# Parks and Reciprocation Solution

The following is a solution to Puzzle #18 – Parks and Reciprocation. Take a look if you haven’t already.

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We are asked in the first part of the question to find the height of the platform above the ground as a function of the angle $\varphi$. One approach is to split the triangle formed by the two rods and the vertical beam into two by dropping a perpendicular from the hinge to the beam. This construction is shown in the diagram to the right.

The platform height $h$ is split into the two lengths $y$ and $z$. From simple trigonometry, $y$ is related to $\varphi$ by

$y=a\cos\varphi$

where $a$ is the length of the shorter rod. Distance $z$ we obtain with Pythagoras’ theorem:

$z=\sqrt{b^2-x^2}$

where

$x=a\sin\varphi.$

Combining the results, we have

$h=a\cos\varphi+\sqrt{b^2-a^2\sin^2\varphi};$

$\displaystyle h(\varphi)=a\left(\cos\varphi+\left[\left(\frac{b}{a}\right)^2-\sin^2\varphi\right]^{\frac{1}{2}}\right).$

Evaluation of this height at the particular angles $\varphi=0,\frac{\pi}{2},\pi$ provides a quick sanity check.

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We are next asked to evaluate the acceleration of the platform at the top and bottom of its cycle. To do so, we need to take derivatives of $h$ with respect to $\varphi$. The general expression will be a mess, but we needn’t worry about its functional form too much, as we’re only interested in the values it assumes at particular angles.

Differentiating once yields

$\displaystyle \frac{dh}{d\varphi}=-a\left(\sin\varphi+\frac{\sin\varphi\cos\varphi}{\left[\left(\frac{b}{a}\right)^2-\sin^2\varphi\right]^{\frac{1}{2}}}\right),$

twice

$\displaystyle \frac{d^2h}{d\varphi^2}=-a\left(\cos\varphi+\frac{\cos^2\varphi-\sin^2\varphi}{\left[\left(\frac{b}{a}\right)^2-\sin^2\varphi\right]^{\frac{1}{2}}}+\frac{\sin^2\varphi\cos^2\varphi}{\left[\left(\frac{b}{a}\right)^2-\sin^2\varphi\right]^{\frac{3}{2}}}\right).$

The true acceleration, $d^2h/dt^2$, is found by multiplying the above by $\omega$2, where $\omega\equiv d\varphi/dt$. Therefore, at the top and bottom of the platform’s oscillation,

$\frac{d^2h(0)}{dt^2}=\ddot{h}_{t}=-\left(\frac{a}{b}+1\right)a\omega^2;$

$\frac{d^2h(\pi)}{dt^2}=\ddot{h}_b=-\left(\frac{a}{b}-1\right)a\omega^2.$

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The final part of the question asks us to find the ratio $\frac{a}{b}$ on the basis that:

• the passengers on the platform feel 1.5 times their usual weight at the bottom of the oscillation;
• the passengers feel weightless at the top of the oscillation.

Where does the sensation of weight come from? The answer is the reaction force, the force exerted on you by whatever you’re sitting on. It is the generic term for the force which is preventing you from falling through your seat; though the Earth’s mass is attempting to pull me downwards with force $mg$, my sofa is responding with an equal and opposite force, such that my acceleration is zero.

If the passengers on the platform experience acceleration $\ddot{h}$, the reaction force $R$ that acts on them must satisfy

$m\ddot{h}=R-mg$

from Newton’s second law. In other words,

$R=m\left(\ddot{h}+g\right).$

If the passengers are accelerating upwards $(\ddot{h}>0)$, $R>mg$, and the passengers feel heavier than usual. This is because the platform must push a little harder on them than it does when they are at rest to overcome their weight. Conversely, if the passengers are accelerating downwards $(\ddot{h}<0)$, $R, and the passengers feel lighter than usual. Indeed, in the particular case that $\ddot{h}=-g$, the reaction force $R$ vanishes because the passengers are in free-fall, and the platform needn’t exert any force on them at all!

Let’s now apply the constraints given by the question. That the passengers feel 1.5 times their usual weight at the bottom of the oscillation means

$\frac{3}{2}mg=m\left(\ddot{h}_b+g\right);$

$\ddot{h}_b=\frac{1}{2}g=-\left(\frac{a}{b}-1\right)a\omega^2.$

Meanwhile, if the passengers are weightless at the top of the oscillation,

$0=m\left(\ddot{h}_t+g\right);$

$\ddot{h}_t=-g=-\left(\frac{a}{b}-1\right)a\omega^2.$

Eliminating $g$ between these two equations yields

$-\left(\frac{a}{b}-1\right)a\omega^2=\frac{1}{2}\left(\frac{a}{b}+1\right)a\omega^2,$

from which we obtain the dimensionless equation

$1-\frac{a}{b}=\frac{1}{2}\left(\frac{a}{b}+1\right).$

A simple rearrangement yields the answer we’re looking for:

$\displaystyle \frac{a}{b}=\frac{1}{3}.$

That is, the long rod is three times longer than the short rod.

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