# cylinder rolling down an inclined plane

by PhMichael
Tags: cylinder, inclined, plane, rolling
 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: $$\vec{u}=\vec{v}+\vec{V}=(vcos \beta -V) \hat{x} - (vsin \beta) \hat{y}$$ Momentum is conserved conserved in the $$\hat{x}$$ direction so that: $$0=-MV+m(v cos \beta -V) \to V=\frac{m}{M+m} v cos \beta$$ so the acceleration of the plane relative to the ground is: $$A=\frac{dV}{dt}=\frac{m}{m+M}a cos \beta$$ About point P, we have: $$\vec{\tau}_{P}=mR(g sin \beta + A cos \beta) \hat{z}$$ $$I_{P}\vec{\alpha}=(0.5MR^{2}+MR^{2})\vec{\alpha}=1.5MR^{2} \frac{a}{R}\hat{z}$$ Equating the last two equations while using the equation of A yields: $$a=gsin\beta\left [ \frac{3}{2}-\left ( \frac{m}{m+M} \right )cos^{2}\beta \right ]$$ 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 $$\hat{x}$$ 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?
P: 802
 Quote by PhMichael to be more specific, is linear momentum conserved along the $$\hat{x}$$ 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?
It depends on the reference frame and the system you consider. Whenever there is no external force, the linear momentum is conserved. Remember how to derive the law from F=dp/dt?

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.
 P: 1,877 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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