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kev0
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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]