A cylinder rolls without slipping- find the angle

AI Thread Summary
To determine the maximum angle \(\theta\) for a cylinder to roll without slipping down an incline, energy conservation must be applied, leading to the equation \(Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}Iw^2\). The moment of inertia for the cylinder is given as \(I = \frac{1}{2}MR^2\), and the relationship between angular and linear velocity is \(w = \frac{v}{r}\). The force equations along the incline yield \(F = mg(\cos\theta - \mu) = ma\) and the normal force \(N = mg\cos\theta\). Clarification is needed regarding the direction of forces, particularly that the normal force is \(N = mg\cos\theta\) and not \(N = mg\sin\theta\). The discussion emphasizes the importance of maximum static friction in determining the angle.
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Homework Statement



A cylinder with radius R and mass M rolls without slipping down an incline plane with angle\theta. The coeff. of friction is \mu. Find the maximum value for the angle for the cylinder to roll without slipping.

Homework Equations



The moment of inertia for a cylinder is I=(MR^2)/2.
w=v/r

The Attempt at a Solution



If the cylinder rolls without slipping, energy must be conserved. So Mgh=1/2Mv^2+1/2Iw^2. If you plug in the equations from above and divide all terms by M, gh=3/4v^2.
So we need to find v.
The equation for the x-component of force (where x is along the plane of the incline), F=mg(cos\theta-\mu)=ma
and the y-component
F=0=N-mgsin\theta or N=mgsin\theta

I have no idea where to go from here and I'm not quite sure I did all of this right. Thank you so much for any help!
 
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You should think when you will have the minimum value of angle.When you have the maximum of static friction.Also i think you have a mistake.In the direction of the slope the forces are Wy=mgsinθ and T<=μN
 
Also N=mgcosθ not mgsinθ
 

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