(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Cylinder rolling down an inclined plane

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