Momentum

3.1 Momentum and Impulse

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

\[ \begin{aligned} \vec{p} = m\vec{v}. \end{aligned} \]

According to Newton’s second law,

\[ \begin{aligned} \vec{F} = \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}, \end{aligned} \]

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

\[ \begin{aligned} \vec{J} = \Delta\vec{p} = \vec{p}(t_2) - \vec{p}(t_1) = \int_{t_1}^{t_2} \vec{F} \,\mathrm{d}t. \end{aligned} \]

• Average Force:

\[ \begin{aligned} \bar{\vec{F}} = m\frac{\Delta\vec{v}}{\Delta t} = \frac{\Delta\vec{p}}{\Delta t}. \end{aligned} \]

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

\[ \begin{aligned} \vec{F}^{ext} = 0 \Rightarrow \vec{p}_{sys} = \text{const}. \end{aligned} \]

The law works in inertial reference frame.

3.3 Applications of Momentum in Aerospace Engineering

TipRocket 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, \(\vec{F}_{ext}\).

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

TipGravity-free Space

\[ \begin{aligned} -\frac{\mathrm{d}m_r}{\mathrm{d}t}u = m_r(t)\frac{\mathrm{d}v(t)}{\mathrm{d}t}. \end{aligned} \]

\[ \begin{aligned} v_f = v_i + u\ln\left(\frac{m_{r,i}}{m_{r,f}}\right). \end{aligned} \]

TipConstant Gravitational Field

\[ \begin{aligned} v_r(t) = -u \ln\left(\frac{m_r(t)}{m_{r,i}}\right) - gt = u \ln\left(\frac{m_{r,i}}{m_r(t)}\right) - gt. \end{aligned} \]

3.4 Center of Mass

TipCenter 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 \[ \begin{aligned} m_{sys} = \sum_{i=1}^{N} m_i, \end{aligned} \]

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

\[ \begin{aligned} \vec{r}_c = \frac{1}{m_{sys}} \sum_{i=1}^{N} m_i\vec{r}_i. \end{aligned} \]

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,

\[ \begin{aligned} \vec{r}_c = \frac{\int_{\text{body}} \mathrm{d}m\,\vec{r}}{\int_{\text{body}} \mathrm{d}m}. \end{aligned} \]

TipTranslational Motion of the Center of Mass

• Velocity of the center of mass

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

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

\[ \begin{aligned} \vec{P}_c = m_{sys}\vec{v}_c = \sum_{i=1}^{N} m_i\vec{v}_i = \sum_{i=1}^{N} \vec{p}_i = \vec{P}_{sys}. \end{aligned} \]

• Acceleration

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

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

\[ \begin{aligned} \vec{F}^{ext} = \frac{\mathrm{d}\vec{P}_c}{\mathrm{d}t} = m_{sys}\frac{\mathrm{d}\vec{v}_c}{\mathrm{d}t} = m_{sys}\vec{a}_c.\ \end{aligned} \]

TipCenter-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:

\[ \begin{aligned} \vec{r}_c &= \sum_{i=1}^{N} m_i\vec{r}_i = 0, \\ \vec{v}_c &= \sum_{i=1}^{N} m_i\vec{v}_i = 0. \end{aligned} \]

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:

\[ \begin{aligned} \vec{P}_c = \vec{P}_{sys} = 0. \end{aligned} \]

TipTwo-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

\[ \begin{aligned} \vec{F}^{ext} = \vec{F}_1 + \vec{F}_2. \end{aligned} \]

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

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

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

\[ \begin{aligned} \vec{F}_1' &= \vec{F}_1 + \vec{F}_1^{\text{eff}} = \frac{m_2\vec{F}_1 - m_1\vec{F}_2}{m_1 + m_2} = m_1\vec{a}_1', \\ \vec{F}_2' &= \vec{F}_2 + \vec{F}_2^{\text{eff}} = \frac{m_1\vec{F}_2 - m_2\vec{F}_1}{m_1 + m_2} = m_2\vec{a}_2'. \end{aligned} \]

Then if we define

\[ \begin{aligned} \vec{r}' &\equiv \vec{r}_1' - \vec{r}_2', \\ \vec{v}' &\equiv \vec{v}_1' - \vec{v}_2', \\ \vec{a}' &\equiv \vec{a}_1' - \vec{a}_2', \\ \vec{f} &\equiv \vec{F}_1' = -\vec{F}_2', \end{aligned} \]

we can calculate

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

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

\[ \begin{aligned} \frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2}, \end{aligned} \]

then one can figure out,

\[ \begin{aligned} \vec{f} = m_1\vec{a}_1' = -m_2\vec{a}_2' = \mu\vec{a}, \end{aligned} \]

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.