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Cylinder rolling down an inclined plane 
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#1
Sep610, 04:24 PM

P: 125

1. The problem statement, all variables and given/known data
So I have this standard problem of a cylinder rolling down an inclined plane, however, this time the plane itself is free to slide on the ground. I need to find the acceleration of that cylinder relative to the plane. 2. The attempt at a solution V / A  the velocity / acceleration of the plane relative to the ground. v / a  the velocity / acceleration of the cylinder relative to the plane. The velocity of the cylinder relative to the ground is: [tex] \vec{u}=\vec{v}+\vec{V}=(vcos \beta V) \hat{x}  (vsin \beta) \hat{y} [/tex] Momentum is conserved conserved in the [tex]\hat{x}[/tex] direction so that: [tex] 0=MV+m(v cos \beta V) \to V=\frac{m}{M+m} v cos \beta [/tex] so the acceleration of the plane relative to the ground is: [tex] A=\frac{dV}{dt}=\frac{m}{m+M}a cos \beta [/tex] About point P, we have: [tex] \vec{\tau}_{P}=mR(g sin \beta + A cos \beta) \hat{z} [/tex] [tex] I_{P}\vec{\alpha}=(0.5MR^{2}+MR^{2})\vec{\alpha}=1.5MR^{2} \frac{a}{R}\hat{z} [/tex] Equating the last two equations while using the equation of A yields: [tex]a=gsin\beta\left [ \frac{3}{2}\left ( \frac{m}{m+M} \right )cos^{2}\beta \right ] [/tex] This answer is correct, however, I'm not sure whether what I did is kosher; to be more specific, is linear momentum conserved along the [tex]\hat{x}[/tex] direction? are fictitious forces, like the one we have here on the cylinder as a result of the plane's acceleration, not treated as real forces so that eventhough they exist, we may still use the conservation of momentum principle? 


#2
Sep710, 07:29 AM

P: 799

In the frame of the ground, if you consider the system of the wedge & the cylinder, the momentum of the system is conserved. With the same system, but in the frame of the wedge, there is fictitious force, i.e. F = dp/dt is not zero, the momentum is not conserved. 


#3
Sep710, 07:45 AM

P: 1,875

Looks good to me! Now if you truly wanted to make this problem difficult you could add in friction, inertia effects, and some good old lagrange multipliers. :p
Also, if you know of Lagrangian dynamics then it could be fun to solve the problem in an alternate way, which I might do just for fun. 


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