A string passing over a pulley has a 3.80 kg mass hanging from one end and a 3.15 kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 0.04 m and mass 0.80 kg.
It is found that if the heavier mass is give a downward speed of 0.20 m/s, it comes to rest in 6.2 s. What is the average frictional torque acting on the pulley?
T = F*r
F = ma
T = I* alpha
The Attempt at a Solution
Newton's second law for masses:
FTA - mag= ma*a (the heavier one)
mb*g -FTB = mb*a
For the pulley:
FTB*rpulley(ccw-positive) - FTA*rpulley(cw-negative) + (+-_____ )= I*alpha
The pulley is rotating ccw, so I thought I had to write minus friction because the friction must be opposite of rotation direction, but then saw this and got really confused.
Determining the sign of frictional torque in this equation, should it be positive or negative?