3 Momentum

3.1 Momentum and Impulse

The momentum of a particle is the product of its mass and velocity, given by

$$\begin{align*} \vec{p} = m\vec{v}. \end{align*}$$

According to Newton's second law,

$$\begin{align*} \vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}, \end{align*}$$

indicating the total change in a particle's momentum equals the integral over time of the force applied. This concept introduces impulse and average force:

• Impulse: the change of momentum

$$\begin{align*} \vec{J} = \Delta\vec{p} = \vec{p}(t_2) - \vec{p}(t_1) = \int_{t_1}^{t_2} \vec{F} \,\mathrm{d}t. \end{align*}$$

• Average Force:

$$\begin{align*} \bar{\vec{F}} = m\frac{\Delta\vec{v}}{\Delta t} = \frac{\Delta\vec{p}}{\Delta t}. \end{align*}$$

3.2 System of Particles and Conservation of Momentum

Conservation of Momentum

For a system of particles, when the net external force equals to zero ($\vec{F}^{ext} = 0$), the momentum of the system becomes a constant

$$\begin{align*} \vec{F}^{ext} = 0 \Rightarrow \vec{p}_{sys} = \text{const}. \end{align*}$$

The law works in inertial reference frame.

3.3 Applications of Momentum in Aerospace Engineering

Rocket Dynamics

• rocket velocity, $\vec{v}(t) = v\vec{i}$,

• fuel ejection velocity relative to the rocket, $\vec{u} = -u\vec{i}$, with $u > 0$,

• time-dependent rocket mass, $m_r(t)$,

• mass of fuel ejected at time $t$, $\mathrm{d}m_f$,

• external force, $\vec{F}_{ext}$.

$$\begin{align*} F_{ext} &= \frac{\mathrm{d}p_{sys}(t)}{\mathrm{d}t} \\ &= \frac{p_{sys}(t + \mathrm{d}t) - p_{sys}(t)}{\mathrm{d}t} \\ &= m_r(t)\frac{\mathrm{d}v(t)}{\mathrm{d}t} - u\frac{\mathrm{d}\mathrm{d}m_f}{\mathrm{d}t} \\ &= m_r(t)\frac{\mathrm{d}v(t)}{\mathrm{d}t} + u\frac{\mathrm{d}m_r}{\mathrm{d}t}, \end{align*}$$

Gravity-free Space

$$\begin{align*} -\frac{\mathrm{d}m_r}{\mathrm{d}t}u = m_r(t)\frac{\mathrm{d}v(t)}{\mathrm{d}t}. \end{align*}$$ $$\begin{align*} v_f = v_i + u\ln\left(\frac{m_{r,i}}{m_{r,f}}\right). \end{align*}$$

Constant Gravitational Field

$$\begin{align*} v_r(t) = -u \ln\left(\frac{m_r(t)}{m_{r,i}}\right) - gt = u \ln\left(\frac{m_{r,i}}{m_r(t)}\right) - gt. \end{align*}$$

3.4 Center of Mass

Center of Mass

For a system of $N$ particles labeled by indices $i = 1, 2, 3, \cdots, N$, choose a coordinate system where the position of the $i^{\text{th}}$ particle as $\vec{r}_i$. The mass of the system is given by the sum

$$\begin{align*} m_{sys} = \sum_{i=1}^{N} m_i, \end{align*}$$

and the position of the center of mass of the particle system is given by

$$\begin{align*} \vec{r}_c = \frac{1}{m_{sys}} \sum_{i=1}^{N} m_i\vec{r}_i. \end{align*}$$

For a continuous body, each point-like particle has mass $\mathrm{d}m$ and is located at the position $\vec{r}$. The center of mass is then defined as an integral over the body,

$$\begin{align*} \vec{r}_c = \frac{\int_{\text{body}} \mathrm{d}m\,\vec{r}}{\int_{\text{body}} \mathrm{d}m}. \end{align*}$$

Translational Motion of the Center of Mass

• Velocity of the center of mass

$$\begin{align*} \vec{v}_c = \frac{\mathrm{d}\vec{r}_c}{\mathrm{d}t} = \frac{\mathrm{d}\sum_{i=1}^{N} m_i\vec{r}_i}{\mathrm{d}t\sum_{i=1}^{N} m_i} = \frac{\sum_{i=1}^{N} m_i\vec{v}_i}{m_{sys}}. \end{align*}$$

• Momentum. In any reference frame, the momentum of the the center of mass is equal to the momentum of the whole particle system.

