Chapter 8, Problem 8.3P

(a)

To determine

Identify the positions of two particles at any instant t and explain about its motion.

Answer to Problem 8.3P

Equation to find the position of particles at any instant t are $r_1=L+\frac{m_2}{m_1+m_2}\left[\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right]+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2$ and $r_2=\frac{m_1}{m_1+m_2}{v}_{0}t-\frac{1}{2}g{t}^2-\left(\frac{m_1{v}_i}{m_1+m_2}\right)\sin \left(\omega t\right)$.

Explanation of Solution

Write the general expression for position coordinate of centre of mass of system.

    $R=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$        (I)

Here¸$m_1,m_2$ are the masses, $r_1$ is the position vector of 1^st mass, $r_2$ is the position vector of 2^nd mass and $R$ is the position vector of centre of mass.

Rearrange the above equation as follows.

  $\left(m_1+m_2\right)R=m_1{r}_1+m_2{r}_2$        (II)

Express $r_1$ in terms of $r_2$

    $r_1=r+r_2$

Here, $r$ is the displacement vector.

Rewrite equation (II) by adding $m_2{r}_1$ on both sides.

    $\begin{array}{c}\left(m_1+m_2\right)R+m_2{r}_1=m_1{r}_1+m_2{r}_2+m_2{r}_1\\ \left(m_1+m_2\right)R+m_2{r}_1-m_2{r}_2=m_1{r}_1+m_2{r}_1\\ \left(m_1+m_2\right)R+m_2\left(r_1-r_2\right)=\left(m_1+m_2\right)r_1\\ R+m_{2}r=\left(m_1+m_2\right)r_1\\ r_1=\frac{R+m_{2}r}{m_1+m_2}\end{array}$        (III)

Write the equation for linear velocity of 1^st mass from the above equation.

      ${\dot{r}}_1=\dot{R}+\frac{m_2}{m_1+m}\dot{r}$

Here, ${\dot{r}}_1$ is the linear velocity of 1^st mass.

Rewrite equation (II) by adding $m_1{r}_2$ on both sides.

    $\begin{array}{c}\left(m_1+m_2\right)R+m_1{r}_2=m_1{r}_1+m_2{r}_2+m_1{r}_2\\ \left(m_1+m_2\right)R+m_1{r}_2-m_1{r}_1=m_2{r}_2+m_1{r}_2\\ \left(m_1+m_2\right)R-m_1\left(r_1-r_2\right)=\left(m_1+m_2\right)r_2\\ R-m_{1}r=\left(m_1+m_2\right)r_2\\ r_2=\frac{\left(m_1+m_2\right)R-m_{1}r}{\left(m_1+m_2\right)}\\ =R-\frac{m_1}{\left(m_1+m_2\right)}r\end{array}$

Draw the diagram showing the spring- mass system.

Write the law of conservation of momentum for spring-mass system.

    $\left(m_1+m_2\right)v_f=m_1{v}_i$

Here, $m_1,m_2$ are the masses attached to the spring, $v_f$ is the final combined velocity pf two masses and $v_i$ is the initial velocity of mass $m_1$.

Rewrite the expression for $v_f$.

    $v_f=\frac{m_1{v}_i}{m_1+m_2}$        (IV)

Write the equation for position of initial centre of mass.

    $R_i=\frac{m_{i}L}{m_1+m_2}$        (V)

Write the equation of motion for coordinate of centre of mass.

    $R=R_i+v_{i}t+\frac{1}{2}\left(-g\right)t^2$

Rewrite the above equation by substituting equations (IV) and (V).

$R=\frac{m_{i}L}{m_1+m_2}+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2$

Write the Lagrange equation in relative coordinates.

    $L_{\text{rel}}=\frac{1}{2}\mu {\dot{r}}^2-\frac{1}{2}k{\left(r-L\right)}^2$

Replace $\left(r-L\right)$ with $r^{\prime }$.

    $L_{\text{rel}}=\frac{1}{2}\mu {\dot{r}}^2-\frac{1}{2}k{r^{\prime }}^2$

Write the Lagrange equation with respect to $r^{\prime }$.

    $\frac{d}{dt}\left(\frac{\partial L_{rel}}{\partial {\dot{r}}^{\prime }}\right)-\frac{\partial L_{rel}}{\partial r^{\prime }}=0$

Rewrite the above equation by substituting $\frac{1}{2}\mu {\dot{r}}^2-\frac{1}{2}k{r^{\prime }}^2$ for $L_{rel}$.

