# Find coefficient of friction

1. Oct 28, 2007

### camherokid

1. The problem statement, all variables and given/known data

The car's wheel with mass 1kg and radius 0.3m has initial speed 40rad/s.
When braked, the force push to the center of the wheel is 10N.
The car stops after 2s.
Find coefficient.

2. Relevant equations

3. The attempt at a solution

This is what I did
Fc= Us.N= Us*m*g
so Us= Fc/mg

But why the problem gives so many information..about velocity..?

2. Oct 28, 2007

### rocomath

you have "initial speed", find final velocity

you must convert 40rad/s to m/s

then from there, plug in

your equations are wrong unless i screwed up

i have

$$\mu_{k}=\frac{1}{g}(\frac{v^{2}}{R}-\frac{F}{m})$$

i simplified it so yours may look different.

EDIT: well i confirmed my equation by simplifying the units and all the units cancel out so seems like it's good, so i hope that helps.

Last edited: Oct 29, 2007
3. Oct 29, 2007

### camherokid

why they the time and final velocity, when stop?

4. Oct 29, 2007

### saket

How did you derive such relation? Would you mind explaining your steps?

@ camherokid:
I have a different line of thought:
Let initial angular speed be w = 40rad/s, radius of the wheel be r = 0.3m, mass of the wheel be M = 1kg, push-force be P = 10N, time be t = 2sec.

Speed of the centre of the wheel, v = w*r = 12m/s. {Assuming, no slipping.}

It has to stop in t = 2sec. Thus, final speed is zero.
Assuming uniform deceleration, a = (0 - v)/t = -v/t = -6m/s^2.

This deceleration will require an average force of magnitude, F = M*a = -M*v/t = -6N .

Now assuming this force comes only by friction, µ*P = -F. (As friction is in opposite direction; considering P to be positive.)

Thus, µ = M*v/(P.t) = (M*w*r)/(P*t) = 0.6

Therefore coefficient of friction, µ = 0.6

5. Oct 29, 2007

### rocomath

actually, it should have been. sign was backwards.

$$\mu_{k}=\frac{1}{g}(\frac{F}{m}-\frac{v^{2}}{R})$$

drew a force body diagram to get Normal force, but i never plugged in numbers.

gotta go to class, i'll check in 3-4 hrs.

Last edited: Oct 29, 2007