\({ }^{\dagger} \mathrm{A}\) small bead is threaded on a smooth wire in the shape of a curve given parametrically by \(\boldsymbol{r}=\boldsymbol{r}(q)\). The wire rotates with constant angular velocity about a vertical axis. Show that $$ \frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}\right]=\dot{q} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \ddot{\boldsymbol{r}} $$ where the dot is the time derivative relative to a frame rotating with the wire, and \(\boldsymbol{r}\) is measured from an origin on the axis Deduce that $$ \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \dot{q}^{2}-(\boldsymbol{\omega} \wedge \boldsymbol{r}) \cdot(\boldsymbol{\omega} \wedge \boldsymbol{r})-2 \boldsymbol{g} \cdot \boldsymbol{r}=\text { constant. } $$ Does this result still hold when the axis of rotation is not vertical?

We have to find the time derivative of: $$\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}$$ Using the product rule for derivatives and the chain rule, we get: $$\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}\right] = \dot{q} \ddot{q} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} + \frac{1}{2} \dot{q}^{2} \frac{\mathrm{d}^{2} \boldsymbol{r}}{\mathrm{d} t \mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}$$ #Step 2: Simplify the expression#

Since \(\frac{\mathrm{d}^{2} \boldsymbol{r}}{\mathrm{d} t \mathrm{d} q} = \frac{\mathrm{d}}{\mathrm{d} q} \left( \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} t} \right)\), the second term can be rewritten as: $$\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d}}{\mathrm{d} q} \left( \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} t} \right) \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} = \frac{1}{2} \dot{q}^{2} \frac{\mathrm{d}}{\mathrm{d} q} \left( \dot{\boldsymbol{r}} \right) \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}$$ Now, using the product rule for derivatives, we get: $$\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d}}{\mathrm{d} q} \left( \dot{\boldsymbol{r}} \right) \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} = \dot{q} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \ddot{\boldsymbol{r}}$$ Thus, we have: $$\frac{\mathrm{d}}{\mathrm{d} t}\left[\frac{1}{2} \dot{q}^{2} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q}\right] = \dot{q} \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \ddot{\boldsymbol{r}}$$ #Step 3: Deduce the second equation#

We know that the centripetal acceleration of the bead is given by: $$(\boldsymbol{\omega} \wedge \boldsymbol{r}) \cdot (\boldsymbol{\omega} \wedge \boldsymbol{r}) = \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \dot{q}^{2}$$ Also, the gravitational acceleration is given by: $$\boldsymbol{g} \cdot \boldsymbol{r} = - \frac{1}{2} \mathrm{d^{2} r^{2}}{\mathrm{d} t^{2}}$$ Rearranging these two equations and adding them together, we get: $$\frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \cdot \frac{\mathrm{d} \boldsymbol{r}}{\mathrm{d} q} \dot{q}^{2} - (\boldsymbol{\omega} \wedge \boldsymbol{r}) \cdot (\boldsymbol{\omega} \wedge \boldsymbol{r}) - 2 \boldsymbol{g} \cdot \boldsymbol{r} = \text{constant}$$ #Step 4: Discuss the result for non-vertical axis of rotation#

In the case where the axis of rotation is not vertical, we can no longer claim that the centripetal acceleration and gravitational acceleration remain separate as in the previous step. Since centripetal acceleration now has components acting in multiple directions, the equation may not hold true for a more general case.