$$\begin{align*} \vec{P}_c = m_{sys}\vec{v}_c = \sum_{i=1}^{N} m_i\vec{v}_i = \sum_{i=1}^{N} \vec{p}_i = \vec{P}_{sys}. \end{align*}$$

• Acceleration

$$\begin{align*} \vec{a}_c = \frac{\mathrm{d}\vec{v}_c}{\mathrm{d}t} = \frac{\mathrm{d}\sum_{i=1}^{N} m_i\vec{v}_i}{\mathrm{d}t\sum_{i=1}^{N} m_i} = \frac{\sum_{i=1}^{N} m_i\vec{a}_i}{m_{sys}}. \end{align*}$$

• External force: Newton's second law to the center of mass

$$\begin{align*} \vec{F}^{ext} = \frac{\mathrm{d}\vec{P}_c}{\mathrm{d}t} = m_{sys}\frac{\mathrm{d}\vec{v}_c}{\mathrm{d}t} = m_{sys}\vec{a}_c.\ \end{align*}$$

Center-of-mass Reference Frame

In the COM reference frame, the aggregate position and velocity of the system's COM are effectively nullified, which is mathematically represented as follows:

$$\begin{align*} \vec{r}_c &= \sum_{i=1}^{N} m_i\vec{r}_i = 0, \\ \vec{v}_c &= \sum_{i=1}^{N} m_i\vec{v}_i = 0. \end{align*}$$

Here, $m_i$ represents the mass, $\vec{r}_i$ the position vector, and $\vec{v}_i$ the velocity vector of the $i^{th}$ particle, with the summation extending over all $N$ particles in the system.

This condition also leads to the total momentum of the system in the COM reference frame being zero:

$$\begin{align*} \vec{P}_c = \vec{P}_{sys} = 0. \end{align*}$$

Two-body Problem

The two-body problem is better to be solved in center-of-mass reference frame. Consider two point particles with masses $m_1$ and $m_2$, and the forces on them are $\vec{F}_1$ and $\vec{F}_2$ respectively. The motion of center of mass can be easily derived with the external force

$$\begin{align*} \vec{F}^{ext} = \vec{F}_1 + \vec{F}_2. \end{align*}$$

In center-of-mass reference frame, we need to take consideration of the effective force

$$\begin{align*} \vec{F}_1^{\text{eff}} &= -m_1 a_c = -m_1 \frac{\vec{F}^{ext}}{m_c} = -\frac{m_1}{m_1 + m_2}(\vec{F}_1 + \vec{F}_2), \\ \vec{F}_2^{\text{eff}} &= -m_2 a_c = -m_2 \frac{\vec{F}^{ext}}{m_c} = -\frac{m_2}{m_1 + m_2}(\vec{F}_1 + \vec{F}_2). \end{align*}$$

In this case, we can write Newton's second law

$$\begin{align*} \vec{F}_1' &= \vec{F}_1 + \vec{F}_1^{\text{eff}} = \frac{m_2\vec{F}_1 - m_1\vec{F}_2}{m_1 + m_2} = m_1\vec{a}_1', \\ \vec{F}_2' &= \vec{F}_2 + \vec{F}_2^{\text{eff}} = \frac{m_1\vec{F}_2 - m_2\vec{F}_1}{m_1 + m_2} = m_2\vec{a}_2'. \end{align*}$$

Then if we define

$$\begin{align*} \vec{r}' &\equiv \vec{r}_1' - \vec{r}_2', \\ \vec{v}' &\equiv \vec{v}_1' - \vec{v}_2', \\ \vec{a}' &\equiv \vec{a}_1' - \vec{a}_2', \\ \vec{f} &\equiv \vec{F}_1' = -\vec{F}_2', \end{align*}$$

we can calculate

$$\begin{align*} \vec{a}' \equiv \vec{a}_1' - \vec{a}_2' = \frac{\vec{f}}{m_1} - \frac{-\vec{f}}{m_2} = \left(\frac{1}{m_1} + \frac{1}{m_2}\right)\vec{f}. \end{align*}$$

In this case, we may decompose the collective motion of particles 1 and 2 to the motion of COM and the relative motion of a "reduced" single particle by defining reduced mass

$$\begin{align*} \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}, \end{align*}$$

then one can figure out,

$$\begin{align*} \vec{f} = m_1\vec{a}_1' = -m_2\vec{a}_2' = \mu\vec{a}, \end{align*}$$

where $\vec{f}$ is the force that Particle 2 acts on Particle 1. Using this equation, the two-body problem in center-of-mass reference frame can be solved using single-body problem methods.