# Find frictional torque acting on the pulley?

## 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

Doc Al
Mentor
T1*R-T2R- frictional torque=I*alpha
That's the torque equation--good! But you'll need other equations to solve for the tensions. (Analyze the forces on each mass.)

T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?

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

Doc Al
Mentor
T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?
Good.

do i use the same "a" for each one?
Yes--they are connected by a string so they must have the same acceleration.

... 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

Doc Al
Mentor
do we use time for some thing???
Sure. You need it to compute the acceleration. (Note that the acceleration of the heavier mass is upward, so we need to double check the signs in your equations.)

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 >.<

Doc Al
Mentor
clarification

T1=M1*g-M1*a

T2=M2*a+M2*g
Assuming T1 is the heavier mass, let's try it again. Write the equations this way:
T1 - M1*g = + M1*a
T2 - M2*g = - M2*a

Now "a" is just the magnitude of the acceleration, which you can calculate directly.

Doc Al
Mentor
T1-M1*g=M1*a
Good.

how we use time to find acceleration??
What's the definition of acceleration?

Doc Al
Mentor
T1*R-T2R- frictional torque=I*alpha
We need to fix the sign of the acceleration in this one also:

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

Now alpha is a positive quantity.

where can i get T1 or T2

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

Doc Al
Mentor
You have 3 equations and 3 unknowns. The acceleration is not an unknown. (It's easy to calculate from the given data.)

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

acceleration = change in v / change in time

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

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

Doc Al
Mentor
You are most welcome. 