# 19.5: Motion on a Cycloid, Cusps Up

We shall imagine either a particle sliding down the inside of a smooth cycloidal bowl, or a bead sliding down a smooth cycloidal wire, Figure XIX.6.

We shall work in intrinsic coordinates to obtain the tangential and normal Equations of motion. These Equations are, respectively:

\[ \ddot{s} = -g \sin \psi \label{19.5.1}\tag{19.5.1}\]

and

\[ \dfrac{mv^2}{ \rho} = R - mg \cos \psi. \label{19.5.2}\tag{19.5.2}\]

Here *R* is the normal (and only) reaction of the bowl or wire on the particle and \( \rho \) is the radius of curvature. The radius of curvature is \(ds/d \psi \) , which, from Equation 19.3.1, (or Equations 19.4.3 and 19.4.5) is

\[ \rho = 4 a \cos \psi \label{19.5.3}\tag{19.5.3}\]

From Equations 19.3.1 and \(\ref{19.5.1}\) we see that the tangential Equation of motion can be written, without approximation:

\[\ddot{s} = -\dfrac{g}{4a}s. \label{19.5.4}\tag{19.5.4}\]

This is simple harmonic motion of period \( 4 \pi \sqrt{a/g} \), independent of the amplitude of the motion. This is the *isochronous* property of the cycloid. Likewise, if the particle is released from rest, it will reach the bottom of the cycloid in a time \( \pi \sqrt{a/g} \) whatever the starting position.

Let us see if we can find the value of *R* where the generating angle is \( \psi \). Let us suppose that the particle is released from rest at a height \(y_0\) above the \(x\)-axis (generating angle = \( \psi_0\) ); what is its speed \(v\) when it has reached a height \(y\) (generating angle \( \psi \) )? Clearly this is given by

\[ \dfrac{1}{2} mv^2 = mg(y_0 - y), \label{19.5.5}\tag{19.5.5}\]

and, following Equation 19.3.2, and recalling that \( \theta = \psi \), this is

\[ v^2 = 2ga(cos2 \psi - cos2 \psi_0 ). \label{19.5.6}\tag{19.5.6}\]

On substituting this and Equation \( \ref{19.5.3} \) into Equation \( \ref{19.5.2}\), we find for *R*:

\[ R = \frac{mg}{2 \cos \psi }(1+ 2 \cos 2 \psi - \cos 2 \psi_0) \label{19.5.7}\tag{19.5.7}\]