# Dynamic cylinder + pulley question

1. Nov 21, 2008

### twofish

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

http://members.shaw.ca/code/cylinder.JPG

2. Relevant equations
I denote the top of the ceiling to the top of cylinder A la.
I denote the top most small pulley above A to the larger pulley bottom directly above A as Xa
I denote the distance from the pulley that changes the direction of the rope to the ceiling la'
I denote the top most pulley above B to the smaller pulley bottom directly above B as Xb
I denote the top of the ceiling to the top of cylinder B as lb.
I call upward motion -ve and downward +ve in the calculations.

3. The attempt at a solution

(la-2xa) + (xa-la') + (lb-2xb) = C
then I say l* is a constant so..
-2xa+xa-2xb=C
now I differentiate to attain velocity
-dXa/dt -2dxb/dt = const/dt = 0
-Va -2Vb=0 therefore Vb = -0.4m/s or 0.4m/s UP

now i differentiate again to attain acceleration
-dVa/dt -2dVb/dt =0
-Aa-2Ab=0 therefore Ab =-1m/s2 or 1m/s^2 UP

Vba = Vb-Va = (-0.4 - 0.8m/s) = -1.2m/s or 1.2m/s UP
Aba = Ab -Aa = (-1.2m/s^2 - 2m/s^2) = -3m/s^2 or 3m/s^2 UP

Can anyone confirm that my initial labeling is correct, and also hopefully that my final answer is correct?

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