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Friction force on disk

  1. Nov 9, 2009 #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. Nov 9, 2009 #2
    How do you get 0.24 here?
     
  4. Nov 9, 2009 #3
    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. Nov 9, 2009 #4

    ideasrule

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    Except if you plug in the radius and mass, you don't get 0.24.
     
  6. Nov 10, 2009 #5
    sorry calc mistake i=.037
     
  7. Nov 10, 2009 #6

    ideasrule

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    If you plug in i=0.037 you should get the right answer. Do you?
     
  8. Nov 10, 2009 #7
    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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