- #1

- 5

- 0

## Homework Statement

I have taken this image from http://farside.ph.utexas.edu/teaching/336k/lectures/node79.html

Hello, I am currently doing a project on Lagrangian mechanics. As part of my project, it was advised that I should redo each of the problems done using the Lagrangian method in the Newtonian format.

So essentially I am requesting help on

**determining the acceleration of the system**

## Homework Equations

[itex]F=m\ddot{x}[/itex]

[itex]τ=Iα[/itex]

[itex]τ=FR[/itex]

## The Attempt at a Solution

I am getting increasingly confused especially when the signs don't match :(

(a = radius of the pulley, not acceleration)

(acceleration is denoted as [itex]\ddot{x}[/itex]

[itex]m_{1}\ddot{x_{1}}=T_{1}-m_{1}g[/itex]

[itex]m_{2}\ddot{x_{2}}=T_{2}-m_{2}g[/itex]

[itex]τ=Iα[/itex]

[itex]τ=(T_{1}-T_{2})a[/itex]

Tangential acceleration of the pulley is the same as the acceleration of the mass, assuming that there is absolute grip of the rope to the pulley.

[itex]\ddot{x}=αa[/itex]

However we can see that

[itex]\ddot{x_{1}}=-\ddot{x_{2}}[/itex]

Therefore,

[itex]m_{1}\ddot{x_{1}}=T_{1}-m_{1}g[/itex]

[itex]m_{2}\ddot{x_{1}}=m_{2}g-T_{2}[/itex]

and,

[itex]α=(T_{1}-T_{2})a/I[/itex]

[itex]\ddot{x}=(T_{1}-T_{2})a^{2}/I[/itex]

Adding the first 2 equations

[itex]m_{1}\ddot{x_{1}}+m_{2}\ddot{x_{1}}=T_{1}-T_{2}+m_{2}g-m_{1}g[/itex]

Replacing the Tensions

[itex]m_{1}\ddot{x_{1}}+m_{2}\ddot{x_{1}}=\ddot{x}I/a^{2}+m_{2}g-m_{1}g[/itex]

Finally rearranging it

[itex](m_{1}+m_{2}-I/a^{2})\ddot{x_{1}}=(m_{2}-m_{1})g[/itex]

[itex]\ddot{x_{1}}=\frac{(m_{2}-m_{1})g}{(m_{1}+m_{2}-I/a^{2})}[/itex]

The signs are everywhere and completely different to the derivation obtained through the Lagrangian method. :I Can anyone explain what my issue is?

EDIT: Included the solution obtained via Lagrangian method

[itex]\ddot{x_{1}}=\frac{(m_{1}-m{2})g}{m_{1}+m_{2}+I/a^{2})}[/itex]