# Friction force on disk

1. Nov 9, 2009

### holtvg

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
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. Nov 9, 2009

### willem2

How do you get 0.24 here?

3. Nov 9, 2009

### holtvg

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. Nov 9, 2009

### ideasrule

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

5. Nov 10, 2009

### holtvg

sorry calc mistake i=.037

6. Nov 10, 2009

### ideasrule

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

7. Nov 10, 2009

### holtvg

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