friction force on disk


by holtvg
Tags: disk, force, friction
holtvg
holtvg is offline
#1
Nov9-09, 03:24 PM
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.
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willem2
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#2
Nov9-09, 05:28 PM
P: 1,351
Quote Quote by holtvg View Post

i=1/2mr^2=.24
How do you get 0.24 here?
holtvg
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#3
Nov9-09, 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.

ideasrule
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#4
Nov9-09, 09:41 PM
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friction force on disk


Except if you plug in the radius and mass, you don't get 0.24.
holtvg
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#5
Nov10-09, 10:10 AM
P: 18
sorry calc mistake i=.037
ideasrule
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#6
Nov10-09, 01:05 PM
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If you plug in i=0.037 you should get the right answer. Do you?
holtvg
holtvg is offline
#7
Nov10-09, 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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