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Frictional force of a disk

by aligass2004
Tags: disk, force, frictional
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aligass2004
#1
Oct15-07, 09:10 PM
P: 236
1. The problem statement, all variables and given/known data

The 1.6kg, 20cm diameter disk in the figure below is spinning at 240rpm. How much friction force must the brake apply to the rim to bring the disk to a halt in 3.5s?

http://i241.photobucket.com/albums/f...045/p13-69.gif

2. Relevant equations



3. The attempt at a solution

I know a free body diagram should be drawn for the disk to take all of the forces into account. I know there's the weight pulling down and the frictional force is acting to the right, but I don't know how to set up a solvable equation.
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Astronuc
#2
Oct15-07, 09:44 PM
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Use 1.6kg, 20cm diameter disk to determine the moment of inertia.

The friction force behaves as a torque with moment r and force f.

Torque/(moment of inertia) = angular acceleration

then

use the appropriate equation of motion for rotation to determine the time to decelerate from the initial angular velocity to stop in 3.5 s.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/mi
aligass2004
#3
Oct15-07, 09:48 PM
P: 236
Ok, I get that, but how do I find the frictional force?

aligass2004
#4
Oct15-07, 11:34 PM
P: 236
Frictional force of a disk

The question isn't asking for time.
Astronuc
#5
Oct16-07, 07:26 AM
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One is given time, and ask what magnitude of force is require to bring the disk to standstill ([itex]\omega[/itex] = 0) in that time. Use the change in angular velocity and time to find the constant angular acceleration.

Applying an external force (friction) will cause the rotational mass to decelerate.

One must apply the appropriate equation(s) of motion, e.g.

[itex]0 = \omega_0\,+\,\alpha\,t[/itex], where [itex]\omega_0[/itex] is the initial angular velocity, and [itex]\alpha[/itex] is the angular acceleration (or deceleration if negative).

With the angular acceleration (or deceleration), use the relationship between torque and moment of intertia.

Then knowing the net torque required to decelerate the disk, then find the necessary friction force applied at the appropriate moment arm (radius of disk).
aligass2004
#6
Oct16-07, 03:11 PM
P: 236
I did the following... I=m(r^2)=.016. I then converted revolutions per minute into radians per second = 25.133. Then I found angular acceleration by using delta w/delta t = 7.181, and finally I found the torque = I(alpha) = .115. I'm still unsure about finding the friction.
Doc Al
#7
Oct16-07, 03:26 PM
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Quote Quote by aligass2004 View Post
I did the following... I=m(r^2)=.016.
That formula is incorrect for a solid disk.

To relate torque to friction force, realize that the friction force acting with a moment arm = r creates the given torque (as Astronuc had stated).
aligass2004
#8
Oct16-07, 03:31 PM
P: 236
Ok so instead I = .032 and T = .23. Then I use T = rF (I think) and I solved for F to get 2.3, but it wasn't right.
Doc Al
#9
Oct16-07, 03:42 PM
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What formula are you using to calculate I? (It's still not right.) But yes, use T = rF.
aligass2004
#10
Oct16-07, 10:22 PM
P: 236
I'm using I = .5m(r^2) = .5(1.6)(.1^2) = .032
Doc Al
#11
Oct17-07, 04:33 AM
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Quote Quote by aligass2004 View Post
I'm using I = .5m(r^2) = .5(1.6)(.1^2) = .032
You're using the correct formula, but recheck your calculation.


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