Dynamics | Part IA, 2006

A motorcycle of mass MM moves on a bowl-shaped surface specified by its height h(r)h(r) where r=x2+y2r=\sqrt{x^{2}+y^{2}} is the radius in cylindrical polar coordinates (r,ϕ,z)(r, \phi, z). The torque exerted by the motorcycle engine on the rear wheel results in a force F(t)\mathbf{F}(t) pushing the motorcycle forward. Assuming F(t)\mathbf{F}(t) is directed along the motorcycle's velocity and that the motorcycle's vertical velocity and acceleration are small, show that the motion is described by

r¨rϕ˙2=gdhdr+F(t)Mr˙r˙2+r2ϕ˙2rϕ¨+2r˙ϕ˙=F(t)Mrϕ˙r˙2+r2ϕ˙2\begin{aligned} \ddot{r}-r \dot{\phi}^{2} &=-g \frac{d h}{d r}+\frac{F(t)}{M} \frac{\dot{r}}{\sqrt{\dot{r}^{2}+r^{2} \dot{\phi}^{2}}} \\ r \ddot{\phi}+2 \dot{r} \dot{\phi} &=\frac{F(t)}{M} \frac{r \dot{\phi}}{\sqrt{\dot{r}^{2}+r^{2} \dot{\phi}^{2}}} \end{aligned}

where dots denote time derivatives, F(t)=F(t)F(t)=|\mathbf{F}(t)| and gg is the acceleration due to gravity.

The motorcycle rider can adjust F(t)F(t) to produce the desired trajectory. If the rider wants to move on a curve r(ϕ)r(\phi), show that ϕ(t)\phi(t) must obey

ϕ˙2=gdhdr/(r+2r(drdϕ)2d2rdϕ2)\dot{\phi}^{2}=g \frac{d h}{d r} /\left(r+\frac{2}{r}\left(\frac{d r}{d \phi}\right)^{2}-\frac{d^{2} r}{d \phi^{2}}\right)

Now assume that h(r)=r2/h(r)=r^{2} / \ell, with \ell a constant, and r(ϕ)=ϵϕr(\phi)=\epsilon \phi with ϵ\epsilon a positive constant, and 0ϕ<0 \leqslant \phi<\infty so that the desired trajectory is a spiral curve. Assuming that ϕ(t)\phi(t) tends to infinity as tt tends to infinity, show that ϕ˙(t)\dot{\phi}(t) tends to 2g/\sqrt{2 g / \ell} and F(t)F(t) tends to 4ϵMg/4 \epsilon M g / \ell as tt tends to infinity.

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