- #1

- 27

- 2

## Homework Statement

I uploaded the homework question. This is #1.

## Homework Equations

None directly given

## The Attempt at a Solution

My main difficulty with the problem is that I am convinced it is much easier than my classmates make it out to be. This is graduate mechanics so I'm pretty sure that the solution can not be this easy, but I don't know where I am wrong. All I am after is a discussion of the starting point so I can fix my intuition and hopefully get better. As a note, one classmate somehow calculated an inertia tensor in his solution which seems unnecessary. He argued symmetry made the solution calculable by examining only one rod with the CoM in the middle of the rod so there would be rotation and translation of the CoM.

The two rods connected to the hinge are perfectly symmetric. Thus the CoM of the system will be on the y-axis at ##y=\frac{l}{4}## (each rod's CoM is half way down its length so the two combined would be on the y-axis). The only force acting on the system is gravity which is acting on the CoM of the system. Thus the problem is essentially a particle (the hinge) being pulled straight down by gravity a distance ##\frac{l}{2}## where ##U=mgy, U(min) = mg\frac{-l}{4} ##and ##U(max)=mg\frac{l}{4}##. ##T = \frac{1}{2}m\dot y^2##. And the rods are relegated to being part of initial conditions to solve the equation of motion. So the Lagrangian is $$\frac{1}{2}m\dot y^2 - mgy$$, the Lagrange equation is trivially $$\ddot y = -g$$ and the solution goes from there to get $$t=\sqrt\frac{3l}{2g}$$ and $$\dot y = -g\sqrt\frac{l}{g} = -\sqrt\frac{3lg}{2}$$

#### Attachments

Last edited: