Rigid Body Lagrangian Formulation
Pivoted Bar With a Moving Support
The power of the Lagrangian formulation lies in its elegant and systematic approach to dynamics. While we initially explored its application to point particles here, extending it to rigid bodies is a natural progression and straightforward.
Recall that the core of the Lagrangian approach is the Lagrangian, defined as the difference between the kinetic energy T and the potential energy V of the system: \mathcal L = T - V. The equations of motion are then derived from the Euler-Lagrange equations:
\frac{d}{dt}\left(\frac{\partial \mathcal L}{\partial \dot{q}_i}\right) - \frac{\partial \mathcal L}{\partial q_i} = 0
where q_i are the generalized coordinates describing the system’s configuration.
For a rigid body, the key difference lies in properly accounting for its kinetic energy. Unlike a point particle which only has translational kinetic energy, a rigid body also possesses rotational kinetic energy due to its extended nature and possible rotation about its center of mass.
To apply the Lagrangian formulation to a rigid body (either 2 dimensional or 3 dimensional), we need to express its total kinetic energy as the sum of its translational and rotational kinetic energy. The translational kinetic energy is the same as that of a point particle located at the body’s center of mass. The rotational kinetic energy, however, depends on the body’s moment of inertia and its angular velocity.
Once the total kinetic energy (and the potential energy, which depends on the forces acting on the body) is correctly expressed in terms of generalized coordinates describing both the body’s position and orientation, the Euler-Lagrange equations can be applied directly. The resulting equations of motion will then naturally incorporate both translational and rotational dynamics, providing a complete description of the rigid body’s motion.
Cylinder on an incline
We’re considering a cylinder of mass M, radius R, and moment of inertia I_\Delta about its central axis. This cylinder is placed on a plane inclined at an angle \varphi to the horizontal. The cylinder rolls down without slipping.
System parameters:
- m: mass of the cylinder.
- R: radius of the cylinder.
- I: moment of inertia about the cylinder’s symmetry axis.
- \varphi: angle of inclination of the plane.
- x: coordinate of the center of mass (CM) along the incline, positive downwards.
- \theta: angle of rotation of the cylinder.
The kinetic energy T is the sum of translational and rotational components:
T = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2
The gravitational potential energy V, with V=0 at x=0, is:
V = -mgx \sin\varphi
The Lagrangian \mathcal L = T - V is:
\mathcal L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 + mgx \sin\varphi
The condition for rolling without slipping is x = R\theta. Its time derivative is \dot{x} = R\dot{\theta}, so \dot{\theta} = \frac{\dot{x}}{R}.
Using x as the generalized coordinate, substitute \dot{\theta} into \mathcal L:
\mathcal L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \left(\frac{\dot{x}}{R}\right)^2 + mgx \sin\varphi
\mathcal L = \frac{1}{2} \left(m + \frac{I}{R^2}\right) \dot{x}^2 + mgx \sin\varphi
This can also be expressed as:
\mathcal L = \frac{1}{2} m \left(1 + \frac{I}{mR^2}\right) \dot{x}^2 + mgx \sin\varphi
Lagrange’s equation for x is:
\frac{\mathrm{d}}{\mathrm{d}t} \left( \frac{\partial \mathcal L}{\partial \dot{x}} \right) - \frac{\partial \mathcal L}{\partial x} = 0
The partial derivatives are:
\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
Substituting these into Lagrange’s equation yields the equation of motion:
\frac{\mathrm{d}}{\mathrm{d}t} \left[ \left(m + \frac{I}{R^2}\right) \dot{x} \right] - mg \sin\varphi = 0
\left(m + \frac{I}{R^2}\right) \ddot{x} = mg \sin\varphi
The acceleration \ddot{x} of the cylinder’s CM is:
\ddot{x} = \frac{mg \sin\varphi}{m + \frac{I}{R^2}} = \frac{g \sin\varphi}{1 + \frac{I}{mR^2}}
The term \frac{I}{mR^2} is dimensionless.
