# General Atwood system with moment of inertia

## 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

$F=m\ddot{x}$

$τ=Iα$

$τ=FR$

## 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 $\ddot{x}$

$m_{1}\ddot{x_{1}}=T_{1}-m_{1}g$
$m_{2}\ddot{x_{2}}=T_{2}-m_{2}g$

$τ=Iα$
$τ=(T_{1}-T_{2})a$

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.

$\ddot{x}=αa$

However we can see that

$\ddot{x_{1}}=-\ddot{x_{2}}$

Therefore,

$m_{1}\ddot{x_{1}}=T_{1}-m_{1}g$
$m_{2}\ddot{x_{1}}=m_{2}g-T_{2}$

and,

$α=(T_{1}-T_{2})a/I$
$\ddot{x}=(T_{1}-T_{2})a^{2}/I$

$m_{1}\ddot{x_{1}}+m_{2}\ddot{x_{1}}=T_{1}-T_{2}+m_{2}g-m_{1}g$

Replacing the Tensions

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

Finally rearranging it

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

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

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

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

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Spinnor
Gold Member
The minus sign difference in the numerator is not a problem but the minus sign in the detonator of your solution is a problem, the inertia of the pulley should have a plus sign, its inertia tends to reduce the acceleration of the system.