What is the acceleration of masses in an Atwood machine with a rotating pulley?

[dumbperson]

Homework Statement

I'm trying to find the acceleration of the masses in an atwood machine, the pulley has a rotational inertia I. The pulley has a radius R

Picture : http://en.wikipedia.org/wiki/File:Atwood.svg
But I made it accelerate the other way, so the equations from Newtons 2nd law are (a bit) different for this picture.

Homework Equations

$$m_1 \cdot a = T_1 - m_1 \cdot g$$
$$m_2 \cdot a = m_2 \cdot g - T_2$$
$$Ʃ \tau = \frac{dL}{dt}$$

The Attempt at a Solution

I know you can do it another way, but I wan't to do this with angular momentum L(with respect to midpoint of the pulley). The pulley has a radius R.

$$L= I_{pulley} \cdot \omega + (m_1 + m_2) \cdot R^2 \cdot \omega = I_{pulley} \cdot \frac{v}{R} + (m_1 +m_2)vR$$

So:
$$\frac{dL}{dt} = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R$$
$$Ʃ \tau = T_2 \cdot R - T_1 \cdot R$$
$$T_2 = m_2 \cdot g -m_2 \cdot a$$
$$T_1 = m_1 \cdot a + m_2 \cdot g$$
So
$$Ʃ \tau = g \cdot R (m_2 - m_1) - a(m_2 +m_1)$$
$$Ʃ \tau = \frac{dL}{dt}$$
So
$$g \cdot R (m_2 - m_1) - a(m_2 +m_1)R = I_{pulley} \cdot \frac{a}{R} + (m_1 +m_2)a \cdot R$$

Solve for a and i'll get:
$$a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+2(m_1+m_2)}$$

The right answer should be:
$$a=\frac{g \cdot (m_2 - m_1)}{\frac{I}{R^2}+m_1+m_2}$$

I notice that if i say:
$$Ʃ \tau = gR(m_2-m_1)$$ , I do get the right answer. But this is not true is it? this would only be true if the masses are not accelerating? is the angular momentum wrong ?

Where do I go wrong? thank you!

 

[I like Serena]

Welcome to PF!

I've included your picture for reference.

You've treated the system as a whole.
This means you should only look at the external forces.

The external force applied on the left side is the weight of m₁, which is m₁g.
And on the right side you have m₂g.
So the sum of all moments is:
$$\Sigma \tau = m_1g R - m_2g R$$
You have looked at the internal forces, but doing so, you are effectively counting things double.

 

[dumbperson]

Welcome to PF!

I've included your picture for reference.

You've treated the system as a whole.
This means you should only look at the external forces.

The external force applied on the left side is the weight of m₁, which is m₁g.
And on the right side you have m₂g.
So the sum of all moments is:
$$\Sigma \tau = m_1g R - m_2g R$$
You have looked at the internal forces, but doing so, you are effectively counting things double.

Oh that makes sense! So if I would not look at the whole system but just the pulley, I would get the right answer too if I say:
$$Ʃ \tau = T_2R-T_1R$$
But then L would be different
$$L= I \cdot \frac{v}{R} $$

That would be right too?
Thanks!

 

[I like Serena]

Yep.

 

[dumbperson]

thanks alot