Friction force on disk

  1. 1. The problem statement, all variables and given/known data
    A 3.10 kg, 31.0-cm-diameter disk in the figure is spinning at 300 rpm, how much force must a brake apply that is perpendicular to the rim of the disk to bring the disc to a halt in 2.4 s.


    2. Relevant equations
    t=i*alpha wf=0+alpha*t t=fr*sin(theta)


    3. The attempt at a solution

    i=1/2mr^2=.24
    wf=0+alpha*t wf=0 w0=300 rpm=31.4 rad/s t=2.4 s alpha=-13.1 rad/s^2
    t=i*alpha=-3.1 n*m
    t=fr*sin(theta) theta=90 degress
    t=fr
    -3.1=f*.155 r=15.5 cm=.155 m
    f=-20 n

    Don't know what i'm doing wrong but it's not the correct answer.
     
    Last edited: Nov 9, 2009
  2. jcsd
  3. How do you get 0.24 here?
     
  4. The i or moment of inertia for a thin disk is given by i=1/2mr^2 where m is the mass of the disk and r is the radius of the disk both in meters.
     
  5. ideasrule

    ideasrule 2,323
    Homework Helper

    Except if you plug in the radius and mass, you don't get 0.24.
     
  6. sorry calc mistake i=.037
     
  7. ideasrule

    ideasrule 2,323
    Homework Helper

    If you plug in i=0.037 you should get the right answer. Do you?
     
  8. I get -3.27 n but the correct answer is -3.14 n, probably just a rounding error. But yes i do get the right answer as i messed up in i.

    Thankyou
     
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