# friction force on disk

by holtvg
Tags: disk, force, friction
 P: 18 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.
P: 1,331
 Quote by holtvg i=1/2mr^2=.24
How do you get 0.24 here?
 P: 18 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.
HW Helper
P: 2,324

## friction force on disk

Except if you plug in the radius and mass, you don't get 0.24.
 P: 18 sorry calc mistake i=.037
 HW Helper P: 2,324 If you plug in i=0.037 you should get the right answer. Do you?
 P: 18 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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