Particle Dynamics

Rectilinear motion is motion along a straight line. Take \(\mathbf{E}_x\) to be along the particle’s line of motion. Then,

\begin{align} \mathbf{r} &= x\mathbf{E}_x + y\mathbf{E}_y + z\mathbf{E}_z \notag \\ \mathbf{v} &= \dot{x}\mathbf{E}_x \notag \\ \mathbf{a} &= \ddot{x}\mathbf{E}_x \notag \end{align}

If \(s\) is arc length, then \(\dot{s}\) and \(\dot{x}\) may not always be the same: they are only the same if the particle’s trajectory direction \(\mathbf{e}_x = \mathbf{E}_x\). Assuming that this is the case, we have:

\begin{align} \mathbf{a} &= \frac{\text{d}v}{\text{d}s}\frac{\text{d}s}{\text{d}t} \notag \\ a\mathbf{E}_x &= \frac{\text{d}v}{\text{d}s}v\mathbf{E}_x \notag \end{align}

Projecting this equation along \(\mathbf{E}_x\) to turn it into a scalar equation, we have:

\begin{align} \boxed{a\text{ d}s = v\text{ d}v} \end{align}

A particle is endowed with an inertial measure \(m\) called its mass. The linear momentum \(\mathbf{G}\) of a particle is defined to be the product of its mass and velocity:

\begin{align} \boxed{\mathbf{G} = m\mathbf{v}} \end{align}

Then, we can postulate the following balance law for particles:

\begin{align} \boxed{\mathbf{F} = \dot{\mathbf{G}}} \end{align}

where \(\mathbf{F}\) is the sum of all external forces acting on the particle. This law is known as Newton’s 2nd Law, Euler’s 1st Law (extended for rigid bodies), or the balance of linear momentum. Note that if the mass of a system is conserved, \(\dot{\mathbf{G}} = m\dot{\mathbf{v}} = m\mathbf{a}\), which gives rise to the \(\mathbf{F} = m\mathbf{a}\) form that is commonly seen.

Realize that linear momentum is conserved if \(\mathbf{F} = 0\), since this implies that \(\dot{\mathbf{G}}=0\).

If we write \(\mathbf{F} = m\mathbf{a}\) in the Cartesian basis \(\{\mathbf{E}_x, \mathbf{E}_y, \mathbf{E}_z\}\), we get:

\begin{align} F_x\mathbf{E}_x + F_y\mathbf{E}_y + F_z\mathbf{E}_z = m(a_x\mathbf{E}_x + a_y\mathbf{E}_y + a_z\mathbf{E}_z) \notag \end{align}

Projecting along each direction, we can get three scalar equations of motion (EOM). We can then solve the three differential equations they represent for the motion of the particle.

4 Steps to Solve Particle Kinetics

1. Kinematics: choose the system, basis, and write expressions for \(\mathbf{r}, \mathbf{v}, \mathbf{a}\) if they are easy.
2. Draw the free body diagram and write expressions for the forces if applicable.
3. Write the vector equation of the balance of linear momentum.
4. Project the vector equation and analyze for the solution.