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Friction force on disk 
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#1
Nov909, 03:24 PM

P: 18

1. The problem statement, all variables and given/known data
A 3.10 kg, 31.0cmdiameter 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. 


#2
Nov909, 05:28 PM

P: 1,397




#3
Nov909, 09:37 PM

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.



#4
Nov909, 09:41 PM

HW Helper
P: 2,322

Friction force on disk
Except if you plug in the radius and mass, you don't get 0.24.



#5
Nov1009, 10:10 AM

P: 18

sorry calc mistake i=.037



#6
Nov1009, 01:05 PM

HW Helper
P: 2,322

If you plug in i=0.037 you should get the right answer. Do you?



#7
Nov1009, 03:42 PM

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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