Consider a two degree of freedom system as shown in the figure, where PQ is a rigid uniform rod of length, 𝒃 and mass, 𝒎.

Assume that the spring deflects only horizontally and force F is applied horizontally at Q. For this system, the Lagrangian, L is

Step-by-Step Derivation and Analysis:

Based on the system depicted in the diagram:
- There is a cart of mass M moving horizontally with displacement x, attached to a spring of stiffness k.
- A uniform rigid rod PQ of mass m and length b is hinged to the cart at point P.
- The angular displacement of the rod with the vertical is θ.
- The acceleration due to gravity g acts vertically downwards.

1. Kinetic Energy of the System (T):

The total kinetic energy of the system is the sum of the kinetic energy of the cart ($T_M$) and the kinetic energy of the uniform rod ($T_{\mathrm{rod}}$).

The kinetic energy of the cart is:

$T_M=\frac{1}{2}M{\dot{x}}^2$

To find the kinetic energy of the uniform rod, let us consider an infinitesimal mass element dm at a distance y from the pivot point P along the rod (where 0 ≤ y ≤ b). Since the rod is uniform:

$dm=\frac{m}{b}dy$

The position coordinates of this element are:

$x_y=x+y\sin \theta$

$y_y=-y\cos \theta$

Differentiating these coordinates with respect to time gives the velocity components of the mass element:

${\dot{x}}_y=\dot{x}+y\dot{\theta }\cos \theta$

${\dot{y}}_y=y\dot{\theta }\sin \theta$

The square of the velocity of the element is:

$v_y^2={\dot{x}}_y^2+{\dot{y}}_y^2={(\dot{x}+y\dot{\theta }\cos \theta )}^2+{\left(y\dot{\theta }\sin \theta \right)}^2$

Expanding and simplifying using the identity ${\sin }^2\theta +{\cos }^2\theta =1$:

$v_y^2={\dot{x}}^2+y^2{\dot{\theta }}^2+2y\dot{x}\dot{\theta }\cos \theta$

The kinetic energy of the rod is found by integration:

$T_{\mathrm{rod}}={\int }_0^b\frac{1}{2}dm{v}_y^2=\frac{m}{2b}{\int }_0^b({\dot{x}}^2+y^2{\dot{\theta }}^2+2y\dot{x}\dot{\theta }\cos \theta )dy$

Integrating with respect to y over the limits 0 to b:

$T_{\mathrm{rod}}=\frac{m}{2b}{[{\dot{x}}^{2}y+\frac{y^3}{3}{\dot{\theta }}^2+y^2\dot{x}\dot{\theta }\cos \theta ]}_0^b$

$T_{\mathrm{rod}}=\frac{1}{2}m{\dot{x}}^2+\frac{1}{6}m{b}^2{\dot{\theta }}^2+\frac{1}{2}mb\dot{x}\dot{\theta }\cos \theta$

Combining the kinetic energy of the cart and the rod yields the total kinetic energy of the system:

$T=\frac{1}{2}(M+m){\dot{x}}^2+\frac{1}{2}mb\dot{\theta }\dot{x}\cos \theta +\frac{1}{6}m{b}^2{\dot{\theta }}^2$

2. Potential Energy of the System (V):

The total potential energy is the sum of the potential energy stored in the spring and the gravitational potential energy of the rod.

Using the line of the guide track (hinge point P) as the reference line for zero potential energy:
- The potential energy of the spring is:

$V_{\mathrm{spring}}=\frac{1}{2}k{x}^2$
- The center of mass of the uniform rod is at a distance of b/2 from the pivot P. Its vertical height relative to the reference line is:

$y_{\mathrm{cm}}=-\frac{b}{2}\cos \theta$
- The gravitational potential energy of the rod is:

$V_{\mathrm{rod}}=-mg\frac{b}{2}\cos \theta$

Thus, the total potential energy is:

$V=\frac{1}{2}k{x}^2-mg\frac{b}{2}\cos \theta$

3. Formulation of the Lagrangian (L):

The Lagrangian is defined as the difference between the total kinetic and potential energy:

$L=T-V$

Substituting the expressions for T and V:

$L=\left[\frac{1}{2}\right(M+m){\dot{x}}^2+\frac{1}{2}mb\dot{\theta }\dot{x}\cos \theta +\frac{1}{6}m{b}^2{\dot{\theta }}^2]-[\frac{1}{2}k{x}^2-mg\frac{b}{2}\cos \theta ]$

Simplifying:

$L=\frac{1}{2}(M+m){\dot{x}}^2+\frac{1}{2}mb\dot{\theta }\dot{x}\cos \theta +\frac{1}{6}m{b}^2{\dot{\theta }}^2-\frac{1}{2}k{x}^2+mg\frac{b}{2}\cos \theta$