Finding acceleration of a mass being pulled by another mass over two pulleys

[cener]

Homework Statement

A mass m₁ is pulled along a frictionless horizontal surface by a rope of tension T₁, which passes over a pulley of mass m_P1 to a hanging pulley of mass m_P2, after which it is fixed to the ceiling, as shown in Fig. 1. Labeling the
tension between pulleys 1 and 2 T₂, and that between pulley 2 and the ceiling as T₃, find the acceleration of the mass
m1. Note that the acceleration of pulley 2 is half the acceleration of m₁
The acceleration of m1 is:
(a) 0
(b) \frac{m(P2)g}{2m1+mP1+(3/4)mP2}
(c) \frac{m(P2)g}{2m1+mP2}
(d) g
(e) \frac{m(P2)g}{2m1+mP1+mP2}

Homework Equations

I have no clue

The Attempt at a Solution

when I did it, I got a=\frac{m2g}{2m1+(m2)/2}
But, this is not right. Does it depend on if pulley1 has a mass?
http://www.freeimagehosting.net/wph3q
(if for some reason the diagram doesn't show)
http://www.freeimagehosting.net/wph3q

 

[grzz]

Yes! You noticed the trouble in your solution.

The masses of the two pulleys MUST be taken into account. The reason for this is that the pulleys are assumed to be rotating. That was the reason that the tensions in all parts of the string were named differently.The string is not just slipping over smooth motionless pulleys.

Hence one must use
F = ma ............for the linear motion and
Torque = moment of inertia x angular acceleration..for the angular motion.

 

[cener]

So how do I start?

What I have:
τ_mp1=Iω
τ_mp2=Iω
F_m1=ma
-where all of the accelerations are the same
F_mp2=ma
-where the acceleration is half of the other acceleration

Is this correct? If so, how do I put them together?

 

[haruspex]

Try to stick to the given variable names where appropriate: m₁, T₁, T₂, T₃, m_p1, m_p2. Of course you'll also need new variables for acceleration: maybe a for the mass m₁ (and therefore a/2 for the vertical accn of pulley 2), and α for the angular accn (ω is usually for angular velocity).
I find it surprising that the radii of the pulleys don't come into it. I would let those be r₁ and r₂ and see what happens.
With all that in place, list the forces acting on each component and write out corresponding acceleration equations.

 

[grzz]

The moment of inertia for the pulley may be assumed to be that for a uniform disc
i.e. I = \frac{M_{p}R^{2}}{2}.

The angular acceleration \alpha is best put in terms of the linear acceleration a
using a = R\alpha

Then when the equation for the torque τ is used it will be noticed that the radius R of the pulley will be eliminated from the equation.