# Find frictional torque acting on the pulley?

1. May 3, 2008

### saturn67

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

2. May 3, 2008

### Staff: Mentor

That's the torque equation--good! But you'll need other equations to solve for the tensions. (Analyze the forces on each mass.)

3. May 3, 2008

### saturn67

T1=M1*g-M1*a

T2=M2*a+M2*g

this is right?

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

4. May 3, 2008

### Staff: Mentor

Good.

Yes--they are connected by a string so they must have the same acceleration.

5. May 3, 2008

### saturn67

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

do we use time for some thing???

6. May 3, 2008

### saturn67

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

7. May 3, 2008

### Staff: Mentor

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.)

8. May 3, 2008

### saturn67

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 >.<

9. May 3, 2008

### Staff: Mentor

clarification

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.

10. May 3, 2008

### Staff: Mentor

Good.

What's the definition of acceleration?

11. May 3, 2008

### Staff: Mentor

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.

12. May 3, 2008

### saturn67

where can i get T1 or T2

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

13. May 3, 2008

### Staff: Mentor

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

14. May 3, 2008

### saturn67

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

acceleration = change in v / change in time

15. May 3, 2008

### saturn67

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

16. May 3, 2008

### Staff: Mentor

Right!

17. May 3, 2008

### saturn67

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

18. May 3, 2008

### saturn67

WOOT it work

BIG thank!!! Doc Al for help me

19. May 3, 2008

### Staff: Mentor

You are most welcome.