Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 11A (Dynamics and Relativity)

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

The original PDF measures $\theta$ at the rim point $O$, from the positive $x^{\prime }$-axis to $OP$; it is not the angle at the centre of the circle. The converted TeX omits this entire subpart. In rotating Cartesian coordinate system, the centre is $(a,0)$, and the nonzero intersection of the ray $\mathbf{r}=\rho (\cos \theta ,\sin \theta )$ with $(x^{\prime }-a{)}^2+y^{\prime 2}=a^2$ has $\rho =2a\cos \theta$. ThereforeThe second form stays regular even at the rim point $O$, where the polar radius is zero. Define the outward unit normal from the centre and the increasing-angle unit tangent vector byThe first two derivatives in the rotating reference frame are ${\mathbf{v}}^{\prime }=2a\dot{\theta }\mathbf{\tau }$ and ${\mathbf{a}}^{\prime }=2a\ddot{\theta }\mathbf{\tau }-4a{\dot{\theta }}^2\mathbf{n}$. These describe a bead on a horizontal hoop rotating about its rim.

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 $\boxed{\ddot{\theta }=-({\omega }^2/2)\sin 2\theta }$.

For nonzero $\omega$, the equilibrium points satisfy $\sin 2\theta =0$. The equilibrium point $\theta =0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta =\eta$ small in the tangential ordinary differential equation givesThus the small oscillation is simple harmonic motion, with angular frequency $|\omega |$ and ordinary frequency $|\omega |/(2\pi )$.

The endpoints $\theta =\pm \pi /2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot{\eta }=+{\omega }^2\eta$ and is unstable. Equivalently, the centrifugal potential along the hoop is $-\frac{1}{2}{\omega }^2|\mathbf{r}{|}^2=-2a^2{\omega }^2{\cos }^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega =0$, every bead position with zero relative velocity is an equilibrium point, but perturbing that velocity produces unrestored drift; there is no nonzero restoring frequency.