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The following system of two particles connected by a massless rod rotates about an axis perpendicular to the plane of the system.enter image description here

The top mass $m_{r}$ is at a distance $L_{r}$ from the rotation point and the bottom mass $m_{l}$ is at a distance $L_{l}$ from the rotation point. $L_{r}+L_{l}=L$ where $L$ is the length of the rod.

If I apply Newton's second law along the $\theta$ direction to each mass I get $$-m_{r}g\cos\theta=m_{r}L_{r}\frac{d^{2}\theta}{dt^{2}}\tag{1}$$ $$m_{l}g\cos\theta=m_{l}L_{l}\frac{d^{2}\theta}{dt^{2}}\tag{2}$$

Usually though Newton's law for rotational motion, $\tau_{external}=Id^{2}\theta/dt^{2}$ where $\tau_{external}$ is the torque due to the gravitational forces and $I$ is the momentum of inertia of the two particles, is applied which leads to $$g\cos\theta[m_{l}L_{l}-m_{r}L_{r}]=(m_{l}L_{l}^{2}+m_{r}L_{r}^{2})\frac{d^{2}\theta}{dt^{2}}\tag{3}$$

I can see how to get (3) by adding (1) and (2) but I can not see how to get (1) and (2) from (3). Are equations (1) and (2) correct?

For example they seem to imply that

$$g\cos\theta=L_{l}\frac{d^{2}\theta}{dt^{2}}=L_{r}\frac{d^{2}\theta}{dt^{2}}$$ i.e. that $L_{l}=L_{r}$?

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    $\begingroup$ Careful, if you consider each mass separately, there will be a normal force of the rod acting on the mass. Therefore, your equation (1) and (2) are incomplete. $\endgroup$ – Frotaur Dec 25 '16 at 17:14
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The rod will exert a tangential force on the particles. Your equations (1) and (2) then become

$$N_{r}-m_{r}g\cos\theta=m_{r}L_{r}\frac{d^{2}\theta}{dt^{2}}$$

and

$$-N_{l}+m_{l}g\cos\theta=m_{l}L_{l}\frac{d^{2}\theta}{dt^{2}}$$

If you now add these two equations and use the fact that $N_{r}L_{r}-N_{l}L_{l}=0$ you get equation (3).

How do we know that $N_{r}L_{r}-N_{l}L_{l}=0$? One argument is because (3) is the equation you get if you use a Lagrangian formalism and (3) implies that $N_{r}L_{r}-N_{l}L_{l}=0$.

The Lagrangian is

$$L=\frac{1}{2}m_{r}L_{r}^{2}\left(\frac{d\theta}{dt}\right)^{2}+\frac{1}{2}m_{l}L_{l}^{2}\left(\frac{d\theta}{dt}\right)^{2}-m_{r}gL_{r}\sin\theta+m_{l}gL_{l}\sin\theta$$

and from Lagrange equation $\partial L/\partial \theta - d/dt\partial L/\partial \dot{\theta}$ you get exactly (3).

A more "physical" explanation to why $N_{l}L_{l}=N_{r}L_{r}$ is that because the rod is massless, the torque due to the forces $N_{l},N_{r}$ acting on the rod must be zero.

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  • $\begingroup$ How do you know that $N_{r}-N_{l}=0$? $\endgroup$ – JennyToy Dec 25 '16 at 21:14
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    $\begingroup$ I edited my answer to add one argument that uses the Lagrangian for this system. $\endgroup$ – user2175783 Dec 25 '16 at 21:42
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As Frotaur states, you cannot consider the two masses separately. They are rigidly connected through the rod. Because of this they cannot move independently. They must have the same angular acceleration.

If you try to move one of the masses through a different angle than the other, the rod will bend and a bending moment will resist what you are trying to do. So your equations for each mass should include this bending moment.

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