UPSC Maths Optional · Paper II · Full Proofs + PYQs 2000–2026
For a system with \(n\) generalized coordinates \(q_i\), the equations of motion are \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot{q}_i}\right)-\dfrac{\partial L}{\partial q_i}=0\) where \(L=T-V\).
Step 1 Hamilton's principle: \(\delta\int_{t_1}^{t_2}L(q_i,\dot q_i,t)\,dt=0\).
Step 2 Vary: \(\delta\int L\,dt=\int\!\left(\sum_i\dfrac{\partial L}{\partial q_i}\delta q_i+\dfrac{\partial L}{\partial\dot q_i}\delta\dot q_i\right)dt\).
Step 3 Integrate second term by parts (\(\delta\dot q_i=\frac{d}{dt}\delta q_i\)): \(\int\dfrac{\partial L}{\partial\dot q_i}\delta\dot q_i\,dt=\left[\dfrac{\partial L}{\partial\dot q_i}\delta q_i\right]_{t_1}^{t_2}-\int\!\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot q_i}\right)\delta q_i\,dt\). Boundary: \(\delta q_i=0\) at endpoints.
Step 4 So \(\delta\int L\,dt=\int\sum_i\!\left[\dfrac{\partial L}{\partial q_i}-\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot q_i}\right)\right]\delta q_i\,dt=0\). Since \(\delta q_i\) are arbitrary: \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot q_i}\right)-\dfrac{\partial L}{\partial q_i}=0\). \(\blacksquare\)
With \(p_i=\partial L/\partial\dot q_i\) and \(H=\sum p_i\dot q_i-L\): \(\dot q_i=\partial H/\partial p_i\) and \(\dot p_i=-\partial H/\partial q_i\).
\(dH=\sum(p_i\,d\dot q_i+\dot q_i\,dp_i)-\dfrac{\partial L}{\partial q_i}\,dq_i-\dfrac{\partial L}{\partial\dot q_i}\,d\dot q_i\). Since \(p_i=\partial L/\partial\dot q_i\): \(p_i\,d\dot q_i-\dfrac{\partial L}{\partial\dot q_i}d\dot q_i=0\). So \(dH=\sum(\dot q_i\,dp_i-\dot p_i\,dq_i)\) (using Lagrange eqs: \(\dot p_i=\partial L/\partial q_i\)). Comparing with \(dH=\sum\!\left(\dfrac{\partial H}{\partial p_i}dp_i+\dfrac{\partial H}{\partial q_i}dq_i\right)\): \(\dot q_i=\partial H/\partial p_i\) and \(\dot p_i=-\partial H/\partial q_i\). \(\blacksquare\)
If \(L\) does not depend explicitly on time, then \(H=T+V=\) total energy is conserved.
\(\dfrac{dH}{dt}=\sum\!\left(\dfrac{\partial H}{\partial q_i}\dot q_i+\dfrac{\partial H}{\partial p_i}\dot p_i\right)+\dfrac{\partial H}{\partial t}=\sum(-\dot p_i\dot q_i+\dot q_i\dot p_i)+\dfrac{\partial H}{\partial t}=\dfrac{\partial H}{\partial t}\). If \(L\) (and hence \(H\)) does not depend on \(t\): \(dH/dt=0\Rightarrow H=\text{const}\). For holonomic systems with time-independent constraints: \(H=T+V\). \(\blacksquare\)
For incompressible fluid: \(\text{div}(\mathbf{v})=0\), i.e., \(\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0\).
Consider a small parallelepiped \([x,x+\delta x]\times[y,y+\delta y]\times[z,z+\delta z]\). Mass flux out in \(x\)-direction: \(\rho(u+\partial_x(\rho u)\delta x)\delta y\delta z-\rho u\delta y\delta z=\partial_x(\rho u)\delta x\delta y\delta z\). Similarly for \(y,z\). Total rate of mass increase inside: \(-\dfrac{\partial\rho}{\partial t}\delta x\delta y\delta z\). Conservation: \(\dfrac{\partial\rho}{\partial t}+\text{div}(\rho\mathbf{v})=0\). For incompressible (\(\rho=\text{const}\)): \(\text{div}(\mathbf{v})=\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}+\dfrac{\partial w}{\partial z}=0\). \(\blacksquare\)
Flow is irrotational (\(\nabla\times\mathbf{v}=0\)) iff there exists a scalar \(\phi\) (velocity potential) such that \(\mathbf{v}=\nabla\phi\).
