1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find frictional torque acting on the pulley?

  1. May 3, 2008 #1
    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. jcsd
  3. May 3, 2008 #2

    Doc Al

    User Avatar

    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.)
     
  4. May 3, 2008 #3
    T1=M1*g-M1*a

    T2=M2*a+M2*g

    this is right?

    do i use the same "a" for each one?
     
  5. May 3, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Yes--they are connected by a string so they must have the same acceleration.
     
  6. May 3, 2008 #5
    ... i got 2.95873mN for friction ... still wrong

    do we use time for some thing???
     
  7. May 3, 2008 #6
    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
     
  8. May 3, 2008 #7

    Doc Al

    User Avatar

    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.)
     
  9. May 3, 2008 #8
    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 >.<
     
  10. May 3, 2008 #9

    Doc Al

    User Avatar

    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.
     
  11. May 3, 2008 #10

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    What's the definition of acceleration?
     
  12. May 3, 2008 #11

    Doc Al

    User Avatar

    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.
     
  13. May 3, 2008 #12
    where can i get T1 or T2

    since we have 2 equation and 3 unknow
    T1,T2 and a
     
  14. May 3, 2008 #13

    Doc Al

    User Avatar

    Staff: Mentor

    You have 3 equations and 3 unknowns. The acceleration is not an unknown. (It's easy to calculate from the given data.)
     
  15. May 3, 2008 #14
    a= R* alpha
    alpha=a/R??

    acceleration = change in v / change in time
     
  16. May 3, 2008 #15
    Vf=Vo+at
    a=0.034328m/s^2
    right?
     
  17. May 3, 2008 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Right!
     
  18. May 3, 2008 #17
    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
     
  19. May 3, 2008 #18
    WOOT it work

    BIG thank!!! Doc Al for help me
     
  20. May 3, 2008 #19

    Doc Al

    User Avatar

    Staff: Mentor

    You are most welcome. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find frictional torque acting on the pulley?
Loading...