Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 11A / b / ii

Because the contact is frictionless, the horizontal normal force is ${\mathbf{F}}_h=N\mathbf{n}$, with $N$ a signed outward component. Any vertical contact component balances Newtonian gravity and does not enter the horizontal equations. With $\mathbf{\omega }=\omega \hat{\mathbf{z}}$, the rotating reference frame equation givesSince $\hat{\mathbf{z}}\times \mathbf{\tau }=-\mathbf{n}$, $\mathbf{r}\cdot \mathbf{n}=2a{\cos }^2\theta$, and $\mathbf{r}\cdot \mathbf{\tau }=-a\sin 2\theta$, its normal and tangential components areHence the requested horizontal force vector isIts magnitude is the absolute value of the scalar coefficient. That coefficient need not always be negative: the constraint can exert either sign of normal force, depending on the bead's relative velocity. The tangential equation, which does not contain the reaction, is $\boxed{\ddot{\theta }=-({\omega }^2/2)\sin 2\theta }$.