    $\begin{array}{c}\frac{d}{dt}\frac{\partial }{\partial {\dot{r}}^{\prime }}\left[\frac{1}{2}\mu {\dot{r}}^2-\frac{1}{2}k{r^{\prime }}^2\right]-\frac{\partial \left[\frac{1}{2}\mu {\dot{r}}^2-\frac{1}{2}k{r^{\prime }}^2\right]}{\partial r^{\prime }}=0\\ \frac{d}{dt}\left(\frac{2}{2}\mu \left({\dot{r}}^{\prime }\right)-0\right)-\left(0-\frac{2}{2}k\left(r^{\prime }\right)\right)=0\\ \mu \ddot{r}+k{r}^{\prime }=0\end{array}$

Write the solution for above equation.

    $r^{\prime }=A\sin \left(\omega t\right)+B\cos \left(\omega t\right)$

Replace $r^{\prime }$ by $r-L$

  $\begin{array}{c}r-L=A\sin \left(\omega t\right)+B\cos \left(\omega t\right)\\ r=L+A\sin \left(\omega t\right)+B\cos \left(\omega t\right)\end{array}$

At time t=0, r=L.

    $\begin{array}{c}L=L+A\sin \left(\omega \left(0\right)\right)+B\cos \left(\omega \left(0\right)\right)\\ B=0\end{array}$

Differentiate $r=L+A\sin \left(\omega t\right)+B\cos \left(\omega t\right)$ with respect to t.

    $\dot{r}=0+A\omega \cos \left(\omega t\right)+B\omega \cos \left(\omega t\right)$

At t=0, $\omega ={\omega }_0\text{and}\dot{r}=v_0$.

Rewrite the previous equation by substituting

    $\begin{array}{c}{v}_0=0+A{\omega }_0\cos \left({\omega }_0\left(0\right)\right)+B{\omega }_0\cos \left({\omega }_0\left(0\right)\right)\\ =A{\omega }_0\\ A=\frac{v_0}{{\omega }_0}\end{array}$

Rewrite the equation $r=L+A\sin \left(\omega t\right)+B\cos \left(\omega t\right)$ by substituting 0 for B and $\frac{v_0}{{\omega }_0}$ for $A$.

    $\begin{array}{c}r=L+\frac{v_0}{{\omega }_0}\sin \left(\omega t\right)+\left(0\right)\cos \left(\omega t\right)\\ =L+\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\end{array}$

Substitute $L+\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)$ for $r$ and $\frac{m_{i}L}{m_1+m_2}+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2$ for $R$ in equation (III).

  $\begin{array}{c}{r}_1=\frac{m_{1}L}{m_1+m_2}+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2+\frac{m_2}{m_1+m}\left(L+\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right)\\ =L\left(\frac{m_1}{m_1+m_2}+\frac{m_2}{m_1+m_2}\right)+\frac{m_2}{m_1+m_2}\left(\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right)+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2\\ =L+\frac{m_2}{m_1+m_2}\left[\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right]+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2\end{array}$

Replace $m_1+m_2$ with $M$. Write the equation to find the position of 2^nd mass after time t.

  $r_2=R-\frac{m_1}{\left(m_1+m_2\right)}r$

Substitute $L+\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)$ for $r$ and $\frac{m_{i}L}{m_1+m_2}+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2$ for $R$ in the above equation.

  $\begin{array}{c}{r}_2=\frac{m_{1}L}{m_1+m_2}+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2-\frac{m_1}{\left(m_1+m_2\right)}\left(L+\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right)\\ =-\frac{m_1}{m_1+m_2}\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2\\ =\frac{m_1}{m_1+m_2}{v}_{0}t-\frac{1}{2}g{t}^2-\left(\frac{m_1{v}_i}{m_1+m_2}\right)\sin \left(\omega t\right)\end{array}$

Conclusion:

Therefore, the equation to find the position of particles at any instant t are $r_1=L+\frac{m_2}{m_1+m_2}\left[\left(\frac{v_0}{{\omega }_0}\right)\sin \left(\omega t\right)\right]+\left(\frac{m_1{v}_i}{m_1+m_2}\right)t-\frac{1}{2}g{t}^2$ and $r_2=\frac{m_1}{m_1+m_2}{v}_{0}t-\frac{1}{2}g{t}^2-\left(\frac{m_1{v}_i}{m_1+m_2}\right)\sin \left(\omega t\right)$.