If \(\mathbf{v}=\nabla\phi\): \(\nabla\times\mathbf{v}=\nabla\times\nabla\phi=0\) (curl of gradient is zero). Conversely, if \(\nabla\times\mathbf{v}=0\) in a simply-connected domain: by Poincaré lemma there exists \(\phi\) with \(\mathbf{v}=\nabla\phi\). \(\blacksquare\)
For incompressible irrotational: \(\nabla\cdot(\nabla\phi)=0\Rightarrow\nabla^2\phi=0\) (Laplace equation). So \(\phi\) is harmonic.
Along a streamline in steady, irrotational, incompressible flow: \(p+\dfrac{1}{2}\rho v^2+\rho gz=\text{constant}\).
Step 1 Euler's equation of motion for inviscid flow: \(\rho\dfrac{D\mathbf{v}}{Dt}=-\nabla p+\rho\mathbf{g}\). For steady flow: \(\dfrac{D\mathbf{v}}{Dt}=(\mathbf{v}\cdot\nabla)\mathbf{v}\).
Step 2 Use identity: \((\mathbf{v}\cdot\nabla)\mathbf{v}=\nabla(v^2/2)-\mathbf{v}\times(\nabla\times\mathbf{v})\). For irrotational flow: \(\nabla\times\mathbf{v}=0\). So Euler's equation: \(\rho\nabla(v^2/2)=-\nabla p-\rho\nabla(gz)\) (taking \(\mathbf{g}=-g\hat{z}\)).
Step 3 \(\nabla\!\left(\dfrac{p}{\rho}+\dfrac{v^2}{2}+gz\right)=0\Rightarrow\dfrac{p}{\rho}+\dfrac{v^2}{2}+gz=\text{const}\). Multiply by \(\rho\): \(p+\dfrac{1}{2}\rho v^2+\rho gz=C\). \(\blacksquare\)
For 2D incompressible flow with continuity \(\partial u/\partial x+\partial v/\partial y=0\): there exists \(\psi\) (stream function) with \(u=\partial\psi/\partial y,v=-\partial\psi/\partial x\). Streamlines: \(\psi=\text{const}\).
The continuity equation \(\partial u/\partial x+\partial v/\partial y=0\) is the compatibility condition for existence of \(\psi\) with \(u=\psi_y,v=-\psi_x\). Check: \(\psi_{yx}-(-\psi_{xy})=0\) automatically (mixed partials equal). For irrotational flow additionally: \(\partial v/\partial x-\partial u/\partial y=0\Rightarrow-\psi_{xx}-\psi_{yy}=0\Rightarrow\nabla^2\psi=0\). \(\blacksquare\)
Complex potential: \(w(z)=\phi+i\psi\) is analytic, \(dw/dz=u-iv\) (complex velocity).
Set up Lagrange's equation for an Atwood machine (masses \(m_1,m_2\) connected by inextensible string over pulley).
Generalized coordinate: \(q=x\) (descent of \(m_1\)). \(T=\dfrac{1}{2}(m_1+m_2)\dot x^2\). \(V=-m_1gx+m_2gx=(m_2-m_1)gx\). \(L=T-V=\dfrac{1}{2}(m_1+m_2)\dot x^2-(m_2-m_1)gx\). Lagrange's equation: \((m_1+m_2)\ddot x=(m_1-m_2)g\Rightarrow\ddot x=\dfrac{(m_1-m_2)g}{m_1+m_2}\). \(\blacksquare\)
Derive the equation of motion of a simple pendulum of length \(l\) using Lagrange's method.
Coordinate: \(\theta\) (angle from vertical). \(T=\dfrac{1}{2}ml^2\dot\theta^2\), \(V=-mgl\cos\theta\). \(L=\dfrac{ml^2\dot\theta^2}{2}+mgl\cos\theta\). \(\dfrac{\partial L}{\partial\dot\theta}=ml^2\dot\theta\), \(\dfrac{\partial L}{\partial\theta}=-mgl\sin\theta\). Lagrange: \(ml^2\ddot\theta+mgl\sin\theta=0\Rightarrow\ddot\theta+\dfrac{g}{l}\sin\theta=0\). For small \(\theta\): \(\ddot\theta+\dfrac{g}{l}\theta=0\) (SHM with \(\omega^2=g/l\), period \(T=2\pi\sqrt{l/g}\)). \(\blacksquare\)
For a particle moving in a central field, show that angular momentum is conserved using Lagrangian mechanics.
