tire rolling down slope - angle - friction


by mybrohshi5
Tags: angle, friction, rolling, slope, tire
mybrohshi5
mybrohshi5 is offline
#1
Apr11-10, 03:46 PM
P: 365
1. The problem statement, all variables and given/known data

Suppose the hoop were a tire. A typical coefficient of static friction between tire rubber and dry pavement is 0.88. If the angle of the slope were variable,

what would be the steepest slope down which the hoop could roll without slipping?


3. The attempt at a solution

using the x components i came up with

mgsin([latex]\theta[/latex]) - ffric = m*a

since [latex] \mu = \frac{f_{fric}}{N} [/latex] then ffric = [latex]\mu[/latex]*(mgcos[latex]\theta[/latex])

so i put those together and came up with

mgsin[latex]\theta[/latex] - 0.88(mgcos[latex]\theta[/latex]) = ma

mass cancels out so

gsin[latex]\theta[/latex] - 0.88(gcos[latex]\theta[/latex]) = a

This is where i am stuck.

Any help would be great :) thank you
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cjpgconman
cjpgconman is offline
#2
Apr11-10, 05:01 PM
P: 5
You're on the right track and close!
You just have to find the force of friction
so from the sum of forces in the x direction and the torque equation, respectively, we have this:
mg sinθ - friction = m a friction * R = I a / R
Since the moment of inertia for a hoop is mr^2 and we replace that in the torque equation and solve for acceleration we get:
Torque:
friction*R=(mR^2)*(a/R)
Friction=m*a
so a=Friction/m

Now we substitute that in the sum of forces in the x direction
we get:
mg sinθ-friction = friction
friction = (m*g*sinθ)/2
now we just need a second equation for friction which is:
friction=Normal*mu
and the equation for normal is:
N=m*g*cosθ
solve for θ (to get it you'll have to take the inverse tangent)

Hope that helps it worked for me
mybrohshi5
mybrohshi5 is offline
#3
Apr11-10, 05:10 PM
P: 365
Thank you :)

That worked perfectly


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