Solving for Tension in Masses Connected by String Over Pulley

[quick]

Blocks of mass m_1 and m_2 are connected by a massless string that passes over the frictionless pulley in the figure. Mass m_1 slides on a horizontal frictionless surface. Mass m_2 is released while the blocks are at rest.

http://s93755476.onlinehome.us/stuff/knight.Figure.13.68.jpg

i know the acceleration of the masses is m_2*g/(m_1+m_2)
i need to find:
A. the tension of the string
B. supposing the pulley has mass m_p and radius R. Find the acceleration of m_1 and the tensions in the upper and lower portions of the string.
C. find the tension in the upper portion of the string.
D. find the tension in the lower portion of the string.

i know T_1 (upper portion of string) is T_1 = m_1*a
T_2(lower portion) is T_2 - m_2*g = m_2*a
torque = T_2*R - T_1*R = I*alpha

i still don't know how to solve for the tension of the string since there are two different tensions on the string depending on the pulley. i tried adding the two tensions but that didn't work. I am not really sure how to incorporate mass of the pulley. any help is appreciated.

 

[quantumdude]

i know T_1 (upper portion of string) is T_1 = m_1*a
T_2(lower portion) is T_2 - m_2*g = m_2*a

You need to be careful here, because these two equations contradict each other. The first says that block 1 is moving to the right (and therefore block 2 is moving down), and the second equation says the exact opposite. I'd rewrite the second equation to read:

m₂g-T₂=m₂a.

i still don't know how to solve for the tension of the string since there are two different tensions on the string depending on the pulley. i tried adding the two tensions but that didn't work.

You're not supposed to add the tensions. You're supposed to solve the 3x3 system of equations that you have. Your 3 unknowns are T₁, T₂, and a.

im not really sure how to incorporate mass of the pulley. any help is appreciated.

Look at your torque equation:

torque = T_2*R - T_1*R = I*alpha

The mass of the pulley appears in I, the rotational inertia. Also make the substitution: alpha=a/R (the no-slip condition).

 

[gnome]

I assume you have no problem solving for the tension in Part A, right? (Part A apparently assumes that the pulley is frictionless AND massless.)

For the rest, you need to express I as a function of mass and alpha as a function of a, so you end up with 3 equations and 3 unknowns. (You do realize that a for part B will be different than the a you got for part A?)