# Newtonian formalism exercises

 Exercise 1 A particle of mass ${m}$ moves in the plane ${xy}$. Its position vector is ${\vec{r}=a\cos \omega t \vec{i}+b\sin \omega t \vec{j}}$ with ${a,b,\omega}$ positive constants and ${a>b}$. Show that: The particle’s trajectory is an ellipse. For ${x=a\cos\omega\ t}$ and ${y=b\sin\omega\ t}$ it is $\displaystyle \left( \frac{x}{a} \right)^2+\left( \frac{y}{b} \right)^2=\cos^2\omega t+\sin^2\omega t=1$ Which is the equation of an ellipse The particle is subject to a central force oriented to the origin It is {\begin{aligned} \vec{F} &= m\dfrac{d\vec{v}}{dt}\\ &= m\dfrac{d^2\vec{r}}{dt^2}\\ &= m\dfrac{d^2}{dt^2}a\cos \omega t \vec{i}+b\sin \omega t \vec{j}\\ &=-m\omega^2\vec{r} \end{aligned}} Which is a central force oriented to the origin (note the minus sign).

 Exercise 2 A particle of constant mass ${m}$ is subject to to a force ${F}$. Suppose that for ${t_1}$ e ${t_2}$ the velocities are ${\vec{v}_1}$ and ${\vec{v}_2}$, respectively. Show that the work the force does on the particle equals its change in kinetic energy. {\begin{aligned} W &= \int_{t_1}^{t_2}\vec{F}\cdot d\vec{r}\\ &= \int_{t_1}^{t_2}\vec{F}\cdot\dfrac{d\vec{r}}{dt}dt \\ &= \int_{t_1}^{t_2}\vec{F}\cdot\vec{v}dt\\ &= \int_{t_1}^{t_2}m\dfrac{d\vec{v}}{dt}\cdot\vec{v}dt\\ &=m\int_{t_1}^{t_2}\vec{v}\cdot d\vec{v}\\ &=\dfrac{1}{2} m\int_{t_1}^{t_2}d(\vec{v}\cdot\vec{v})\\ &=1/2m(v_2^2-v_1^2) \end{aligned}}

 Exercise 3 For the conditions of exercise 1 calculate: The particle’s kinetic energy in the positive semi-major axis and on the positive semi-minor axis. First note that ${\vec{v}=-\omega a\sin \omega t \vec{i}+\omega b \cos \omega t \vec{j}}$. Now ${K=1/2m(\omega^2a^2\sin^2\omega t+\omega^2b^2\cos^2\omega t)}$. Let ${A}$ denote the semi-major axis extension and ${B}$ the semi-minor axis extension. Hence ${K_A=1/2m\omega ^2b^2}$ and ${K_B=1/2m\omega ^2a^2}$ The work done by the force field when it moves the particle from the positive semi-major axis to the positive semi-minor axis. {\begin{aligned} W &= \int_A^B\vec{F}\cdot d\vec{r}\\ &= \int_A^B(-m\omega ^2\vec{r})\cdot d\vec{r}\\ &= -m\omega ^2\int_A^B\vec{r}\cdot d\vec{r}\\ &= -1/2m\omega ^2\int_A^B d(\vec{r}\cdot\vec{r})\\ &=1/2m\omega ^2(a^2-b^2) \end{aligned}} The total work done by the force field by moving the particle along the ellipse. ${W=0}$. Do you see why?

 Exercise 4 Show that if a particle is subject to ${\vec{F}}$ and ${\vec{v}}$ is its instantaneous velocity then the instantaneous power is $\displaystyle \mathcal{P}=\vec{F}\cdot\vec{v}$ By definition it is ${dW=\vec{F}\cdot d\vec{r}}$. Hence {\begin{aligned} \mathcal{P}&=\dfrac{dW}{dt}\\ &= \vec{F}\cdot\dfrac{d\vec{r}}{dt}\\ &= \vec{F}\cdot\vec{v} \end{aligned}}

 Exercise 5 Show that the integral $\int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}$ is independent of a particle’s trajectory if and only if ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Let ${\Gamma=P_1AP_2BP_1}$ denote a closed curve and admit that $\int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}$ is path independent. Then {\begin{aligned} \oint \vec{F}\cdot d\vec{r}&=\int_\Gamma\vec{F}\cdot d\vec{r} \\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}} Where the last equality follows from our assumption that $\int _{P_1}^{P_2} \vec{F}\cdot d\vec{r}$ is path independent. Suppose now that ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. {\begin{aligned} \int_\Gamma\vec{F}\cdot d\vec{r}&=\int_{P_1AP_2}\vec{F}\cdot d\vec{r}+\int_{P_2BP_1}\vec{F}\cdot d\vec{r}\\ &= \int_{P_1AP_2}\vec{F}\cdot d\vec{r}-\int_{P_1BP_2}\vec{F}\cdot d\vec{r}\\ &=0 \end{aligned}} Where the last equality follows from our ${ \displaystyle\oint\vec{F}\cdot d\vec{r}=0 }$. Hence ${\displaystyle\int_{P_1AP_2}\vec{F}\cdot d\vec{r}=\int_{P_1BP_2}\vec{F}\cdot d\vec{r}}$.

 Exercise 6 A particle of mass ${m}$ moves along ${x}$ subject to to a conservative force field ${V(x)}$. If the particle’s positions are ${x_1}$ and ${x_2}$ on ${t_1}$ and ${t_2}$, respectively, show that, if ${E}$ is the total energy then $\displaystyle t_2-t_1=\sqrt{\frac{m}{2}}\int_{x_1} ^{x_2}\frac{dx}{\sqrt{E-V(x)}}$ Write ${1/2m\left( \dfrac{dx}{dt} \right)^2+V(x)=E}$ solve in order to ${dt}$ and integrate.

 Exercise 7 Consider a particle of mass ${m}$ that moves vertically on a resistive medium where the retarding force is proportional to the particle’s velocity. Consider that particle initially moves in the downward direction with an initial velocity ${v_0}$ from an height ${h}$. Derive the particle’s equation of motion ${z=z(t)}$. It is ${F=m\dfrac{dv}{dt}=-mgg-kmv}$ with ${v<0}$ and ${-kmv>0}$. Then solve in order to ${dv}$ and integrate to find ${v=\dfrac{dz}{dt}=-\dfrac{g}{k}+\dfrac{kv_0+g}{k}e^{-kt}}$ (the terminal velocity ${v_t}$ is ${v_t=\displaystyle \lim_{t \rightarrow +\infty} v=-g/k}$). Solving in order to ${dz}$ and integrating and it is ${z=h-\dfrac{gt}{k}+\dfrac{kv_0+g}{k^2}(1-e^{-kt})}$.

 Exercise 8 A particle is shot vertically on a region where exists a constant gravitational field with a constant initial velocity ${v_0}$. Show that, in the presence of resistive force proportional to the square of the particle’s instantaneous velocity, the velocity of the particle when it returns to its initial position is: $\displaystyle \frac{v_0 v_t}{\sqrt{v_0^2+v_t^2}}$ Where ${v_t}$ denotes the particle’s terminal velocity.