Past exam of the physics B course of the University of Cambridge

The 2017 release includes selected final answers, rather than a complete set of worked solutions. Those papers are retained.

The transfer is a Hohmann transfer orbit. Write $r_0$ for the original circular Kepler orbit radius and $a_t$ for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving $r_1=1.2r_0$, and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period $T_0$,Using one sidereal day, $T_0=86164\mathrm{s}$, gives $t=4.9703\times 10^4\mathrm{s}=13.806\mathrm{h}$; the usual $24\mathrm{h}$ approximation gives $13.844\mathrm{h}$. The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.

There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be Earth's radius. Thenusing $R_E\simeq 6371\mathrm{km}$ and $r_0\simeq 42164\mathrm{km}$. Thus $13.8\mathrm{h}$ answers the explicitly labelled geometry; $13.5\mathrm{h}$ answers the literal altitude wording.

To find the two-mass three-spring normal modes, choose both displacements positive to the right, measured from mechanical equilibrium. The middle spring's extension changes by $x_2-x_1$. Hooke's law and Newton's second law therefore giveIn a normal mode both displacements have the same angular frequency: take $x_j=A_j\cos (\omega t+\varphi )$. A nonzero amplitude vector exists when the determinant vanishes:Writing $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$, the two angular frequencies areThe corresponding eigenvectors can be specified bySince $D>|m_2-m_1|$ for positive masses, the lower normal mode ${\omega }_{-}$ has $A_2/A_1>0$: both bodies move in the same direction at any instant. The higher normal mode ${\omega }_{+}$ has $A_2/A_1<0$: they move in opposite directions. Their amplitudes are generally unequal. For the useful check $m_1=m_2=m$, these reduce to ${\omega }_{-}=\sqrt{k/m}$ with $A_2=A_1$, and ${\omega }_{+}=\sqrt{3k/m}$ with $A_2=-A_1$. Each angular frequency has units $\mathrm{rad}{\mathrm{s}}^{-1}$.

An Electron has electric charge $-e$, so its gain of kinetic energy through an accelerating electric potential difference of magnitude $V$ is $eV$. Starting from rest, conservation of energy givesWith $e=1.60218\times 10^{-19}\mathrm{C}$, electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$, and $V=150\mathrm{V}$,This is a speed; the velocity points along the accelerating path, opposite the electric field. Here $v/c\simeq 0.0242$, where $c$ is the speed of light, so the classical kinetic energy approximation is accurate. As a check, special relativity gives $\gamma =1+eV/(m_e{c}^2)$ and $v=c\sqrt{1-{\gamma }^{-2}}\simeq 7.262\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, agreeing to the quoted precision.

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.

For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency $f_c=10\mathrm{GHz}$ and carrier frequency $f=30\mathrm{GHz}$, $\sqrt{1-(f_c/f{)}^2}=\sqrt{8}/3$. ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check $v_p{v}_g=c^2$ follows directly from the dispersion relation. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.

Use orthonormal basis vectors ${\mathbf{e}}_i(t)$ fixed in the rotating reference frame, and expand the angular momentum as $\mathbf{J}={\sum }_i{J}_i{\mathbf{e}}_i$. Over time $dt$, the basis undergoes infinitesimal rotation $\mathbf{\Omega }dt$, soThe product rule gives the rotating-frame derivative formula:Since the applied couple is the rate of change of angular momentum in the inertial frame,The second term accounts for the changing basis, even when the components of angular momentum are constant in the rotating reference frame.

One application is the slow steady precession of a rapidly spinning gyroscope supported at one point. Let its centre of mass be a distance $\ell$ from the pivot along its spin axis, at angle $\alpha$ to the upward vertical. Write $J\simeq I_1\omega$ for its dominant spin angular momentum. In a reference frame precessing steadily about the vertical with angular velocity ${\mathbf{\Omega }}_p$, the spin angular momentum is approximately stationary, soThe gravitational torque has magnitude $mg\ell \sin \alpha$. The cross product on the right has magnitude ${\Omega }_{p}J\sin \alpha$, giving, for a nonvertical axis,Its direction follows from the torque equation. This is the slow precession approximation: the additional angular momentum associated with precession is small compared with the spin contribution.

For this gyrocompass, let $\mathbf{n},\mathbf{w},\mathbf{z}$ be local north, west and upward unit vectors, forming a right-handed basis. Earth's rotation has local componentsLet the spin axis point along $\mathbf{u}=\cos \theta \mathbf{n}+\sin \theta \mathbf{w}$, and write its dominant angular momentum as $J_s\mathbf{u}$, where $J_s=I_1\omega >0$. In mechanical equilibrium relative to Earth, this leading angular momentum is fixed in the terrestrial rotating reference frame. The bearings can exert horizontal torques to keep the axle horizontal, but free rotation about the vertical implies $G_z=0$. The vertical component of the rotating-frame derivative formula therefore requiresAway from either geographical pole, $\cos \lambda \ne 0$, henceThis is a necessary equilibrium condition; the next part distinguishes its stable and unstable spin orientations. The rapid-spin approximation neglects angular momentum terms of order $I_2{\Omega }_E$ compared with $I_1\omega$. At either pole the horizontal component of Earth's rotation vanishes, and this north-seeking mechanism gives no preferred horizontal direction. The paper explicitly excludes $90^{\circ }$; if latitude is signed, $-90^{\circ }$ must likewise be excluded.

