Recent content by anuttarasammyak

After transformation of ##X_2=3-x_2## the two partial differential equations are $$-2(4-x_1^2 -X_2^2)x_1+(x_1+X_2-3)=0$$ $$-2(4-x_1^2 -X_2^2)X_2+(x_1+X_2-3)=0$$ This pair of eqution is obviously symmetric between ##x_1## and ##X_2##. This may help you to prove the features you find that have...

I am sorry for my careless mistakes on L. I corrected #4. It seems in good accord with your result.

continued from #2: With the help of Wolfram I get one real solution. https://www.wolframalpha.com/input?i=Solve+(4-x^2-(y-3)^2)(-2x)+(x-y)=0+and++(4-x^2-(y-3)^2)2(y-3))+(x-y)=0.+ where ##x=x_1,y=x_2## Let us confirm it. It is easily observed that the two curve formula coincide when...

Square of the distance is $$L^2=(y_1-y_2)^2+(x_1-x_2)^2=(4-x_1^2-(x_2-3)^2)^2+(x_1-x_2)^2$$ The condition of minimum is $$\frac{\partial L^2}{\partial x_1}=\frac{\partial L^2}{\partial x_2}=0$$ Intersection of these two curves would give an answer.

In the c limit two OP disks and this disk both have mass 2M. KE of the former is finite 2M, that of the latter diverges. If we don’t provide material to the disk of mass M, it would become a ring KE of which diverges. When density of the disk is homgeneous in IFR where COM of the rotating disk...

From OP Say the material of the disk is incompressible fluid the density of which is $$ \rho = \frac{M}{\pi R^2 d}$$ where d is thickness of the disk. As ##\omega r## approaches c, space metric of the rotating system provides another ##\gamma## factor. Say we pour the material to fill the...

In your formula K=M for ##\omega R=1##. I think that even the rim part speed reaches light speed, its element contribution is infinitesimal or at least finite in the volume integral. Supplemental stress energy would take place but I do not think it matters essentially in this discussion.

@gleem When we regard O as a point in the massless rigid body composed of the arms y, z, and d, only internal forces act on O. Their total is zero and they produce no momentum of force. This applies to all the other parts of the rigid body except the three points, i.e., p,q and the z arm end...

Graphene is not a single-atom lattice. It has two sublattices (A and B). At a Dirac point (K point), the wavefunction is not a simple scalar — it has two components. That makes difference. Please check this layman’s conjecture by yourself.

In theory melting point of heavier mono isotope is higher. $$T_{28}<T_{29}< T_{30} $$ For an example D2O has melting point 3.82 degree Celsius. We need an equation of melting point that includes mass of molecule to investigate the difference in quantity.

$$\ln 10 e^{y\ln 10 }dy = \frac{dx}{x}$$ $$e^{y\ln 10 } = \ln |x |+ C$$ $$y= \frac{\ln(\ln |x| + C)}{\ln 10}$$defined for $$|x| >e^{-C}$$ x,C < 0 cases are added to your result.

The small angle between the wall and the diagonal rod, blue line, in the figure of my #90 you quoted. We can make it infinitesimal.

I know that, without support, L-shaped components experience significant stress concentration at the inner corner due to bending. This corner acts as a structural discontinuity, increasing the stress by up to about 1.6 times compared with straight sections. Therefore, having no support is not a...

@kuruman thank you for a good resume of your thought. I observe different results between us on Fy, vertical component of the force from the wall to the rod. @kuruman ##F_y < 0##． ##\theta## dependent. $$F_y=W(1-\frac{L_2}{L_1})=W(1-\frac{L_2}{h}\cot\theta)$$ where h is distance of the two...

Thank you @Steve4Physics . By your advice I will restate #68 refrain from using the word subsystem. Red forces act on HR. They produce minus torque. Blue forces act on DR. They produce plus torque. These torques cancel each other. I hope it is correct and we can alalyse the system by...