Use polar coordinates: \(T=\dfrac{m}{2}(\dot r^2+r^2\dot\theta^2)\), \(V=V(r)\). \(L=\dfrac{m}{2}(\dot r^2+r^2\dot\theta^2)-V(r)\). Since \(\theta\) does not appear in \(L\) (cyclic): \(\dfrac{\partial L}{\partial\theta}=0\Rightarrow\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial\dot\theta}\right)=0\Rightarrow\dfrac{d}{dt}(mr^2\dot\theta)=0\Rightarrow L_z=mr^2\dot\theta=\text{const}\) (angular momentum). \(\blacksquare\)
Find Hamilton's equations for a harmonic oscillator with Lagrangian \(L=\dfrac{m\dot x^2}{2}-\dfrac{kx^2}{2}\).
Generalized momentum: \(p=\partial L/\partial\dot x=m\dot x\Rightarrow\dot x=p/m\). Hamiltonian: \(H=p\dot x-L=\dfrac{p^2}{m}-\left(\dfrac{p^2}{2m}-\dfrac{kx^2}{2}\right)=\dfrac{p^2}{2m}+\dfrac{kx^2}{2}=T+V\). Hamilton's equations: \(\dot x=\partial H/\partial p=p/m\), \(\dot p=-\partial H/\partial x=-kx\). Eliminating \(p\): \(\ddot x=-kx/m\Rightarrow\) SHM. \(\blacksquare\)
Show that \(\phi=x^2-y^2\) is a velocity potential for 2D flow. Find the velocity and stream function.
Check: \(\nabla^2\phi=2-2=0\). \(\checkmark\) (\(\phi\) is harmonic). Velocity: \(u=\phi_x=2x,v=\phi_y=-2y\). Stream function: \(\psi_y=u=2x\Rightarrow\psi=2xy+f(x)\). \(\psi_x=-v=2y\Rightarrow2y+f'(x)=2y\Rightarrow f'=0\Rightarrow f=\text{const}\). Stream function: \(\psi=2xy\). Complex potential: \(w=\phi+i\psi=(x^2-y^2)+2ixy=(x+iy)^2=z^2\). Streamlines: \(2xy=C\) (rectangular hyperbolas). \(\blacksquare\)
For the flow \(u=ax,v=-ay\) (where \(a>0\)), verify irrotationality, find velocity potential and stream function.
Continuity: \(\partial u/\partial x+\partial v/\partial y=a-a=0\). \(\checkmark\) Irrotationality: \(\partial v/\partial x-\partial u/\partial y=0-0=0\). \(\checkmark\) Velocity potential: \(\phi_x=u=ax\Rightarrow\phi=ax^2/2+g(y)\). \(\phi_y=g'(y)=v=-ay\Rightarrow g=-ay^2/2\). \(\phi=a(x^2-y^2)/2\). Stream function: \(\psi=axy\) (by analogous calculation, or since \(w=az^2/2\)). \(\blacksquare\)
A source of strength \(m\) is at the origin in a uniform stream with velocity \(U\) in \(x\)-direction. Find the complex potential and stagnation point.
Complex potential: \(w=Uz+\dfrac{m}{2\pi}\ln z\). Velocity: \(\dfrac{dw}{dz}=U+\dfrac{m}{2\pi z}=u-iv\). Stagnation point: \(dw/dz=0\Rightarrow U=-m/(2\pi z)\Rightarrow z=-m/(2\pi U)\). Stagnation point at \((-m/(2\pi U),0)\) on negative \(x\)-axis. \(\blacksquare\)
A venturimeter has inlet diameter 20 cm and throat 10 cm. If the pressure difference is \(h=25\) cm of mercury, find the flow rate. (Take \(C_d=0.98\).)
