec \phi \ \ sin (\theta - \phi) $$ where $$\theta = \Omega t + \frac{x}{r} + \theta_0-\frac{x_0}{r} ; \theta_0=0$$ $$\phi = \tan^{-1}\frac{x}{r}$$ Lagrangian is made of kinetic energy...

Thanks.　 For x/r ##\rightarrow \infty##, the purple force decreases so ##\dot{x}## would also decrease. For x=0 wire tension cannot generate centripetal force. The singularity would come from it. For x=+0 wire tension is +##\infty## thus the purple force is also +##\infty## which would cause...

I see. Tension from the wire (blue) generates centripetal force (green) for circulation of m. The rest component (purple) causes unwrapping the wire. Thus x increases with time. Am I correct ?

Does the wire wrapping direction wrt ##\Omega## matter in your result ? It seems that one tighten and the other loosen the wire.

https://www.feynmanlectures.caltech.edu/II_29.html is a good one to know the fundamentals.

Applied magnetic field should be perpendicular to the velocity of the particle so that its track is a circle not a spiral.

Some alpha particles could be on circle track in constant kinetic energy and come back to the heavy source with a designed homogeneous magnetic field.

You may find some classic books of your taste at "Books in Astronomy" in Project Gutenberg, free, though I am not good at astronomy.

I would like to expalin it in the case of ellipse orbit of planets as shown in the figure below. You see that F has components that is parallel and opposite to v. Therefore v is reduced. The work done in small time dt by pulling force to the planet is $$ dW=\mathbf{F} \cdot \mathbf{v} dt <...

At least energy, momentum, angular momentum of the system is conserved. I hesitate to refer spin or helicity or other pure quantum states due to my poor understanding. We can change volume of photon gas. On the other hand, I hesitate to say that electron gas is incompressible.

An electron and a positron meet and become two photons. The system quantum state is conserved and photons are compressible. The reverse process is possible.

The angular momentum before and after the change in the moment of inertia is conserved: $$M=I_1\omega_1=I_2\omega_2$$ Assuming ##I_1 > I_2## , it follows that ##\omega_1<\omega_2##. As for the rotational energy, $$E_1=\frac{M^2}{2I_1} < \frac{M^2}{2I_2} =E_2$$ Therefore, the skaters must exert...

As for the special case ##x=\frac{\pi}{2}## for n=0 (mod 4) $$a_n=-\frac{1}{n^b}\sin c_n$$ for n=1 (mod 4) $$a_n=\frac{1}{n^b}\cos c_n$$ for n=2 (mod 4) $$a_n=\frac{1}{n^b}\sin c_n$$ for n=3 (mod 4) $$a_n=-\frac{1}{n^b}\cos c_n$$ where $$c_n=\frac{1}{5+\frac{1}{n}}\frac{\pi}{2} =...

Thanks you all for good teachings. I add would add a rough hand drawing explaining the event.