# Gravitational potential energy of two bars

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Consider two cylindrical and homogeneous rods that touch at a common pivot (see figure). One of the rods has length $L_{r}$ and mass $m_{r}$, the other has length $L_{l}$ and mass $m_{l}$.
What is their gravitational potential energy?

Is it

$$U=\frac{L_{r}}{2}m_{r}g\cos\theta-\frac{L_{l}}{2}m_{l}g\cos\theta$$ or

$$U=\frac{L_{l}}{2}m_{l}g\cos\theta-\frac{L_{r}}{2}m_{r}g\cos\theta$$ or

$$U=\frac{L_{r}}{2}m_{r}g\cos\theta+\frac{L_{l}}{2}m_{l}g\cos\theta$$?

edited Mar 30, 2017
what do you think ? Atleast show some effort
The 3 answers are my effort. The question just asks what is the potential energy of the two rods.  Since the gravitational force on each rod in each case is $-mg$ I would assume it is the second option because $-mg$ should be the derivative of the potential energy with respect to $y$. But I somehow need to write $U$ in terms of $y$.
Please can you explain your difficulty. What do you mean by "I somehow need to write $U$ in terms of $y$?
$F_{l}=-m_{l}g=-\frac{\partial U_{l}}{\partial y}$ but how do I write $U_{l}(y)$? Similarly $F_{r}=-m_{r}g=-\frac{\partial U_{r}}{\partial y}$.

Okay, You are doing right so far,
Just take the components of length, take x axis as the refrence and w.r.t it take out potential energy of system.

so, that would be $m_r(L_r/2) cos\theta$ for the rod above ,

From vertically opposite angles,

and $- m_l(L_l/2)cos\theta$

total energy hence = $P_r +P_l$

$=m_r(L_r/2) cos\theta - m_l(L_l/2)cos\theta$

answered Mar 31, 2017 by (2,320 points)
selected Mar 31, 2017
Just to be sure, you are saying that for the left rod, the center the mass is at $L_{r}/2\cos(\theta+\pi)=-L_{r}/2\cos\theta$?