Cylinder within a cylinder
Let’s consider a solid cylinder rolls without slipping along the inner surface of a larger, fixed hollow cylinder. This setup elegantly captures coupled rotation and translation and is well-suited for a rigorous Lagrangian analysis.
Let the larger hollow cylinder have inner radius R, fixed in space. Inside sits a smaller solid cylinder, radius r, mass m, and with moment of inertia I about its center G. The small cylinder is constrained to roll without slipping inside the larger one.
We describe the configuration with two angular coordinates. The first is \theta, the angle between the vertical and the line OG joining the center O of the fixed cylinder and the moving center G of the small cylinder. The second is \alpha, which measures the rotation of the small cylinder about its axis—that is, if \dot\alpha is the angular velocity of the cylinder about G relative to the rotating OG, then its total angular velocity in the inertial frame is \Omega_s = \dot \theta + \dot\alpha. This separation is especially useful for systems where rolling occurs on a moving reference.
The kinetic energy T contains two contributions. The center of mass G moves on a circle of radius R-r about O with speed (R-r)\dot\theta, so there is a translational part \frac{1}{2}m(R-r)^2\dot\theta^2. Additionally, the small cylinder rotates about its own center with angular velocity \Omega_s, yielding a rotational part \frac{1}{2}I(\dot\theta+\dot\alpha)^2. Thus the kinetic energy takes the form
T = \frac{1}{2}m(R-r)^2\dot\theta^2 + \frac{1}{2}I(\dot\theta + \dot\alpha)^2
Gravitational potential energy V is naturally referenced with V=0 when the centers O and G are horizontally aligned (i.e., when \theta = \pm \frac{\pi}{2}). The center G lies a vertical distance (R-r)\cos(\theta) below O, yielding:
V = -mg(R-r)\cos(\theta)
A key ingredient is the rolling without slipping constraint. The contact point P (at the interface) must have zero velocity in the lab frame. Analyzing velocities, one finds:
(R-r)\dot\theta + r(\dot\theta + \dot\alpha) = 0
which simplifies to:
R\dot\theta + r\dot\alpha = 0 And therefore:
\dot\alpha = -\frac{R}{r}\dot\theta
The absolute angular velocity becomes:
\Omega_s = \dot\theta + \dot\alpha = \frac{r-R}{r} \dot\theta
Plug this constraint into the energies, expressing everything in terms of the single variable \theta. The kinetic energy reduces to:
T = \frac{1}{2}m(R-r)^2\dot\theta^2 + \frac{1}{2}I\left(\frac{r - R}{r}\right)^2\dot\theta^2
or, factoring,
T = \frac{1}{2}\left[m(R - r)^2 + I \left(\frac{R - r}{r}\right)^2\right]\dot\theta^2
The Lagrangian is then:
\mathcal L = T - V = \frac{1}{2}\left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\dot\theta^2 + mg(R - r)\cos(\theta)
We may now derive the equation of motion. The Euler-Lagrange equation in \theta is:
\frac{d}{dt}\left(\frac{\partial \mathcal L}{\partial \dot\theta}\right) - \frac{\partial \mathcal L}{\partial \theta} = 0
A direct calculation yields:
\frac{\partial \mathcal L}{\partial \dot\theta} = \left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\dot\theta
and:
\frac{\partial \mathcal L}{\partial \theta} = -mg(R - r)\sin(\theta)
As a result, the equation of motion governing the system reads:
\left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\ddot\theta + mg(R - r)\sin(\theta) = 0
