Problems

Solution:

1. 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 \(\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}\)
2. 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*}