# Cylinder rolling down an inclined plane

1. Sep 6, 2010

### PhMichael

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:

$$A=\frac{dV}{dt}=\frac{m}{m+M}a cos \beta$$

$$\vec{\tau}_{P}=mR(g sin \beta + A cos \beta) \hat{z}$$

$$I_{P}\vec{\alpha}=(0.5MR^{2}+MR^{2})\vec{\alpha}=1.5MR^{2} \frac{a}{R}\hat{z}$$

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?

2. Sep 7, 2010

### hikaru1221

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.

3. Sep 7, 2010

### Mindscrape

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.