Examples, exercises & solutions

We set up the equation of motion for the particles:

Add these two equations:

As expected, we see that the repelling mutual force has no effect on the center of mass. We can solve this equation, using the initial condition the $MV(0) = m_1v_1(0) + m_2 v_2(0) \rightarrow V(0) = \frac{mv_0 + 2mv_0}{m+2m} = v_0$

As the next step we calculate $R(t)$:

The initial condition is: $R(0) = \frac{m_1 x_1(0) + m_2 x_2(0)}{m_1+m_2} = \frac{1}{3}x_{10} + \frac{2}{3}x_{20}$.

This gives

The center of mass of two point masses is on the line connecting $m_1$ and $m_2$. We denote this line as the $x$-axis, with $m_1$ as the origin.

• The center of mass is than given by (with $m_1$ = 3kg, $m_2$ = 2kg, $x_1$=0 and $x_2 = x_1 + L$ = 0.5m):

• The reduced mass is given by:

This is a 1-dimensional problem.

• The velocity of the center of mass is:

• The reduced mass is given by:

• In the CM frame the velocities of the cars are:

Cart 1: mass $m_1$ = 2kg, velocity $v_1$ = 4m/s
Cart 2: mass $m_2$ = 3kg, velocity $v_2$ = -2m/s

• The total kinetic energy in the lab frame is

• The velocity of the center of mass is

• The total kinetic energy in the center-of-mass frame is

with

Thus

• The reduced mass is

The relative velocity is

The kinetic energy associated with the motion of the reduced mass (i.e. the kinetic energy in the CM frame) is:

as we expected.

The position of the center of mass is

where $[m]$ indicates that the unit is meters.

Note: $\hat{x}$ and $\hat{y}$ do not carry units!

• Velocity of the center of mass:
  $$\vec{V} = \frac{\sum_i m_i \vec{v}_i}{\sum_i m_i}$$

  (20)
  Since the velocities are all parallel to the $x$-axis, we can drop the vector notation. Substituting the data for mass and velocity, gives:

• Position of the center of mass:
  $$\vec{V} = \frac{d\vec{R}}{dt} \rightarrow \vec{R}(t) = 2vt \hat{x} +\vec{c}$$

  (22)
  At $t=0$ all particles at location $(0,y_0)$. Thus, we find

• Total angular momentum:

• Angular momentum associated with the center of mass:

which is indeed the same as the total angular momentum. This is in this case to be expected as the angular momentum seen from the CM frame is $\vec{L}' = 0$ as in the CM frame the position vector and momentum vector are parallel for all four particles.

We split the kinetic energy in the kinetic energy associated with the center of mass and the kinetic energy as seen from the CM frame:

Due to symmetry, the center of mass velocity is $V$.

In the CM frame, all particles rotate with $\omega$ and thus have a velocity of magnitude $v' = \omega R$. As all particles have the same mass, we have $M = 8m$. The kinetic energy is:

All nitrogen molecules feel gravity and have interaction with each other and with the wall of the container. If we write down the equation of motion for all molecules (labelled $i$) and the container we get:

with $\vec{F}_{molecule \; i \; on \; vessel \; wall}$ the force of molecule $i$ on the vessel wall and $\vec{F}_{ji}$ the force from molecule $j$ on molecule $i$. All these forces are internal forces and when summing over all particles (including the vessel) will cancel each other as they all obey N3.

Thus is we add the equations, we find:

On the left side, we recognize the total momentum which we can write in terms of the center of mass: $M_c \dot{\vec{x}_c} + \sum_i m_i \dot{\vec{x}_i} = M\vec{V}$.

And on the right hand side we see the total mass $M = M_c + \sum m_i$.

Thus, we conclude:

The entire container drops with acceleration $-g$.