Particle Kinetics

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.

Consider a particle with position vector \(\mathbf{r}\) with a force \(F\) applied on it. The moment of \(\mathbf{F}\) about some point \(P\) is defined as:

\begin{align} \mathbf{M}^P = (\mathbf{r}-\mathbf{r}_p)\times\mathbf{F} \end{align}

Now, say this particle has mass \(m\) and velocity \(\mathbf{v}\). The angular momentum of \(m\) about some origin \(O\) is the moment of its linear momentum:

\begin{align} \mathbf{H}^O = \mathbf{r}\times m\mathbf{v} \end{align}

Finally, taking the moment of both sides of the balance of linear momentum, we get the balance of angular momentum:

\begin{align} \boxed{\mathbf{M}^O = \dot{\mathbf{H}}^O} \end{align}

The power of a force \(\mathbf{F}\) acting on a particle of mass \(m\) traveling with a velocity \(\mathbf{v}\) is defined as:

\begin{align} P = \mathbf{F} \cdot \mathbf{v} \end{align}

The work of a force \(\mathbf{F}\) in the interval \([t_1, t_2]\) is the integral of power with respect to time:

\begin{align} W_{\mathbf{F}, 12} = \int_{t_1}^{t_2}\mathbf{F}\cdot\mathbf{v}\text{ d}t = \int_{\mathbf{r}(t_1)}^{\mathbf{r}(t_2)} \mathbf{F} \cdot \text{d}\mathbf{r} \end{align}

The kinetic energy of a particle is defined to be:

\begin{align} T = \frac{1}{2}m\mathbf{v}\cdot\mathbf{v} \end{align}

Consider the collision of two particles. We make the following assumptions:

• no rotation (angular momentum is conserved)
• no change in mass
• the collision time (contact time) is extremely short
• impact is frictionless, so all collision forces are along the normal (linear momentum is conserved along the tangent)

For particles, we also define a coefficient of restitution that measures how elastic the collision is, where \(e=1\) is fully elastic:

\begin{align} e=-\frac{\mathbf{v}_1'\cdot\mathbf{n} - \mathbf{v}_2'\cdot\mathbf{n}}{\mathbf{v}_1\cdot\mathbf{n}-\mathbf{v}_2\cdot\mathbf{n}} \end{align}

where \(\mathbf{v}'\) is the velocity after collision.