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johnschmidt
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Homework Statement
http://essbaum.com/images/Pulley_diagram.png
We have a bonus question where we are asked to find the mass of ##m_{1}## necessary to accelerate ##m_{2}## upward at ##1g##, and we have to include friction in the pulley bearings as well as the pulley inertias. The pulleys are flat disks, and the rope doesn't have mass or stretch.
##m_{2}## = 1kg
##a_{2}## = 1g (positive is upward)
##r_{p1}## (radius of pulley 1) = 10cm
##m_{p1}## (mass of pulley 1) = 1kg
##\tau_{frictionofp1}## (friction torque of pulley 1 bearing) = 0.4Nm
##r_{p2}## (radius of pulley 2) = 20cm
##m_{p2}## (mass of pulley 2) = 2kg
##\tau_{frictionofp2}## (friction torque of pulley 2 bearing) = 0.6Nm
Homework Equations
Torque ##\tau_{3}## is associated with tension ##T_{3}## and torque ##\tau_{2}## is associated with tension ##T_{2}##
##I## = moment of inertia
##\alpha## = angular acceleration
The Attempt at a Solution
First I solve the torques of pulley 2:
##\tau_{2} - \tau_{3} = I_{p2} \alpha_{p2} + \tau_{friction of p2}##
##T_{2} r_{p2} - T_{3} r_{p2} = \frac{m_{p2} r_{p2}^2}{2} \frac{a_{2}}{r_{p2}} + \tau_{frictionofp2}##
##T_{2} - T_{3} = \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}##
Now solve for ##T_{3}##
##T_{3} = m_{2} g + m_{2} a_{2}##
Substituting ##T_{3}## into the previous equation gives
##T_{2} - m_{2} g - m_{2} a_{2} = \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}##
Solve for ##T_{2}##
##T_{2} = m_{2} g + m_{2} a_{2} + \frac{m_{p2} a{2}}{2} + \tau_{frictionofp2}##
Great! Now I have ##T_{2}## in terms of known variables.
But now I get stuck. How do I proceed with pulley 1? I can follow the same procedure as above and find the difference between ##T_{1}## and ##T_{2}##, but that seems to neglect that pulley 1 is in motion downward.
Similarly if I solve for the forces on pulley 1 (for example ##T_{1} + T_{2} = \left(m_{1} + m_{p1}\right) g - \left(m_{1} + m_{p1}\right) a_{1}## then I am ignoring the forces that go into rotating the inertia of pulley 1...
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