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Gravitational potential energy of two bars

2 votes

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


asked Mar 30, 2017 in Physics Problems by member192 (260 points)
edited Mar 30, 2017 by Einstein
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 $F_{r}=-m_{r}g=-\frac{\partial U_{r}}{\partial y}$.

1 Answer

2 votes
Best answer

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 physicsapproval (2,290 points)
selected Mar 31, 2017 by member192
Just to be sure, you are saying that for the left rod, the center the mass is at $L_{r}/2\cos(\theta+\pi)=-L_{r}/2\cos\theta$?