Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 3 / a / Solution

In cylindrical coordinates,Differentiating the position vector $\mathbf{r}=r{\mathbf{e}}_r+z{\mathbf{e}}_z$ twice gives the velocity and accelerationThis derives acceleration in cylindrical coordinates rather than assuming it.

At a fixed time, the wire has parametrization $r\mapsto r{\mathbf{e}}_r+f(r){\mathbf{e}}_z$. Its derivative is the nonzero tangent vectorA smooth wire exerts a constraint force $\mathbf{R}$ with $\mathbf{R}\cdot \mathbf{t}=0$. Because $g$ is the gravitational acceleration, the gravitational force is $-mg{\mathbf{e}}_z$; the printed $-g{\mathbf{e}}_z$ is interpreted per unit mass. Projecting Newton's second law, $m\ddot{\mathbf{r}}=\mathbf{R}-mg{\mathbf{e}}_z$, along $\mathbf{t}$ eliminates $\mathbf{R}$:The chain rule gives $\ddot{z}=f^{\prime }(r)\ddot{r}+f^{\prime \prime }(r){\dot{r}}^2$. Rearrangement yieldsThis is the governing equation for a bead on a uniformly rotating wire. Its azimuthal acceleration contributes nothing to the projection because ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=0$.