Simple Dynamic Problems, Classical Mechanics

Question

I am struggling with some revision I am doing and I would really appreciate your help.

A particle of mass $m$ is projected with an initial speed $u_{0}$ at an angle $\frac{π}{3}$ to the horizontal.

It experiences a gravitational force $mg$ vertically downwards and a air-resistance force $−mc\textbf{v}$, where $\textbf{v}$ is the velocity of the particle.

Choose an inertial frame with origin $O$ at the initial position of the particle, with $\textbf{k}$ vertically up and $\textbf{i}$ horizontal in the plane of the motion, so that the position vector of the particle can be written as: $$r = x(t)\textbf{i} + z(t)\textbf{k}$$

Hence, show that: $$\ddot{x}=-c\dot{x}$$ $$\ddot{z}=-g-c\dot{z}$$

And deduce that: $$x(t)=\frac{u_{0}}{2c}(1-e^{-ct})$$ $$z(t)=\frac{1}{c^2}\bigg(g+\frac{cu_{0}\sqrt{3}}{2}\bigg)(1-e^{-ct})-\frac{g}{c}t$$

Thus, find: $$\lim_{t\rightarrow\infty}x(t)$$ $$\lim_{t\rightarrow\infty}z(t)$$ And what does this tell us about the trajectory?

Thanks.

Answer 1

Let $O$ be the point from which the particle is projected, and let the $i$ and $k$ axis be the horizontal and vertical lines. Newton's equation of motion can be written as: $$ m \ddot{r} = - m g i - m c m \dot{r}, $$ where $r(t) = x(t) i + z(t) k$. Expand this in terms of the basis $\{ i, k \}$ and set the two coefficients to be equal. Thus you get the two second order differential equations for $x(t)$ and $z(t)$.

Answer 2

Let the upwards $z$-direction and rightwards $x$-direction both be positive. The vector equation is given by:

\begin{align}m\ddot{\vec{r}} &= -mc\dot{\vec{r}} - m\vec{g} \\ \implies m(\ddot{x} \ \vec{i} + \ddot{z} \ \vec{k})&= -mc(\dot{x} \ \vec{i} + \dot{z} \ \vec{k}) - mg \vec{k} \end{align}

Writing out Newton's equations in the $x$ and $z$ dimensions, we have:

\begin{align}\ddot{x} &= -c\dot{x} \\ \ddot{z} &= -g -c\dot{z}\end{align}

Observe that we can rewrite the above equations in terms of the respective component velocities:

\begin{align}\dot{v_x} &= -cv_x \\ \dot{v_z} &= -g -cv_z\end{align}

The first equation for $x$ is trivial to solve; integrating from the initial $x$-velocity $v_{x0} = u_0 \cos{\frac{\pi}{3}} = u_0\frac{1}{2}$ to $v_x$ and from initial time $t_0 = 0$ to $t$, we get:

\begin{equation}v_x = \frac{u_0}{2}e^{-ct}\end{equation}

We integrate once more from $0$ to $t$ to obtain our displacement function for $x$:

\begin{equation}x(t) = \frac{u_0}{2c}(1-e^{-ct})\end{equation}

Solving the differential equation for $z$ is not as simple. The insight one must have here is to reduce the equation into some form $a\dot{u}=bu$; the way to go about this is to convince yourself that when $\ddot{z} = 0$, we have the terminal velocity:

\begin{equation}v_{z-ter} = -\frac{g}{c}\end{equation}

Rewriting the equation with $\dot{z} = v_z$ and $\ddot{z} = \dot{v_z}$, it is easy to see that:

\begin{equation}\dot{v_z} = -c(v_z - v_{z-ter})\end{equation}

Which is in the aforementioned form that we want. You can proceed to solve for this differential equation in the same manner we did for the other to derive:

\begin{equation}z(t) = \frac{1}{c^2}\Big[\frac{cu_0 \sqrt{3}}{2} + g \Big](1 - e^{-ct}) - \frac{g}{c}t\end{equation}

Taking the limits as $t \rightarrow \infty$ yields $\lim_{t \rightarrow \infty} x(t) = \frac{u_0}{2c}$ and $\lim_{t \rightarrow \infty} z(t) = - \infty$. Note that $x(t)$ is asymptotically bounded due to the presence of the linear drag force; had there been no drag force, then $\ddot{x}$ would be equal to zero, and by the First Law of Newton, the projectile would continue to persist in its $x$-direction motion with velocity $v_x = \frac{u_0}{2}$.