\[\newcommand{\u}[1]{\boldsymbol{\mathsf{#1}}} \renewcommand{\b}[1]{\boldsymbol{#1}} \newcommand{\t}[1]{\textsf{#1}} \newcommand{\m}[1]{\mathbb{#1}} \def\RR{\bf R} \def\bold#1{\bf #1} \def\mbf#1{\mathbf #1} \def\uv#1{\hat{\usf {#1}}} \def\dl#1{\underline{\underline{#1}}} \newcommand{\usf}[1]{\boldsymbol{\mathsf{#1}}} \def\bs#1{\usf #1}\]

Kinetics: Simple Applications

The force experienced by a body moving through a fluid due to its rotation is called Magnus force.

\(\begin{equation} \bs{F}_{\mathcal{M}}=m B \bs{\omega}_t\times \bar{\bs{v}},\\ \end{equation}\) where $B$ is a positive constant.

Baseball, soccer, or a frisbee

Let’s recall the equations of motion: \(\begin{align} \bs{F}&=\dot{\bs{G}}_t,\\ \bs{M}&=\dot{\bs{H}}_t,\\ \end{align}\) where $\bs{F}$ is the total force acting on the body, $\bs{M}$ is the total moment acting on the body, $\bs{G}_t$ is linear momentum of the solid, and $\bs{H}_t$ is the angular momentum of the solid.

After the ball has been hit, kicked, and the frisbee has been thrown, the net moment acting on the solids is zero. The interaction of the solid with the surrounding air is modeled using the Magnus force.

We take that the $t=0$ at the solid’s center of mass is at the origin. We take that the gravitational force is in the $-\bs{E}_2$ direction. Thus, the total gravitational force on the solid is $-m g \bs{E}_2$, where $m$ is the solid’s total mass, $g$ is acceleration due to gravity. The rate of change of linear momentum is

\[\begin{equation} \dot{\bs{G}}_t=m \dot{\bar{\bs{v}}}_t \end{equation}\]

Cricket, Baseball, and Soccer

We assume that initially, the velocity of the center of mass in $(v_1)_0\uv{E}_1+(v_2)_0\uv{E}_2$. In cricket, the spin of the ball can be modeled as the ball being provided with an initial angular velocity of $(\omega_3)_0\uv{E}_3$, where $(\omega_3)_0$ can be either positive or negative and has units of radians/second. The equations of motion in this case simplify to,

\[\begin{align} m \dot{\bar{\bs{v}}}&=-m g\uv{E}_2+m B (\omega_3)_0 \bs{E}_3\times \bar{\bs{v}}\\ \dot{\bar{\bs{v}}}&= -g\uv{E}_2+ B (\omega_3)_0 \bs{E}_3\times \bar{\bs{v}} \end{align}\]

Let $\bar{x}_i$ be the components of the position vector of the center of mass in the current configuration with respect to the basis $\uv{E}_i$. In terms of $\bar{x}_i$, the velocity of the center of mass is $ \dot{\bar{x}}_i\uv{E}_i$. In terms of the components $\bar{x}_i$, the last equation reads, \(\begin{align} \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= -g\uv{E}_2+ B (\omega_3)_0 \bs{E}_3\times (\dot{\bar{x}}_1\uv{E}_1+\dot{\bar{x}}_2\uv{E}_2+\dot{\bar{x}}_3\uv{E}_3)\\ \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= -g\uv{E}_2+ B (\omega_3)_0 (\dot{\bar{x}}_1\uv{E}_2-\dot{\bar{x}}_2\uv{E}_1)\\ \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= (-g +B (\omega_3)_0 \dot{\bar{x}}_1)\uv{E}_2- B (\omega_3)_0 \dot{\bar{x}}_2\uv{E}_1 \end{align}\)

Taking the dot product on both sides of the above equation w.r.t $\uv{E}_i$ we get,

\[\begin{align} \ddot{\bar{x}}_3 &=0\\ \ddot{\bar{x}}_2&=(-g +B (\omega_3)_0 \dot{\bar{x}}_1)\\ \ddot{\bar{x}}_1&=- B (\omega_3)_0 \dot{\bar{x}}_2 \end{align}\]

From the first of the above set of equations we have that $x_3=constant$.

Mathematica files

• Numerical solution of the equations of motion

Baseball, Soccer, and Frisbee

We assume that initially, the velocity of the center of mass in $(v_1)_0\uv{E}_1+(v_2)_0\uv{E}_2$. In a frisbee throw, the spin can be modeled as the frisbee being provided with an initial angular velocity of $(\omega_2)_0\uv{E}_2$, where $(\omega_2)_0$ can be either positive or negative and has units of radians/second. The equations of motion in this case simplify to,

\[\begin{align} m \dot{\bar{\bs{v}}}&=-m g\uv{E}_2+m B (\omega_2)_0 \bs{E}_2\times \bar{\bs{v}}\\ \dot{\bar{\bs{v}}}&= -g\uv{E}_2+ B (\omega_2)_0 \bs{E}_2\times \bar{\bs{v}} \end{align}\]

In terms of the components $\bar{x}_i$, the last equation reads, \(\begin{align} \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= -g\uv{E}_2+ B (\omega_2)_0 \bs{E}_2\times (\dot{\bar{x}}_1\uv{E}_1+\dot{\bar{x}}_2\uv{E}_2+\dot{\bar{x}}_3\uv{E}_3)\\ \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= -g\uv{E}_2+ B (\omega_2)_0 (-\dot{\bar{x}}_1\uv{E}_3+\dot{\bar{x}}_3\uv{E}_1)\\ \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 \\ \ddot{\bar{x}}_1\uv{E}_1+ \ddot{\bar{x}}_2\uv{E}_2+ \ddot{\bar{x}}_3\uv{E}_3 &= B (\omega_2)_0 \dot{\bar{x}}_3\uv{E}_1 -g\uv{E}_2 -B (\omega_2)_0 \dot{\bar{x}}_1\uv{E}_3+\\ \end{align}\)

Taking the dot product on both sides of the above equation w.r.t $\uv{E}_i$, we get,

\[\begin{align} \ddot{\bar{x}}_3 &=-B (\omega_2)_0 \dot{\bar{x}}_1\\ \ddot{\bar{x}}_2&=-g\\ \ddot{\bar{x}}_1&=B (\omega_2)_0 \dot{\bar{x}}_3 \end{align}\]

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