Putnam 1956 A3

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Putnam 1956 A3

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Source: Putnam 1956

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#1 h Jul 1, 2022, 9:31 PM

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A particle falls in a vertical plane from rest under the influence of gravity and a force perpendicular to and proportional to its velocity. Obtain the equations of the trajectory and identify the curve.

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#2 h Mar 9, 2025, 12:43 PM

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Take the $x-$ axis horizontal and the $y-$ axis vertically downwards. Then the equations of motion are:

$y'' = g - kx'$ ;

$x'' = ky'$ .

If $k = 0$ , then we have the usual constant acceleration downwards $\left(x = 0, y = \frac{gt^2}2\right)$ . So we suppose $k \neq 0$ . By inspection, $x = A[\cos(kt) - 1] + B[\sin(kt) - kt]$ (we have chosen the terms $-A$ and $-kBt$ to give $x(0) = x'(0) = 0$ ). Hence, integrating $x'' = ky'$ yields $y = -A\sin(kt) + B[\cos(kt) - 1]$ . But $y'(0) = 0 \Rightarrow A = 0$ . Finally, to satisfy $y'' = g - kx'$ , we have $B = -\frac{g}{k^2}$ . So $x = \frac{gt}{k} - \frac{g}{k^2}\sin(kt), y = \frac{g}{k^2}[1 - \cos(kt)]$ . This is the equation of a cycloid.

This post has been edited 1 time. Last edited by Mathzeus1024, Mar 9, 2025, 12:44 PM

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#3 h Yesterday at 1:08 AM

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The force of gravity $\mathbf{F} = m \mathbf{g}$ directed vertically downwards. The force perpendicular to the velocity $\mathbf{F} = k \mathbf{v} \times \mathbf{n}$ , where $\mathbf{n}$ is a unit vector perpendicular to the plane of motion
$\begin{cases} \dfrac{dv_x}{dt} = -\dfrac{k}{m} v_y, \\ \dfrac{dv_y}{dt} = -g + \dfrac{k}{m} v_x. \end{cases}\Rightarrow \begin{cases} v_x(t) = \dfrac{g m}{k} \left(1 - \cos\left(\dfrac{k}{m} t\right)\right)\\ v_y(t) = -\dfrac{g m}{k} \sin\left(\frac{k}{m} t\right) \end{cases}$ Integrating velocities to obtain coordinates
$x(t) = \frac{g m}{k} \left( t - \frac{m}{k} \sin\left(\frac{k}{m} t\right) \right),\quad y(t) = -\frac{g m^2}{k^2} \left(1 - \cos\left(\frac{k}{m} t\right)\right)$ Let $\theta = \frac{k}{m} t$ , the trajectory equations take the form of a cycloid
$\begin{cases} x = \dfrac{g m^2}{k^2} (\theta - \sin \theta), \\ y = -\dfrac{g m^2}{k^2} (1 - \cos \theta). \end{cases}$
The particle trajectory is a cycloid reflected relative to the $x$ axis. If we write it in parametric form, then
$x = \frac{g m^2}{k^2} \left( \frac{k}{m} t - \sin\left(\frac{k}{m} t\right) \right),\quad y = -\frac{g m^2}{k^2} \left( 1 - \cos\left(\frac{k}{m} t\right) \right)$

This post has been edited 2 times. Last edited by Tip_pay, Yesterday at 1:09 AM

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