Two particles each of mass m are connected by an (inextensible) string of length I. One particle moves on a horizontal table (assume no friction), The string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line. Find the Lagrange equations of motion of the particles. Hint: Let the coordinates of the particle on the table be r and $\mathit{\theta }$, and let the coordinate of the other particle be z. Eliminate one variable from $\mathit{L}\left(usingr+\left|z\right|=I\right)$and write two Lagrange equations.

Two particles each of mass m are connected by an (inextensible) string of length I. One particle moves on a horizontal table (assume no friction), the string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line. The given equation is$\left(r+\left|z\right|=I\right)$.

The Lagrange equation is used to construct the equations of the motion of a solid mechanics issue in the matrix form, including damping.

For a particle on a flat surface, use polar coordinates, and the kinetic energy in polar coordinates have the following form:

$T_1=\frac{1}{2}\left(r^2+r^2{\varphi }^2\right)$

Add the kinetic energy of the second particle to the total kinetic energy. Since $r+\left|z\right|=I$is a constraint, take a derivative with respect to time to obtain$r=z$(assuming that the z direction is facing away from the table and that particle doesn't climb the table, which would induce a change in the sign in coordinate). From this, the kinetic energy of the hanging particle is calculated as:

$$\begin{gathered} T_2=\frac{1}{2}m{z}^2 \\ =\frac{1}{2}m{r}^2 \end{gathered}$$

So, the total kinetic energy is:

$$\begin{gathered} T=T_1+T_2 \\ =\frac{1}{2}m\left(\dot{2}{r}^2+r^{\dot{2}}{\varphi }^2\right) \end{gathered}$$

Now, for the potential energy, the particle on the table has a constant gravitational potential energy since it is on a flat surface, and in turn would produce a constant term in the Lagrangian equation (which we can simply ignore since taking derivatives in the Euler equations would eliminate a constant term). The potential energy of a hanging particle is:

$$\begin{gathered} V=-mgz \\ =-mg\left(I-r\right) \end{gathered}$$

Since the z-axis points in the bottom direction (away from the bottom of the table), the full Lagrangian equation is:

$$\begin{gathered} L=T-V \\ =\frac{1}{2}m\left(\ddot{2}{r}^2+r^{\dot{2}{\varphi }^2}\right)+mg\left(I-r\right) \end{gathered}$$

Observe the Euler equation fordegree of freedom. The Euler equations reads:

$\frac{d}{dt}\frac{\partial L}{\partial r}=0$.

First, calculate the required derivatives.

$$\begin{gathered} \frac{\partial L}{\partial r}=2\dot{mr}\frac{d}{dt}\frac{\partial L}{\partial r} \\ =2\ddot{m}r\frac{\partial L}{\partial r} \\ =m\dot{r}{\varphi }^2-mg \end{gathered}$$

Using all of the equations above, $\ddot{2}r-\dot{r}{\varphi }^2+g=0$.

Next, move onto thedegree of freedom. The Euler equations reads:

.

First, calculate the required derivatives.

Calculate the time derivative.Since it is a constant with respect to time (it follows from the Euler equation), and it is often written in that form (without taking the derivative with respect to time) because it represents a constant of motion—a quantity which doesn't change with time. So, afterdiving by the factorinsert the calculated derivatives into the Euler equation we obtain:

Therefore,the two Lagrange equations of motion of the particles areand.