Acceleration of a mass hanging from a pulley

[EzequielSeattle]

Homework Statement

Two blocks, of masses m1 and m2 (with m1<m2) are attached by a massless string of fixed length that runs over a pulley which is free to rotate about its fixed center. The pulley has mass m3, radius r3, and rotational moment of inertia I = (2/3) m3 r3^2 about its center (it is a non-uniform disk). The string does not slip on the pulley. Give the downward acceleration of m2 in terms of the quantities above.

Homework Equations

F=ma
α=aR
τ=Iα

The Attempt at a Solution

I drew FBD's for each object. The forces acting on m2 are going to be m1g in the upward direction and m2g in the downward direction. I'm just not sure how to incorporate the pulley in here. The torque exerted on the pulley is Iα, so (2/3)m3r3^2*α. It will also be r x F, so r3 x (m2g-m1g). I'm just not sure where to go from here...

 

[haruspex]

The forces acting on m2 are going to be m1g in the upward direction

Not so.
Create unknowns for the tensions, bearing in mind that the string tensions may be different on the two sides of the pulley.

 

[EzequielSeattle]

OK, I've tried this and here's what I've got (T_2 is the tension in the string connecting m_2, T_1 is the tension in the string connecting m_1)

τ=T_2 * r - T_1 * r
a = αr

a will be the same as the acceleration of mass 2, because mass 2 is on a string connected to the pulley.

Iω = τ

ω = τ/I

ω = gr(m_2 - m_1) / I

Am I correct so far? So to find a, should I find the derivative of ω and then multiply by the radius, because a = αr?

 

[haruspex]

Iω = τ

ω?

 

[EzequielSeattle]

Ok, yeah, I realize that that's not the equation and it ought to be α.

I decided to try a different approach. The torque on the pulley is equal to rg(m_2 - m_1).

The torque is also equal to Iα, or I(a/r)

Combining my two equations for the torque, I get

rg(m_2 - m_1) = (2/3)(m_pulley)r^2 * (a/r)

Solving for a gives

a = ((3/2)g (m_2 - m_1))/(m_pulley)

I'm pretty sure this is correct, but I could be wrong.

 

[haruspex]

Ok, yeah, I realize that that's not the equation and it ought to be α.

I decided to try a different approach. The torque on the pulley is equal to rg(m_2 - m_1).

The torque is also equal to Iα, or I(a/r)

Combining my two equations for the torque, I get

rg(m_2 - m_1) = (2/3)(m_pulley)r^2 * (a/r)

Solving for a gives

a = ((3/2)g (m_2 - m_1))/(m_pulley)

I'm pretty sure this is correct, but I could be wrong.

No, you are neglecting that m1 and m2 are also accelerating.
Write the free body equations for the two hanging masses.