\(A_1=\pi(0.1)^2=0.01\pi\) m², \(A_2=\pi(0.05)^2=0.0025\pi\) m². \(h=0.25\times13.6=3.4\) m of water. Flow rate: \(Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}\). \(\sqrt{2gh}=\sqrt{2\times9.81\times3.4}=\sqrt{66.708}=8.168\) m/s. \(A_1^2-A_2^2=(0.01\pi)^2-(0.0025\pi)^2=\pi^2(10^{-4}-6.25\times10^{-6})=\pi^2\times9.375\times10^{-5}\). \(\sqrt{A_1^2-A_2^2}=\pi\times9.682\times10^{-3}\). \(Q=0.98\times\dfrac{0.01\pi\times0.0025\pi\times8.168}{\pi\times9.682\times10^{-3}}=0.98\times\dfrac{2.5\times10^{-5}\pi\times8.168}{9.682\times10^{-3}}\approx0.98\times0.006624\approx0.00649\) m³/s \(=6.49\) litres/sec. \(\blacksquare\)
Prove Torricelli's theorem: speed of efflux from a small hole at depth \(h\) below the free surface is \(v=\sqrt{2gh}\).
Apply Bernoulli between free surface (point A) and hole (point B). At A: \(p_A=p_0\) (atmosphere), \(v_A\approx0\) (large tank), height \(=h\) above hole. At B: \(p_B=p_0,v_B=v\), height \(=0\). Bernoulli: \(p_0+0+\rho gh=p_0+\dfrac{1}{2}\rho v^2+0\Rightarrow\rho gh=\dfrac{1}{2}\rho v^2\Rightarrow v=\sqrt{2gh}\). \(\blacksquare\)
Water flows through a pipe that narrows from cross-section \(A_1=0.04\) m² to \(A_2=0.01\) m². If velocity at wide end is \(v_1=2\) m/s, find \(v_2\) and the pressure difference (taking \(\rho=1000\) kg/m³).
Continuity: \(A_1v_1=A_2v_2\Rightarrow v_2=\dfrac{A_1v_1}{A_2}=\dfrac{0.04\times2}{0.01}=8\) m/s. Bernoulli (horizontal pipe): \(p_1+\dfrac{1}{2}\rho v_1^2=p_2+\dfrac{1}{2}\rho v_2^2\Rightarrow p_1-p_2=\dfrac{1}{2}\rho(v_2^2-v_1^2)=\dfrac{1000}{2}(64-4)=30000\) Pa \(=30\) kPa. \(\blacksquare\)
Show that the complex potential \(w=Uz+\dfrac{Ua^2}{z}\) represents uniform flow past a cylinder of radius \(a\). Find the pressure distribution on the cylinder.
\(w=U\!\left(z+\dfrac{a^2}{z}\right)=U\!\left(re^{i\theta}+\dfrac{a^2}{r}e^{-i\theta}\right)=U\!\left(r+\dfrac{a^2}{r}\right)\cos\theta+iU\!\left(r-\dfrac{a^2}{r}\right)\sin\theta\). Stream function: \(\psi=U\!\left(r-\dfrac{a^2}{r}\right)\sin\theta\). On \(r=a\): \(\psi=0\). So \(r=a\) is a streamline — the surface of the cylinder. \(\checkmark\) Velocity: \(\dfrac{dw}{dz}=U\!\left(1-\dfrac{a^2}{z^2}\right)\). On \(r=a\): \(|dw/dz|^2=4U^2\sin^2\theta\Rightarrow v^2=4U^2\sin^2\theta\). Bernoulli: \(p+\dfrac{1}{2}\rho v^2=p_\infty+\dfrac{1}{2}\rho U^2\). So \(p=p_\infty+\dfrac{1}{2}\rho U^2(1-4\sin^2\theta)\). \(\blacksquare\) Note: D'Alembert's paradox — net drag = 0 for ideal fluid.
A bead of mass \(m\) slides without friction on a straight wire rotating in a horizontal plane with constant angular velocity \(\omega\). Write the Lagrangian and find the equation of motion.
Let \(r\) = distance of bead from origin (generalized coordinate). \(T=\tfrac{1}{2}m(\dot r^2+r^2\omega^2)\), \(V=0\) (horizontal plane). \(L=\tfrac{1}{2}m(\dot r^2+r^2\omega^2)\). Lagrange equation: \(\ddot r-\omega^2 r=0\). Solution: \(r=Ae^{\omega t}+Be^{-\omega t}\). The bead moves outward exponentially (centrifugal effect). \(\blacksquare\)
A double pendulum has two masses \(m\) each on rods of length \(l\) each. Set up the Lagrangian and find the frequencies of small oscillations.
