Newtonian mechanics and equations of motion
After a first introduction to some concepts of computational physics,
and to the numerical solution of simple differential equations of first
order, we now turn to the analysis of dynamical situations. In classical
mechanics they are usually described by Newton's laws, connecting
forces with the acceleration of the body on which they act. This
typically leads to differential equations of second order.
In the following, we will analyse such situations with the example of
a cannon ball being shot under some angle with a fixed muzzle velocity.
Both enter as parameters fixing the initial conditions of the problem.
The subject of this lecture is also discussed in chapter 2, section 2
of Giordano & Nakanishi. The Runge-Kutta methods are discussed in the
appendices A.1 and A.2 there.
The simple physical problem and its analytical solution
One of the first problems in mechanics is the calculation of the
trajectory of a cannon ball, fired at some (fixed) angle θ
with some (fixed) muzzle velocity $v_0$. In general, the trajectory is
given by a differential equation of second order, namely
\[ m \ddot{\mathrm{\underline r}} =
\frac{\mathrm{d^2{\underline r}}}{{\mathrm d} t^2} =
{\mathrm{\underline F}}({\mathrm{\underline r}},
{\mathrm{\underline v}}, t)
\, , \]
where all vectors are underlined (here, two-dimensional vectors
with x- and y-components are sufficient) and where
F is the force acting on the cannon ball. Ignoring
for the moment effects such as air resistance, the force acting on
the cannon ball is gravity only:
\[ \mathrm{\underline{F}}_{\mathrm{grav}}
(\mathrm{\underline{r}}\, ,\mathrm{\underline{v}} \, , t)
=-mg \mathrm{\underline{e}}_y
= -mg \left( \begin{array}{c} 0\\1 \end{array} \right)\, . \]
Here, the unit vector in y-direction has been made explicit.
Due to this simple form of the force, the differential equations for the
x- and y-components decouple completely, such that
\[\begin{eqnarray*} m \ddot{x}\,\, &=& \,\, 0 \\
m \ddot{y}\,\, &=& \,\, -mg \, . \end{eqnarray*} \]
With $x_0=x(t_0=0)=0$ and $y_0=y(t_0=0)=0$ the solution therefore reads
\[ \begin{eqnarray*}
x(t)\,&=&\,x_0+ v_{0,x}t \,&=&\, v_0\, \cos\theta\,\cdot\,t \\
y(t)\,&=&\,y_0+ v_{0,x}t- \frac{g}{2}t^2\,&=&
\,v_0\, \sin\theta\,\cdot\,t-\frac{g}{2}t ^2
\end{eqnarray*} \]
This implies that the cannon ball will follow a parabolic trajectory,
flying with constant horizontal velocity
\[ v_x =v_0\cos\theta\,.\]
Including drag forces
Obviously, this simple picture is not entirely correct for a true
cannon ball fired in reality. The most prominent feature
distinguishing reality from the simplified picture above is the effect
of air resistance on the trajectory. This "drag force" can be
written as
\[ \mathrm{\underline{F}}_\mathrm{drag}
(\mathrm{\underline{r}}\,,\mathrm{\underline{v}}\, ,t)\,=\,
-B_\mathrm{drag}\,v^2\, \mathrm{\underline{e}}_v\, =\,
-B_\mathrm{drag}\,v^2
\left( \begin{array}{c} v_x/v \\ v_y/v \end{array} \right) \]
where the unit vector in direction of the velocity,
$\underline e_v$ has been replaced by its components,
and where $B_{\rm drag}$ is the drag coefficient, which we
take as a constant for the time being. In the homework assignment we
will take a closer look into the physics of air resistance, and we
will include the effect of varying air pressure due to
different altitudes.
Here, obviously, the velocity is given by
\[ v^2\,=\,v_x^2+v_y^2\]
Next →
Frank Krauss and Daniel Maitre
Last modified: Wed Jul 22 09:30:41 BST 2009