Recent content by Bling Fizikst

It should be 'approximately' true i guess as ##r## is given to be small . Hence , the second order term ##r^2\approx 0## . So , i guess we can only get an approximate answer .

##r## is the radius of orifice . Read the question carefully .

##dV=2\pi x y dx=\frac{\pi\omega^2}{g}x^3\cdot dx## . So , my volume would be $$V=\int_{r}^{R}\frac{\pi\omega^2}{g}x^3\cdot dx=\frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$ So , final answer should be : $$V_{\text{out}}=\pi R^2 h - \frac{\pi\omega^2}{4g}\left(R^4-r^4\right)$$

high school level? yes

I am not sure how i will integrate to find the volume . Tell me how to proceed .

Let's say at the steady state the vertex of the parabola (paraboloid) is at the origin . Then the eqn of the formed parabola would be $$y=\frac{\omega^2x^2}{2g}$$ Now , initial volume of liquid is ##\pi R^2h## . As the liquid flows out of the orifice , the surface would maintain it's structure...

Actually , i already saw the solution using frame of rotating disk , so , i wanted to try it out with lagrangian (if it makes stuff more straightforward) . Also , about the potential energies , can we find them ? for instance if we consider the x axis as the reference for gravitational potential...

Source : JEE Advanced , Physics Sir JEE YT I tried to attempt it using Lagrangian , so according to the coordinate axes given in the diagram , the position of the particle is let's say ##(0,d,-z)## Let ##r## be the distance between the particle and the axis of rotation such that it subtends...

This was a new learning experience for me , so i don't really mind . But i do agree that one should try to solve problems on their own . At the end , it just matters that i learnt to solve it even if i had to look at the solution. I just learnt a new technique and i am happy! Also , i would...

Anyways , i found a really elegant solution in David Morin :

Yes

$$(m+M)\ddot{r} -mr\omega^2- 2mr\omega^2=0$$ $$\implies \omega'=\sqrt{\frac{3m\omega^2}{m+M}}$$

it goes up so +Mgr? please forgive me if i am acting dumb , this is my first time using the potential energy function in this way . It gives me the wrong ans anyways , so it is wrong

Is it -Mgr:nb)

Need help on this , really deadstuck . Writing forces and using ##\frac{-\partial U}{\partial r}=F## doesn't seem to help?