For the gyrocompass oscillation, use the same local basis and define $\theta$ about the stable orientation, with spin angular momentum directed north at $\theta =0$. Because the vertical is perpendicular to the horizontal axle, the extra vertical angular momentum from the slow azimuthal motion is $I_2\dot{\theta }$. To leading order in Earth's small angular velocity,A constant vertical contribution $I_2{\Omega }_E\sin \lambda$ from Earth's rotation has zero terrestrial time derivative; the transverse terrestrial contributions affect the following equation only at order ${\Omega }_E^2$. Taking the vertical component of the torque equation givesThe terrestrial derivative of $J_s\mathbf{u}$ is horizontal, as is ${\mathbf{\Omega }}_E\times (I_2\dot{\theta }\mathbf{z})$, so neither contributes to this component. Thus the small angle satisfies, using the small-angle approximation $\sin \theta \simeq \theta$,This is simple harmonic motion, $\theta =A\cos ({\omega }_{\mathrm{shm}}t+\varphi )$, with positive squared angular frequency away from the poles. The approximation requires rapid spin, small angular displacement and negligible bearing friction. Near the opposite orientation, putting $\theta =\pi +\delta$ instead yields $\ddot{\delta }-{\omega }_{\mathrm{shm}}^2\delta \simeq 0$, which is unstable. Consequently, the paper's oscillation claim applies to the stable equilibrium, not both north--south spin directions. Without damping the ideal gyrocompass continues to oscillate rather than settling.

For a thin gyrocompass flywheel with rotational symmetry, the perpendicular axis theorem gives $I_1=I_x+I_y=2I_2$. This does not require the mass to be concentrated at the rim. Convert the specified spin to angular velocity and use one sidereal day for Earth's rotation:At the given latitude,The associated oscillation period is $T=2\pi /{\omega }_{\mathrm{shm}}\simeq 29.0\mathrm{s}$. This uses the stated neglect of the support assembly's contribution to the moment of inertia; a substantial additional vertical moment of inertia would lengthen the oscillation period.

The bending moment $B(x)$ is the internal torque transmitted across a cut through an elastic beam, about the transverse bending axis. It balances the moments of the external forces on the isolated portion on one side of the cut. Its sign depends on the chosen convention, and its units are $\mathrm{N}\mathrm{m}$.

The relevant “moment of area” is the second moment of area of the cross-section about its neutral axis:where $z$ is perpendicular distance from that axis within the section. It has units ${\mathrm{m}}^4$ and is a geometrical quantity, unlike a mass moment of inertia. For a homogeneous beam under pure bending, the neutral axis passes through the section's centroid, and its longitudinal strain is zero.

Assume a slender homogeneous elastic beam obeying linear elasticity, with plane sections remaining plane and no resultant longitudinal force. A length element of its neutral axis subtending a small angle $d\varphi$ has length $Rd\varphi$. A neighbouring fibre at distance $z$ from that axis has a length change of magnitude $zd\varphi$, giving longitudinal strain $\epsilon =z/R$ with signs chosen consistently for tensile and compressed fibres. By Hooke's law, its longitudinal stress is $\sigma =E\epsilon =Ez/R$.

The torque of these longitudinal forces about the neutral axis is thereforeHere $E$ is Young's modulus and $I$ the second moment of area. The displayed relation uses a signed radius and bending moment; an unsigned geometrical radius is $|R|=EI/|B|$. For consistency with the negative bending moments in the later parts, take the deflection $y$ positive downward and the signed curvature as $1/R\simeq y^{\prime \prime }$. Then the small-slope Euler-Bernoulli beam equation reads $EI{y}^{\prime \prime }=B$. Reversing the vertical coordinate reverses the corresponding sign relation but leaves the physical deflection unchanged.

By vertical force balance and reflection symmetry, each support exerts an upward force $W/2$. Use the paper's convention in which the sagging bending moment is negative. For a cut at $0<x<L/2$, the left support's moment about that cut givesFor $L/2<x<L$, include the central downward force when taking moments of all loads to the left:ThusThe bending moment is continuous at the load, with $B(L/2)=-WL/4$, and vanishes at both freely rotating supports. The jump is in its first derivative, reflecting the concentrated transverse force.

The uniform weight per unit length is $q=\rho abg$, where $\rho$ is the mass density and $g$ the gravitational acceleration. The total weight is $qL$, so each support provides an upward force $qL/2$. The weight on the portion from $0$ to $x$ is $qx$ acting at that portion's midpoint, a distance $x/2$ from the cut. In the same bending moment convention as part (c),The units are $(\mathrm{N}{\mathrm{m}}^{-1}){\mathrm{m}}^2=\mathrm{N}\mathrm{m}$. Its greatest magnitude occurs at $x=L/2$, where $B=-q{L}^2/8$.