# 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?
 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?