# Homework Help: Atwood's Problem (pulley has mass)

1. Apr 26, 2010

### ernay

1. The problem statement, all variables and given/known data

In the Atwood's machine shown below the masses of the blocks are ma = 3kg and mb = 2kg. The moment of inertia of this frictionless pulley about its axis is Ic = .6 kg m2 and its radius is Rc = .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:
EPot f - EPot i + Kfinal - Kinitial = 0
so we know that:

Kinitial = 0 (since its not moving yet)
EPot f = m1(g)(h)
EPot i = m2(g)(h)
Kfinal = (1/2)m2v2 + (1/2)m1v2 + (1/2) Iw2

then using vf2= vi2+2ah
again vii = 0, so we can now replace every vf2 by 2ah

at this point you have:
m1(g)(h) - m2(g)(h) + (1/2)(ma)(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)(ma)(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

2. Apr 27, 2010

### 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

3. Apr 27, 2010

### ernay

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

4. Apr 27, 2010

### 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.