Recent content by Phizyk

\omega is the angular velocity of the disk, v is the translational velocity. v and \omega are constants. Integration extends over the area of the disk with \vec{r} vectors starting at the center.

Hi, How do integrate this? I wish to see it step by step and I'm glad for any help i can get. \int_{ \vec{r}\in{A}} \frac{ \vec{v}+ \vec{\omega}\times\vec{r}}{| \vec{v}+ \vec{\omega}\times\vec{r}|}d^{2}r where A is area of disk with radius R.

[\frac{b}{x}] the floor and ceiling functions. x-1\leq{[x]}\leq{x}

[\frac{b}{x}] it is entier function. I can not solve second case... It is harder than first. Can I do (\frac{x}{a}-1)\frac{b}{x}\leq{[\frac{x}{a}]\frac{b}{x}}\leq{\frac{b}{a}} and use |f(x)-g|\leq{\epsilon} so g=\frac{b}{a}?

Hi, I am in need of very hard physics and maths problems from olympiads best in English/German/Russian. Thanks

Hi, How can I calculate left and right-sided limits? \frac{x}{a}[\frac{b}{x}] \frac{b}{x}[\frac{x}{a}] \frac{x}{\sqrt{|sinx|}} in point x=0. Thanks for help.

This is a brilliant solution. Thanks for all...

So, we can not integrate this... But if A=B, we have \int{\frac{cos^{2}\frac{x}{2}}{|cos\frac{x}{2}|}}dx and it can be simple integrate...

I used to function calculator and I received: \int{\frac{A+Bcosx}{\sqrt{A^{2}+B^{2}+2ABcosx}}}=\frac{(B^{4}-4AB^{3}+4A^{2}B^{2})x^{5}}{5(24B^{4}+96AB^{3}+144A^{2}B^{2}+96A^{3}B+24A^{4})}-\frac{B^{2}x^{3}}{3(2B^{2}+4AB+2A^{2})}+x+O(x^{7})+C I think what I can neglect O(x^{7})... Is it correct?

Hi, How can I integrate this \int_{0}^{2\pi}\frac{A+Bcosx}{\sqrt{A^{2}+B^{2}+2ABcosx}}dx Thanks for help.

But this equation \frac{dx}{dt}=\sqrt{\frac{2k}{m}cosx+C^{'}} can I solve? Can I obtain x(t)? For t=0 x=0. It's a equation of motion.

Great. Thanks Hallsoflvy.

Hi! I have big problem with solve this equation: m\frac{d^{2}x}{dt^{2}}+ksinx=0 I can't go ahead, because I don't know how solve this \frac{dx}{\sqrt{cosx}}=\sqrt{\frac{2k}{m}}dt Phizyk

I apologize to bad equation C=C_{1}+C_{2} It's wrong... You have a good solution, because V=\frac{q}{C} and V_{1}=\frac{q}{C_{1}} V_{2}=\frac{q}{C_{2}} and V=V_{1}+V_{2} so \frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}

\frac{dR}{d\theta}=cos\theta{cos(\phi+\theta)}-sin\theta{sin(\phi+\theta)}=0 but cos(\phi+\theta)=cos\theta{cos\phi}-sin\theta{sin\phi} sin(\phi+\theta)=sin\phi{cos\theta}+cos\phi{sin\theta} to equation \frac{dR}{d\theta}=cos\phi(cos^{2}\theta-sin^{2}\theta)-2sin\theta{cos\theta}sin\phi=0...