
#1
Apr1110, 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])  f_{fric} = m*a since [latex] \mu = \frac{f_{fric}}{N} [/latex] then f_{fric} = [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 



#2
Apr1110, 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 sinθfriction = 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 



#3
Apr1110, 05:10 PM

P: 365

Thank you :)
That worked perfectly 


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