Past exam of the physics B course of the University of Cambridge / 2021 / Paper 2 / Section B / B6

The examination requires three notes. All four options are developed here for completeness.

For torque-free precession of a symmetric top, take principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, with no external torque. In an inertial frame moving with its centre of mass, its angular momentum $\mathbf{L}$ is constant in an inertial frame, and its rotational kinetic energy is conserved.

In the principal body axes, Euler equations for a torque-free rigid body reduce toTherefore ${\omega }_3$ is constant andThus the transverse angular velocity rotates uniformly in the body frame; its signed rate reverses between oblate and prolate inertia tensors. This body-frame wobble rate should be distinguished from the space-frame precession of the symmetry axis.

Let ${\mathbf{e}}_3$ be the body's symmetry axis. Sinceits angle $\theta$ to $\mathbf{L}$ is constant, with $\cos \theta =I_{\parallel }{\omega }_3/L$, andHence the symmetry axis describes a cone about fixed $\mathbf{L}$ at the space-frame precession rateThere is no nutation of this angle for torque-free motion of an exactly symmetric top. In the usual Euler angles, ${\omega }_3=\dot{\psi }+\dot{\varphi }\cos \theta$, so $\dot{\psi }=-{\Omega }_b$. Both rates describe the same motion in different axes. A freely tossed axisymmetric disc illustrates the effect; a top supported at a point under gravity is a different, externally torqued problem.

Lagrangian mechanics expresses dynamics through generalized coordinates $q_i$ that respect the constraints. For a system with conservative, velocity-independent forces and standard kinetic energy, the Lagrangian is $L=T-V$. Its equations follow from the principle of stationary action:for variations that vanish at the endpoints. Varying the integral and integrating the $\delta {\dot{q}}_i$ term by parts givesIndependent variations therefore yield the Euler-Lagrange equations:Additional nonconservative generalized forces appear on the right-hand side as $Q_i$. The action is stationary, not necessarily a minimum. For ideal holonomic constraints, using suitable generalized coordinates removes the need to solve explicitly for constraint forces.

The conjugate momentum is $p_i=\partial L/\partial {\dot{q}}_i$. If a coordinate is cyclic, $\partial L/\partial q_i=0$ and its conjugate momentum is conserved. If $L$ has no explicit time dependence, the energy functionis conserved. It equals $T+V$ for the usual quadratic kinetic energy with time-independent geometry, but that identification is not universal for every Lagrangian or constraint.

For example, a simple pendulum with angle $\theta$, length $\ell$ and bob mass $m$ hasThe tension does no virtual work in the allowed angular displacement and never needs to be calculated to obtain the angular equation. The method scales naturally to coupled systems and coordinates with complicated geometry.

For a slender elastic beam, the Euler-Bernoulli beam equation assumes small strain, small deflection slopes and cross-sections that remain plane and normal to the deflected centreline. A fibre at signed distance $y$ from the neutral axis has longitudinal strain ${ϵ}_{xx}=-y\kappa$, where $\kappa$ is the centreline curvature. Hooke's law gives ${\sigma }_{xx}=-Ey\kappa$, with Young's modulus $E$.

Choose the internal bending-moment sign so that $M=-\int y{\sigma }_{xx}dA$. ThenHere $I_A$ is the second moment of area, with units ${\mathrm{m}}^4$, not the mass moment of inertia of a rotating body. For a uniform material under pure bending, zero axial resultant places the neutral axis through the section's centroid.

For small slopes, $\kappa \simeq w^{\prime \prime }(x)$. Combining the moment-curvature relation with transverse force and moment balance gives, with a consistent load sign convention,For constant stiffness this becomes $E{I}_A{w}^{(4)}=q$. The bending stiffness $E{I}_A$ has units $\mathrm{N}{\mathrm{m}}^2$. Boundary conditions specify displacement and slope at a clamped end, or bending moment and shear force at a free end.

For a cantilever of length $\ell$ carrying a transverse end load $F$, take deflection positive towards the load. The moment magnitude is $M(x)=F(\ell -x)$, and integration with $w(0)=w^{\prime }(0)=0$ givesThe free-end moment vanishes and the tip shear supplies the applied load. For a rectangular section of width $b$ and bending thickness $h$, $I_A=b{h}^3/12$, so increasing thickness strongly increases bending resistance. The model ceases to be reliable for large deflections, plastic strains or short thick beams where shear deformation matters.

The Magnus effect is the transverse force produced when a spinning body moves relative to a fluid. Rotation changes the flow and circulation around the body; the resulting asymmetric pressure and wake deflect the fluid, with the opposite transverse force on the body. Examples include the curved flight of spinning balls and the lift of rotating-cylinder sails.

A useful quantitative model is steady two-dimensional incompressible flow past a cylinder of radius $a$, with fluid velocity $\mathbf{U}=U\hat{\mathbf{x}}$ far away. Let circulation $\Gamma$ be positive anticlockwise as viewed from $+z$. The potential-flow velocity at the cylinder surface isThe Bernoulli equation outside the boundary layer givesIntegrating the pressure force around the cylinder, per unit span, yieldsThus the Kutta–Joukowski theorem givesThe sign here refers to the fluid velocity relative to the cylinder. For a body moving through otherwise stationary fluid, that relative fluid velocity points opposite to the body's translational velocity.

Real viscosity and the no-slip boundary condition transmit the surface rotation to the fluid and help set the circulation and separation pattern. Pure inviscid flow with only an impermeability condition does not determine $\Gamma$ from the cylinder's angular speed. Therefore the circulation must not be assigned a universal spin-rate formula without additional assumptions. The cylinder result illustrates the lift mechanism; quantitative forces on a finite spinning ball also depend on its geometry, Reynolds number, rotation rate, surface roughness and wake.