Find frictional torque acting on the pulley?

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

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?


2. Relevant equations

sum of torque = I*alpha
I = (1/2)mR^2
a=R*alpha
3. 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

 

T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?

do i use the same "a" for each one?

 

... i got 2.95873mN for friction ... still wrong

do we use time for some thing???

 

T1=35.8875N
T1*R=1.7226

T2=-25.839
T2*R=-1.24027

I=0.000864
alpha=4.79167

i use all that and got 2.95873mN

 

T1-M1*g=M1*a

how we use time to find acceleration??

since we have the giving acceleration, time: this is where i have no idea >.<

 

where can i get T1 or T2

since we have 2 equation and 3 unknow
T1,T2 and a

 

a= R* alpha
alpha=a/R??

acceleration = change in v / change in time

 

Vf=Vo+at
a=0.034328m/s^2
right?

 

T1 - M1*g = + M1*a (found T1)
T2 - M2*g = - M2*a (found T2)

I=0.000864

T1*R + T2*R + frictional torque = I*alpha

frictional =0.5mN

 

WOOT it work

BIG thank!!! Doc Al for help me