I used the center of horizontal momentum frame. I just used the y-moment to find the angle of the velocity component, so I can write the velocity in the horizontal direction. I'll edit the equations to make it clearer
This problem I already solved using another resource (just get the coordinate of the center of mass reach and from it, get to the larger mass. R = (3v02) / (4g)). But I'm having some trouble calculating using moment conservation. Here what I've done so far:
$$ 3\vec v_0 = \vec v_1 +2\vec v_2 $$...
"In (1) and (3), what is the initial height of the mass center of each side?"
I was trying to calculate from the end of the string, but really from the center of mass it seems a lot simpler.
"In (2), you have a factor x (l-x) in the GPE term, but it is (l-x)^2 in (4). Do you see the asymmetry...
I solved this problem easily using Newton's second law, but I had problems trying to use mechanical energy conservation to solve it.
How I solved using Newton's second law:
##\text{(part of the rope that is on the left)}\, m_1=x\rho g,\, \text{(part of the rope that is on the right)}\...
I am trying to prove the following expression below:
$$ \int _{0}^{1}p_{l}(x)dx=\frac{p_{l-1}(0)}{l+1} \quad \text{for }l \geq 1 $$
The first thing I did was use the following relation:
$$lp_l(x)+p'_{l-1}-xp_l(x)=0$$
Substituting in integral I get:
$$\frac{1}{l}\left[ \int_0^1 xp'_l(x)dx...
Sorry, I forgot to organize and correct something here. 1 / | r-r '| is a scalar, in my papers I was disregarding the module, forget that part of the divergence theorem.
The integral that I have to solve is as follows:
\oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}, integrating with respect to r'
Then I apply the divergence theorem, resulting in:
\iiint \limits _{v} \nabla \cdot \frac{1}{|r-r'|}dv' =...
Doing R=|r-r'|, i get the expected result: \nabla \frac{1}{|r-r'|} = -\frac{1}{R^2}\hat r=-\frac{(r-r')}{|r-r'|^3}
But doing it this way seems extremely wrong, as I seem to be disregarding the module. So I tried to do it by the chain rule, and I got:
\nabla...