Chapter 7, Problem 7.50P

To determine

Identify the two modified Lagrange equations and solve for $\ddot{x},\ddot{y}\text{and}\lambda$ and verify the result using the Newtonian method also.

Answer to Problem 7.50P

Modified Lagrangian equations are $\frac{\partial L}{\partial x}+\lambda \frac{\partial f}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$, $\frac{\partial L}{\partial y}+\lambda \frac{\partial f}{\partial y}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)$ and required equations are $\ddot{x}=-\frac{m_2}{m_1+m_2}g$, $\ddot{y}=\frac{m_2}{m_1+m_2}g$ and $\lambda =-\frac{m_1{m}_2}{m_1+m_2}g$ and the results are verified using Newtonian method too.

Explanation of Solution

Draw the diagram of given arrangement.

Write the equation for the kinetic energy of the system.

    $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2$

Here, $T$ is the kinetic energy, $m_1$ is the mass, $\dot{x}$ is the velocity of 1^st mass, $m_2$ is the mass and $\dot{y}$ is the velocity of 2^nd mass.

Assume the potential energy at the height of table is zero. The only potential energy is due to the gravity.

    $U=-m_{2}gy$

Write the equation for the Lagrangian of the system.

    $L=T-U$

Rewrite the above equation by substituting the previous two equations.

    $\begin{array}{c}L=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2-\left(-m_{2}gy\right)\\ =\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2+m_{2}gy\end{array}$        (I)

Write the Euler-Lagrange equation with constraints.

    $\frac{\partial L}{\partial q_i}+\lambda \frac{\partial f}{\partial q_i}=\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)$

Here, $q_i$ is the generalized coordinate and ${\dot{q}}_i$ is the generalized velocity.

Write the two modified Lagrange equations.

    $\frac{\partial L}{\partial x}+\lambda \frac{\partial f}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$        (II)

  $\frac{\partial L}{\partial y}+\lambda \frac{\partial f}{\partial y}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)$        (III)

Write the constraint equation.

    $f=x+y$

Differentiate the above function with respect to x.

    $\begin{array}{c}\frac{\partial f}{\partial x}=\frac{\partial \left(x+y\right)}{\partial x}\\ =1\end{array}$        (IV)

Differentiate $f=x+y$ with respect to y.

    $\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial \left(x+y\right)}{\partial y}\\ =1\end{array}$        (V)

Differentiate equation (I) with respect to x.

    $\begin{array}{c}\frac{\partial L}{\partial x}=\frac{\partial \left(\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2+m_{2}gy\right)}{\partial x}\\ =0\end{array}$        (VI)

Differentiate equation (I) with respect to y.

    $\begin{array}{c}\frac{\partial L}{\partial y}=\frac{\partial \left(\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2+m_{2}gy\right)}{\partial y}\\ =m_{2}g\end{array}$        (VII)

Differentiate equation (I) with respect to $\dot{x}$.

    $\begin{array}{c}\frac{\partial L}{\partial \dot{x}}=\frac{\partial \left(\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2+m_{2}gy\right)}{\partial y}\\ =m_1\dot{x}\end{array}$        (VIII)

Differentiate the above equation with respect to t.

    $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)=\frac{d}{dt}\left(m_1\dot{x}\right)\\ =m_1\ddot{x}\end{array}$

Differentiate equation (I) with respect to $\dot{y}$.

    $\begin{array}{c}\frac{\partial L}{\partial \dot{y}}=\frac{\partial \left(\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2+m_{2}gy\right)}{\partial \dot{y}}\\ =m_2\dot{y}\end{array}$        (IX)

Differentiate the above equation with respect to t.

    $\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)=\frac{d}{dt}\left(m_1\dot{y}\right)\\ =m_2\ddot{y}\end{array}$

Rewrite equation (II) by substituting equations (VI) and (VIII).

    $\begin{array}{c}0+\lambda \left(1\right)=m_1\ddot{x}\\ m_1\ddot{x}=\lambda \end{array}$

Rewrite equation (III) by substituting equations (VII) and (IX).

    $\begin{array}{c}{m}_{2}g+\lambda \left(1\right)=m_2\ddot{y}\\ m_2\ddot{y}=m_{2}g+\lambda \end{array}$

$m_1\ddot{x}=\lambda$ and $m_2\ddot{y}=m_{2}g+\lambda$ are the modified Lagrange equations.

The function $f=x+y$ is a constant.

    $x+y=\text{constant}\Rightarrow \ddot{x}+\ddot{y}=0\Rightarrow \ddot{y}=-\ddot{x}$

Substitute $-\ddot{x}$ for $\ddot{y}$ and $m_1\ddot{x}$ for $\lambda$ in $m_2\ddot{y}=m_{2}g+\lambda$.