Linearizing for small displacements, i.e., for |\theta| \ll 1 so that \sin(\theta) \approx \theta, the motion becomes simple harmonic:
\left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\ddot\theta + mg(R - r)\theta = 0
This equation shows the effective moment of inertia and “restoring constant.” The angular frequency for small oscillations is then
\omega_0 = \sqrt{ \frac{mg(R - r)}{m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2} }
and the corresponding frequency is
f = \frac{1}{2\pi}\sqrt{ \frac{mg(R - r)}{m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2} }
Specializing to the case of a homogeneous solid cylinder with I = \frac{1}{2}mr^2, a simple calculation gives
I\left(\frac{R - r}{r}\right)^2 = \frac{1}{2}m(R - r)^2
so the denominator simplifies to:
m(R - r)^2 + \frac{1}{2}m(R - r)^2 = \frac{3}{2}m(R - r)^2
and:
f = \frac{1}{2\pi}\sqrt{\frac{g}{\frac{3}{2}(R - r)}} = \frac{1}{2\pi}\sqrt{\frac{2g}{3(R - r)}}
Pivoted bar with a moving support
Consider a bar of mass m and length L, pivoted at point A and free to rotate in the vertical plane. The pivot A is subjected to a known horizontal trajectory s(t). The center of mass G is located a distance L from A (for a uniform bar, this would be L = \ell/2 with \ell the bar length). The bar’s moment of inertia about G is I_G. Gravity g acts vertically downward. The configuration is parameterized by \theta, defined as the angle between the bar and the downward vertical at A (so \theta=0 when hanging straight down).
The system’s dynamics are captured by the Lagrangian \mathcal L = T - V, where T is the kinetic energy and V is the potential energy.
The vertical position of G relative to A is L\cos(\theta), so the gravitational potential energy is:
V = -mgL\cos(\theta)
The velocity of G combines the imposed motion of A and the contribution from rotation about A:
Horizontal position:
\begin{aligned} x_G & = s(t) + L\sin(\theta) \\ y_G & = L\cos(\theta) \end{aligned}
Velocities (by differentiation):
\begin{aligned} v_{Gx} & = \dot{s} + L\dot{\theta}\cos(\theta)\\ v_{Gy} & = -L\dot{\theta}\sin(\theta) \end{aligned}
The squared speed of G:
\begin{aligned} v_G^2 & = (v_{Gx})^2 + (v_{Gy})^2 = \dot{s}^2 + 2\dot{s}L\dot{\theta}\cos(\theta) + L^2\dot{\theta}^2(\cos^2\theta + \sin^2\theta) \\ & = \dot{s}^2 + 2\dot{s}L\dot{\theta}\cos(\theta) + L^2\dot{\theta}^2 \end{aligned}
Therefore the kinetic energy is:
T = \frac{1}{2}m v_G^2 + \frac{1}{2}I_G \dot{\theta}^2 = \frac{1}{2}m\dot{s}^2 + mL\dot{s}\dot{\theta}\cos(\theta) + \frac{1}{2}\left(mL^2 + I_G\right)\dot{\theta}^2
The total Lagrangian becomes:
\mathcal L = \frac{1}{2}m\dot{s}^2 + mL\dot{s}\dot{\theta}\cos(\theta) + \frac{1}{2}(mL^2 + I_G)\dot{\theta}^2 + mgL\cos(\theta)
\dot{s}(t) is prescribed; \theta(t) is the generalized coordinate. The Euler-Lagrange equation reads:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal L}{\partial \theta} = 0
Partial derivatives yield:
\frac{\partial \mathcal L}{\partial \dot{\theta}} = (mL^2 + I_G)\dot{\theta} + mL\dot{s}\cos(\theta)
Taking the time derivative:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) = (mL^2 + I_G)\ddot{\theta} + mL\ddot{s}\cos(\theta) - mL\dot{s}\dot{\theta}\sin(\theta)
The partial derivative with respect to \theta:
\frac{\partial \mathcal L}{\partial \theta} = -mL\dot{s}\dot{\theta}\sin(\theta) - mgL\sin(\theta)
The equation of motion becomes:
(mL^2 + I_G)\ddot{\theta} + mL\ddot{s}\cos(\theta) + mgL\sin(\theta) = 0