Calculating Acceleration and Tension in a Pulley System

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The discussion focuses on calculating the acceleration and tension in a pulley system involving two blocks and a pulley. Initially, the acceleration was calculated as 0.80 m/s², but this did not consider the pulley's mass or its effect on the system's dynamics. The revised formula for acceleration incorporates the mass of the pulley, resulting in a new value of 0.533 m/s². Participants express confusion about the inclusion of the pulley's mass in the calculations and seek clarification on the reasoning behind it. Understanding the role of the pulley's mass is crucial for accurate calculations in such systems.
Nash77
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Please refer to problem VI here:

http://www.physics.ohio-state.edu/~dongping/phys131_winter2011/phys131/final-exam-practice-2011.pdf

Use 10 for gravity if needed.

Homework Statement



1. When the system is released from rest, what is the magnitude of the blocks' acceleration?
2. What are the tensions in the part of the cord that supports the heavier block and the lighter block, respectively?
3. What is the magnitude of the pulley's angular acceleration?


Homework Equations



F1 = m1a = m1g - T
F2 = m2a = T - m2g



The Attempt at a Solution



So, we used the equations above, added both equations together to eliminate T, then solved for a.

m1a + m2a = m1g - T + T - m2g
m1a + m2a = m1g - m2g
a(m1 + m2) = g(m1 - m2)
a = g(m1 - m2)
-----------
(m1+m2)

a = 0.80 m/s^2

But this gave an acceleration of 0.80 m/s^2 and it never took into account the mass or acceleration of the pulley. How do we incorporate that information? Our TA emailed us, saying the answer should be 0.533 m/s^2??
 
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So, we found a new way to solve for acceleration and it works, just not sure why.

a = g(m1 - m2)
---------------
(m1+m2+1/2MP)

^^ where MP = mass of the pulley.

Why did we add in the 1/2 mass of the pulley to solve?
 

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