Velocity of a Particle

Question

Consider a particle moving in a straight line from the fixed point $(0,2)$ to another $(\pi,0)$. The only force acting on the particle is gravity.

How would we parametrically define the motion of the particle with time?

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From kinematics, I found that

$\hspace{150pt} y(t)=2-\dfrac{gt^2}{2}$

The slope can be found to be

$\hspace{140pt} m=\dfrac{0-2}{\pi-0}=-\dfrac{2}{\pi}$

Since the path the particle moves on is a straight line, we have

$\hspace{132pt}y=mx+b=-\dfrac{2}{\pi}x+2$

so $\hspace{150pt}x(t)=\dfrac{\pi gt^2}{4}$

Therefore the parametric equations for the position of the particle are

$\hspace{150pt}x(t)=\dfrac{\pi gt^2}{4}$

$\hspace{150pt} y(t)=2-\dfrac{gt^2}{2}$

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Does this seem correct? Any suggestions?

Answer 1

The assumption $y=2-\frac 12g\,t^2$ is wrong.

In fact the energy conservation law says that $$v^2-v_0^2=2g\,(y_0-y)$$If $y=f(x)$, then $v^2=\dot x^2+\dot y^2=\dot x^2+f'(x)^2\,\dot x^2\;$ hence $$\dot x^2=\frac {v_0^2+2g\,[f(x_0)-f(x)]}{1+f'(x)^2}$$If $f(x)=kx+q$ and one puts $x_0=0$ and $v_0=0$, after solving the ode it follows that $$x=-\frac 12g\frac{k}{1+k^2}t^2$$and so$$y=-\frac 12g\frac{k^2}{1+k^2}t^2+q$$considering $\dot x>0$.

Note that $$\lim_{k\rightarrow\infty}\,\frac{k^2}{1+k^2}=1$$ as to the vertical motion.

Answer 2

For constrained motion, you cannot take $y=2-\frac 12gt^2$. The acceleration will be gravity projected onto the path. As the slope is $-\frac 2{\pi},$ the acceleration will be $g \frac 2{\sqrt{4+\pi^2}}$ so we have $y=2-\frac 12g \frac 2{\sqrt{4+\pi^2}}t^2$. Then your calculation of $x$ from $y$ is fine, giving $x=\frac 14{\pi g \frac 2{\sqrt{4+\pi^2}}}t^2$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ The Action $S$ is given by $\pars{~\mbox{the motion occurs along the line}\ y = 2\,\pars{1 - x/\pi}~}$ $$ S = \int_{t_{0}}^{t_{1}}\bracks{{1 \over 2}\, m\pars{\dot{x}^{2} + \dot{y}^{2}} - mgy - m\pi\mu\pars{y + {2 \over \pi}\,x - 2}}\,\dd t $$ The equations of motion are: $$ m\ddot{x} = - 2m\mu\,,\qquad m\ddot{y} = -mg - m\pi\mu \quad\imp\quad \pi\ddot{x} - 2\ddot{y} = 2g $$ $$ \pi\,\dot{x}\pars{t} - 2\dot{y}\pars{t} = \overbrace{\bracks{\pi\,\dot{x}\pars{0} - 2\dot{y}\pars{0}}}^{\ds{\equiv \beta}} + 2gt\,,\quad \pi\,x\pars{t} - 2y\pars{t} = \overbrace{\bracks{\pi\,x\pars{0} - 2y\pars{0}}}^{\ds{\equiv \alpha}} + \beta t + gt^{2} $$ Then, we have the equations $$ \left\lbrace% \begin{array}{rcrcl} \pi x\pars{t} & - & 2y\pars{t} & = & \alpha + \beta t + gt^{2} \\ 2 x\pars{t} & + & \pi y\pars{t} & = & 2\pi \end{array}\right. $$ \begin{align} x\pars{t} & = {\pars{\alpha + \beta t + gt^{2}}\pi + 4\pi\over \pi^{2} + 4} = {\pi \over \pi^{2} + 4}\,\pars{\alpha + 4} + {\pi \over \pi^{2} + 4}\,\beta t + {\pi \over \pi^{2} + 4}\,gt^{2} \\[3mm] y\pars{t} & = {2\pi^{2} - 2\pars{\alpha + \beta t + gt^{2}}\pi \over \pi^{2} + 4} = {2\pi \over \pi^{2} + 4}\,\pars{\pi - \alpha} - {2\pi \over \pi^{2} + 4}\,\beta t - {2\pi \over \pi^{2} + 4}\,gt^{2} \end{align}

$$ \color{#0000ff}{\large% \begin{array}{rcl} x\pars{t} & = & x\pars{0} + \dot{x}\pars{0}t + {\pi \over \pi^{2} + 4}\,t^{2} \\[3mm] y\pars{t} & = & y\pars{0} + \dot{y}\pars{0}t - {2\pi \over \pi^{2} + 4}\,t^{2} \end{array}} $$