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