Problem source

A small heavy bead can slide smoothly in a vertical plane on a fixed wire with equation 
\[
y=x-\frac{x^{2}}{4a},
\]
where the $y$-axis points vertically upwards and $a$ is a positive constant. The bead is projected from the origin with initial speed $V$ along the wire. 
\begin{questionparts}
\item Show that for a suitable value of $V$, to be determined, a motion is possible throughout which the bead exerts no pressure on the wire. 
\item Show that $\theta,$ the angle between the particle's velocity at time $t$ and the $x$-axis, satisfies 
\[
\frac{4a^{2}\dot{\theta}^{2}}{\cos^{6}\theta}+2ga(1-\tan^{2}\theta)=V^{2}.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item The condition that the bead exerts no pressure on the wire is equivalent to the condition that the wire exerts no force on the bead. (Newton's Third Law). This is equivalent to the bead being projected under gravity.

Notice that the initial projection is at $45^{\circ}$ since $\frac{dy}{dx}|_{x=0} = 1$.

The position of the particle (under gravity) at time $t$ is $x = \frac{1}{\sqrt{2}}Vt$ and $y = \frac{1}{\sqrt{2}}Vt - \frac12 gt^2 = x - \frac{1}{2}g \frac{2x^2}{V^2} = x - \frac{g}{V^2}x^2$. Therefore they follow the same trajectory if $\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}$

\item First note that the wire does no work as it is perpendicular to the velocity, so it is fine to use conservation of momentum.

If we take our $0$ GPE level to be be $x = 0$, then we notice the initial energy is $\frac12mV^2$.

Secondly, notice that $\tan \theta = \frac{\d y}{\d x} = 1- \frac{x}{2a} \Rightarrow x = 2a - 2a \tan \theta$

\begin{align*}
y &= 2a(1-\tan \theta) - \frac{4a^2(1-\tan \theta)^2}{4a}\\
&= (1-\tan \theta)(2a-a(1-\tan \theta)) \\
&= a(1-\tan \theta)(1+\tan \theta) \\
&= a(1-\tan^2 \theta)
\end{align*}

GPE $mga(1-\tan^2 \theta)$.

To calculate the kinetic energy, notice that $\dot{x} = v \cos \theta \dot{\theta}$ and $\dot{x} = -2a\sec^2 \theta\dot{\theta} \Rightarrow v = -\frac{2a\dot{\theta} }{\cos^{3} \theta}$.

Therefore, energy at time $t$ is:

\begin{align*}
&& \frac12 m V^2 &= \frac12 m \l - \frac{2a\dot{\theta}}{\cos^3 \theta} \r^2 + mga(1-\tan^2 \theta) \\
\Rightarrow && V^2 &= \frac{4a^2\dot{\theta}^2}{\cos^6 \theta} + 2ag(1-\tan^2 \theta)
\end{align*}

\end{questionparts}