Lagrangian Mechanics: Solving Homework Problem on Two Cylinders

[Sang-Hyeon Han]

Homework Statement

A homogeneous hollow cylinder (mass M, radius R) is in the gravitational field and a horizontal axis through the center P rotatably mounted (central axis of the cylinder is fixed and can be rotated). A small, homogeneous solid cylinder (mass m, radius r) is rolling inside of the hollow cylinder without sliding. The two cylinder axis are parallel
O and P are spatially fixed points and A, B, C, S are body-fixed (i.e, on the cylinders) points so that in equilibrium: C to O, B to O, S on P O.
ψ: Deflection of the hollow cylinder of the equilibrium position.
χ: Deflection of the solid cylinder from the Equilibrium.
ϕ: Angular position of the centroid of the solid cylinder.
a) Find the constraints in this two-cylinder system and define the generalized coordinates.
b) Find the Lagrangian function.
c) What are the equations of motion?
d) Determine the natural frequency of oscillation in the case of small displacements.
this is the problem. I have to solve it.

Homework Equations

I know the generalized coordinates and constraints. I found the potential energy , translational kinetic energy and rotational energy for hollow cylinder, but I don't know the rotational energy for the solid cylinder.

The Attempt at a Solution

I thought that the angular velocity of solid cylinder is dχ/dt. so I tried to solve it, but it was not the correct answer. the solution said that dχ/dt-dϕ/dt is the angular velocity. why??

 

[zwierz]

I don't know the rotational energy for the solid cylinder.

Apply the Euler formula: $\boldsymbol v_S=\boldsymbol v_A+\boldsymbol\omega\times\boldsymbol{AS}$ to the solid cylinder. Here $\boldsymbol v_A$ is the velocity of a point of solid cylinder that lies on the hollow cylinder. This velocity is equal to the velocity of the corresponding point of the hollow cylinder. It is because of nonslipping. From this equation you can find angular velocity of the solid cylinder $\boldsymbol \omega$

 

[zwierz]

You do not need so many angles. $\psi,\varphi$ are the good generalized coordinates.
I have got $(R-r)\dot\varphi=-R\dot\psi-r\omega$

 

[Sang-Hyeon Han]

You do not need so many angles. $\psi,\varphi$ are the good generalized coordinates.
I have got $(R-r)\dot\varphi=-R\dot\psi-r\omega$

Thank you so much.