    $\begin{array}{c}{m}_2\left(-\ddot{x}\right)=m_{2}g+m_1\ddot{x}\\ -\ddot{x}\left(m_1+m_2\right)=m_{2}g\\ \ddot{x}=-\frac{m_2}{m_1+m_2}g\end{array}$

Substitute the above equation in $m_1\ddot{x}=\lambda$.

    $\begin{array}{c}{m}_1\left(-\frac{m_2}{m_1+m_2}g\right)=\lambda \\ \lambda =-\frac{m_1{m}_2}{m_1+m_2}g\end{array}$

Rewrite the equation

Substitute $-\frac{m_2}{m_1+m_2}g$ for $\ddot{y}=-\ddot{x}$

    $\begin{array}{c}\ddot{y}=-\left(-\frac{m_2}{m_1+m_2}g\right)\\ =\frac{m_2}{m_1+m_2}g\end{array}$

Thus, the required equations are $\ddot{x}=-\frac{m_2}{m_1+m_2}g$, $\ddot{y}=\frac{m_2}{m_1+m_2}g$ and $\lambda =-\frac{m_1{m}_2}{m_1+m_2}g$.

Write the equation for force on string from equation 7.122.

    $\lambda \frac{\partial f}{\partial x}=F^{\text{str}}$

Write the expression for constraint forces associated with the tension on string.

    $\begin{array}{c}\lambda \frac{\partial f}{\partial x}=F_x\\ \lambda \frac{\partial f}{\partial y}=F_y\end{array}$

Thus, the $F_x$ is the tension on mass $m_1$ and $F_y$ is the tension on mass $m_2$.

Rewrite the equation $\lambda \frac{\partial f}{\partial x}=F_x$ by substituting 1 for $\frac{\partial f}{\partial x}$ and $m_1\ddot{x}$ for $\lambda$.

    $\begin{array}{c}\left(m_1\ddot{x}\right)\left(1\right)=F_x\\ m_1\ddot{x}=F_x\end{array}$

Rewrite the above equation by substituting $-\frac{m_2}{m_1+m_2}g$ for $\ddot{x}$.

    $F_x=m_1\left(-\frac{m_2}{m_1+m_2}g\right)$

Thus, the magnitude of tension on mass $m_1$ is $\frac{m_1{m}_2}{m_1+m_2}g$.

Substitute 1 for $\frac{\partial f}{\partial y}$ and $m_1\ddot{x}$ for $\lambda$ in the equation $\lambda \frac{\partial f}{\partial y}=F_y$.

    $\left(m_1\ddot{x}\right)\left(1\right)=F_y$

Rewrite the above equation by substituting $-\frac{m_2}{m_1+m_2}g$ for $\ddot{x}$.

Thus, the magnitude of tension on mass $m_2$ is $\frac{m_1{m}_2}{m_1+m_2}g$.

Draw the free body diagram of the system.

Write the Newton’s equation for net force.

    $F_{\text{net}}=ma$

Here, $F_{\text{net}}$ is the net force.

Write the equation for net horizontal force on mass $m_1$.

    $F_{\text{net}}=T$

Here, $T$ is the tension.

Write the Newton’s equation for mass $m_1$.

    $F_{\text{net}}=m_{1}a$

Equate the right-hand sides of above two equations.

    $T=m_{1}a$

Write the net vertical force on mass $m_2$.

    $F_{\text{net}}=m_{2}g-T$

Write the Newton’s equation for mass $m_2$.

    $F_{\text{net}}=m_{2}a$

Equate the right-hand sides of above two equations.

    $m_{2}g-T=m_{2}a$

Substitute $m_{1}a$ for $T$ in the above equation.

    $\begin{array}{c}{m}_{2}g-m_{1}a=m_{2}a\\ a=\frac{m_2}{m_1+m_2}g\end{array}$

Substitute the above equation in $T=m_{1}a$.

    $T=\frac{m_1{m}_2}{m_1+m_2}g$

The result is same that of obtained from modified Lagrangian equations.

Thus, the modified Lagrangian equations are $\frac{\partial L}{\partial x}+\lambda \frac{\partial f}{\partial x}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)$, $\frac{\partial L}{\partial y}+\lambda \frac{\partial f}{\partial y}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{y}}\right)$ and required equations are $\ddot{x}=-\frac{m_2}{m_1+m_2}g$, $\ddot{y}=\frac{m_2}{m_1+m_2}g$ and $\lambda =-\frac{m_1{m}_2}{m_1+m_2}g$ and the results are verified using Newtonian method too.