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:

  1. 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

  2. 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:

  1. 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}

  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}}

  3. 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 {\displaystyle \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 {\displaystyle \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 {\displaystyle \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.

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