How to Solve Atwood's Problem with a Massive Pulley?

[ernay]

Homework Statement

In the Atwood's machine shown below the masses of the blocks are m_a = 3kg and m_b = 2kg. The moment of inertia of this frictionless pulley about its axis is I_c = .6 kg m² and its radius is R_c = .2m. Note that there is no slipping between the rope and the pulley.

Find the accelerations of the blocks and the angular acceleration of the pulley

http://commons.wikimedia.org/wiki/File:Atwoodmachine.gif
this is the closest diagram i could find... the distance of the right block (I'll name it block 2 and the left block 1) above the ground is h.

I thought of two ways to do this problem, but I wanted to try and use Newton's laws.
I solved it using conservation of energy:
E_{Pot f} - E_{Pot i} + K_final - K_initial = 0
so we know that:

K_initial = 0 (since its not moving yet)
E_{Pot f} = m₁(g)(h)
E_{Pot i} = m₂(g)(h)
K_final = (1/2)m₂v² + (1/2)m₁v² + (1/2) Iw²

then using v_f²= v_i²+2ah
again vi_i = 0, so we can now replace every v_f² by 2ah

at this point you have:
m₁(g)(h) - m₂(g)(h) + (1/2)(m_a)(2ah) + (1/2)(m₁)(2ah) + (1/2)(I)(2ah/R²)

Now cancel out all the h's and get:
m₁(g) - m₂(g) + (1/2)(m_a)(2a) + (1/2)(m₁)(2a) + (1/2)(I)(2a/R²)

After solving for a I got a = .49 m/s² of the blocks (I'm not concerned with the angular acceleration right now)

I think my answer is right but..
Is there any way to use Newton's Laws... Maybe use the following?
m₂(g) - T₂ = m₂(a)
T₁ - m₁(g) = m₁(a)
Or does this only apply for massless pulleys?
I tried using it, and I get a different Answer... Also if someone could check my answer itd be nice

 

[resaypi]

M1g - T1 = M1a
T2 - M1g = M2a
torque (t) = T1R - T2R = Ia/r

eliminate T1,T2

M1g -M2g - (T1-T2) = (M1 + M2)a

T1 - T2 = Ia/r^2

M1g - M2g - Ia/r^2 = (M1 + M2)a

a = (M1 - M2)g/(M1 + M2 + I/r^2 )

massles pulley implies I = 0

your answer 0.49 is correct using this approach

 

[ernay]

M1g - T1 = M1a
T2 - M1g = M2a
torque (t) = T1R - T2R = Ia/r

eliminate T1,T2

M1g -M2g - (T1-T2) = (M1 + M2)a

T1 - T2 = Ia/r^2

M1g - M2g - Ia/r^2 = (M1 + M2)a

a = (M1 - M2)g/(M1 + M2 + I/r^2 )

massles pulley implies I = 0

your answer 0.49 is correct using this approach

Thanks for the help, I get it now... Just another question though... When we consider the pulley to massless does it mean that the tensions are equal? Since T1-T2=Ia/r² Or am I jumping to a false conclusion? Thanks again

 

[resaypi]

True, if we consider the pulley massless T1 = T2. The difference is due to the fact that there is a friction force that turns the pulley.