Finding acceleration of an Atwood's machine

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santoki
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Homework Statement


Two masses are suspended from a pulley:

[PLAIN]http://i.imgur.com/B4x8pzx.png[/PLAIN]

mass of pulley = 0.20kg
radius = 0.15m
constant torque = 0.35Nm due to friction between the rotating pulley and its axle
m1 = 0.40kg
m2 = 0.80kg

What is the acceleration of the suspended masses?

Homework Equations


Fnet = I[itex]\alpha[/itex]

The Attempt at a Solution


I tried to make sense out of this problem using the equation from above initially, but my acceleration was way too big. I know the acceleration is supposed to be somewhat big anyway but it was too big. So instead I searched for helm on google first and found an equation which told me to do this:

I = (1/2)mr2 = (1/2)(0.20)(0.5)2 - 0.025kgm2


a = [g(m1 - m2) - [itex]\tau[/itex]/r]/(m1 + m2 + I/r2)
a = 9.8(0.80+0.40)-(0.35/0.15)/(0.80+0.40+(0.0025/(0.15)2)
a = 1.23 m/s2

If this is right, how can I clean it up to make it relevant to how Fnet = I[itex]\alpha[/itex] is supposed to be set up?
 
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santoki said:
If this is right, how can I clean it up to make it relevant to how Fnet = I[itex]\alpha[/itex] is supposed to be set up?
Check your formula:
Fnet I[itex]\alpha[/itex]
 
santoki said:
So instead I searched for helm on google first and found an equation which told me to do this:


a = [g(m1 - m2) - [itex]\tau[/itex]/r]/(m1 + m2 + I/r2)
a = 9.8(0.80+0.40)-(0.35/0.15)/(0.80+0.40+(0.0025/(0.15)2)
a = 1.23 m/s2

If this is right, how can I clean it up to make it relevant to how Fnet = I[itex]\alpha[/itex] is supposed to be set up?
You can't arbitrarily pick an equation out of a book and use it without understanding it, even in the unlikely event that it happened to be the right equation. Instead, use free body diagrams of each block and the pulley, and use Newton's laws for translational or rotational motion on each piece.