Lagrangian mechanics: rolling cylinder
The first step is to define the Lagrangian, denoted by L. It is simply the kinetic energy T of the system minus its potential energy V, \mathcal L = T - V.
For my cylinder, of mass m and radius R, rolling down a plane inclined at an angle \varphi, I need to express these energies. Let x be the distance the cylinder’s center of mass has moved along the incline, and \theta be the angle it has rotated. The moment of inertia of the cylinder about its axis is I.
The kinetic energy has two parts: translational (due to the motion of its center of mass) and rotational (due to its spinning):
T = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2
Here, \dot{x} is the velocity of the center of mass, and \dot{\theta} is the angular velocity.
The potential energy V is due to gravity. If I set the potential energy to be zero at the starting height, then as the cylinder moves down a distance x along the incline, its height decreases by x \sin\varphi. So, the potential energy is:
V = -mgx \sin\varphi
Combining these, my Lagrangian becomes:
\mathcal L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 + mgx \sin\varphi
At this point, x and \theta might seem like two independent ways the system can move. However, if the cylinder rolls without slipping, these two motions are linked. The condition for rolling without slipping is x = R\theta. This means the distance covered by the center of mass is directly related to the angle turned by the cylinder.
Taking the time derivative, I get \dot{x} = R\dot{\theta}, which allows me to express \dot{\theta} as \dot{\theta} = \frac{\dot{x}}{R}. I can use this to simplify my Lagrangian, expressing it in terms of a single generalized coordinate, x:
L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2 + mgx \sin\varphi
This can be rearranged to:
\mathcal L = \frac{1}{2} \left(m + \frac{I}{R^2}\right) \dot{x}^2 + mgx \sin\varphi
I can see that the term \left(m + \frac{I}{R^2}\right) acts like an “effective mass” for the translational motion when rotation is considered.
Once I have the Lagrangian in terms of a single coordinate x and its velocity \dot{x}, I use the Euler-Lagrange equation to find the equation of motion:
\frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial \mathcal L}{\partial \dot{x}} \right) - \frac{\partial \mathcal L}{\partial x} = 0
Calculating the partial derivatives:
\begin{aligned} \frac{\partial \mathcal L}{\partial \dot{x}} & = \left(m + \frac{I}{R^2}\right) \dot{x}\\ \frac{\partial \mathcal L}{\partial x} &= mg \sin\varphi \end{aligned}
Plugging these into the Euler-Lagrange equation, I get:
\left(m + \frac{I}{R^2}\right) \ddot{x} - mg \sin\varphi = 0
This gives me the acceleration \ddot{x} of the cylinder’s center of mass:
\ddot{x} = \frac{mg \sin\varphi}{m + \frac{I}{R^2}} = \frac{g \sin\varphi}{1 + \frac{I}{mR^2}}
This result shows how the acceleration depends on gravity, the incline angle, and the cylinder’s mass distribution (through I). This example demonstrates how I can use the Lagrangian method to solve mechanics problems by focusing on energy and constraints, often leading to a more straightforward derivation of the equations of motion compared to a direct force-based analysis.
For more insights into this topic, you can find the details here.