Question:

There is an open cylinder of radius \(R\), which is stationary. A hollow sphere of radius \(\rho\) and mass \(m\) rolls on the surface without slipping. Find the Lagrangian function.

Answer:

The Lagrangian function is defined as

\[ L = KE – PE \]

where \(KE\) and \(PE\) represent kinetic energy and potential energy, respectively.

The mechanical potential energy is the gravitational force \(\times\) height, which is taken from the center. The height varies with the angle \(\theta\). Thus,

\[ PE = -mg(R-\rho)\cos\theta \]

The negative sign indicates that the height is directed from outside to the center.

Now the kinetic energy has two parts, translational and rotational kinetic energies. (Translational means linear.)

\[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]

The translational velocity, \(v\), is along with the trace of circle whose radius is \(R-\rho\). Therefore,

\[ v = (R-\rho)\omega \]

The angular velocity can be expressed by

\[ \omega = \theta’ = \frac{d\theta}{dt} \]

The moment of inertia of hollow sphere is

\[ I = \frac{2}{3}m\rho^2 \]

Plug everything into above.

\[ L = \frac{1}{2}m(R-\rho)^2 \theta ‘^2 + \frac{1}{2}\frac{2}{3}m\rho^2 \theta ‘^2 + mg(R-\rho)\cos\theta \]

We can rewrite it as

\[ L = \frac{m\theta’^2}{6}(3R^2 -6R\rho + 5\rho^2)+ mg(R-\rho)\cos\theta \]

The Lagrangian depicts the mechanical system and derives the equation of motion.