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## Homework Statement

A string passing over a pulley has a 3.75 kg mass hanging from one end and a 2.70 kg mass hanging from the other end. The pulley is a uniform solid cylinder of radius 4.8 cm and mass 0.75 kg.

In fact, it is found that if the heavier mass is given a downward speed of 0.23 m/s, it comes to rest in 6.7 s. What is the average frictional torque acting on the pulley?

## Homework Equations

sum of torque = I*alpha

I = (1/2)mR^2

a=R*alpha

## The Attempt at a Solution

T1*R-T2R- frictional torque=I*alpha

i use that and i got around 0.415mN which is wrong ...

anyone give me some pointer please