Acceleration of spool being pulled by ribbon wound around another spool

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Homework Statement



Red ribbon wound around a red spool (above) is taped to blue ribbon wound around a blue spool (below). The ribbon is essentially massless, but the solid cylindrical spools each have mass m and radius R. If the red spool can freely rotate on a fixed axle and the blue spool is positioned directly underneath with the ribbon taut, what is the downward acceleration of the blue spool? Gravity is downward.
Enter your answer in terms of some or all of the variables m, R and g.


Homework Equations





The Attempt at a Solution



Used F=ma and got mg-T=ma, a=(mg-T)/m, and kind of got stuck there.
 
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Remember that both spools will rotate. Create variables for the two angular accelerations and write down torque equations. There's another equation relating these two angular accelerations to the blue spool's linear acceleration.
 
I came up with 2 torque equations:
(T-mg)*r=I*alpha (alpha is positive)
(mg-T)*r=I*alpha (alpha is negative)

Can you tell me if these are right or not, also can you give me a hint for the equation that relates angular to linear acceleration? Thanks!
 
postfan said:
I came up with 2 torque equations:
(T-mg)*r=I*alpha (alpha is positive)
(mg-T)*r=I*alpha (alpha is negative)
Please use different symbols for different variables, e.g. αred, αblue.
Your torque equation for the red spool is wrong. Have another think.
Can you tell me if these are right or not, also can you give me a hint for the equation that relates angular to linear acceleration?
It's just like the equation that relates angular velocity of a car wheel to the car's linear velocity, except that here you have two wheels affecting the length of ribbon. It might help to think first about the equation that would apply if the ribbon were stuck on the bottom reel so that it could not spin. How would the linear acceleration of the bottom spool relate to the angular acceleration of the top one? Then swap it around and suppose only the bottom spool can spin, then try to put the two together.
 
postfan said:
I came up with 2 torque equations:
(T-mg)*r=I*alpha (alpha is positive)
(mg-T)*r=I*alpha (alpha is negative)
The torques are related to the tension in the ribbons, which is T (not (T-mg) or (mg-T)). Since the ribbons are massless, T is constant (not affected by height).
 
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