Coordinates: \(\theta_1,\theta_2\) (angles of upper, lower rods). For small angles: \(T=\tfrac{1}{2}m[2l^2\dot\theta_1^2+l^2\dot\theta_2^2+2l^2\dot\theta_1\dot\theta_2]\), \(V=mgl[2(1-\cos\theta_1)+(1-\cos\theta_2)]\approx mgl[\theta_1^2+\tfrac{1}{2}\theta_2^2]\times$ (approx)$. More precisely: \(V\approx mgl\theta_1^2+\tfrac{mgl}{2}\theta_2^2\) (small angle). Equations: \(2\ddot\theta_1+\ddot\theta_2+\tfrac{2g}{l}\theta_1=0\), \(\ddot\theta_1+\ddot\theta_2+\tfrac{g}{l}\theta_2=0\). For normal modes \(\theta_i=A_i e^{i\omega t}\): characteristic equation \(\omega^4-4\tfrac{g}{l}\omega^2+2\tfrac{g^2}{l^2}=0\). Frequencies: \(\omega^2=\tfrac{g}{l}(2\pm\sqrt{2})\). \(\blacksquare\)
Obtain the Lagrangian for a spherical pendulum (mass \(m\), length \(l\)) and write the equations of motion.
Generalized coordinates: \(\theta\) (polar), \(\phi\) (azimuthal). \(T=\tfrac{ml^2}{2}(\dot\theta^2+\sin^2\theta\,\dot\phi^2)\), \(V=-mgl\cos\theta\). \(L=\tfrac{ml^2}{2}(\dot\theta^2+\sin^2\theta\,\dot\phi^2)+mgl\cos\theta\). Lagrange equations: \(ml^2\ddot\theta=ml^2\sin\theta\cos\theta\,\dot\phi^2-mgl\sin\theta\); \(\dfrac{d}{dt}(ml^2\sin^2\theta\,\dot\phi)=0\Rightarrow l_z=ml^2\sin^2\theta\,\dot\phi=\text{const}\) (conserved angular momentum about vertical). \(\blacksquare\)
A disk of mass \(m\), radius \(R\) rolls without slipping down an inclined plane of angle \(\alpha\). Find the acceleration using Lagrangian mechanics.
Generalized coordinate: \(x\) (distance down the plane). Rolling constraint: \(\dot x=R\dot\phi\Rightarrow\phi=x/R\). \(T=\tfrac{1}{2}m\dot x^2+\tfrac{1}{2}I\dot\phi^2\) where \(I=\tfrac{1}{2}mR^2\). So \(T=\tfrac{1}{2}m\dot x^2+\tfrac{1}{4}m\dot x^2=\tfrac{3}{4}m\dot x^2\). \(V=-mgx\sin\alpha\). Lagrange: \(\tfrac{3}{2}m\ddot x=mg\sin\alpha\Rightarrow\ddot x=\tfrac{2g\sin\alpha}{3}\). (Slower than free slide, due to rotational inertia.) \(\blacksquare\)
Write the Lagrangian for a particle of charge \(q\), mass \(m\) in electric field \(\mathbf{E}\) and magnetic field \(\mathbf{B}=\nabla\times\mathbf{A}\). Show that Lagrange's equations give the Lorentz force law.
Lagrangian: \(L=\tfrac{1}{2}m\dot{\mathbf{r}}^2-q\phi+q\dot{\mathbf{r}}\cdot\mathbf{A}\) where \(\mathbf{E}=-\nabla\phi-\partial\mathbf{A}/\partial t\), \(\mathbf{B}=\nabla\times\mathbf{A}\). Generalized momentum: \(p_i=m\dot x_i+qA_i\). Lagrange equation for \(x\)-component: \(\dfrac{d}{dt}(m\dot x+qA_x)=q\dfrac{\partial\phi}{\partial x}+q\dot{\mathbf{r}}\cdot\dfrac{\partial\mathbf{A}}{\partial x}\). Expanding \(\dfrac{dA_x}{dt}=\partial_t A_x+(\dot{\mathbf{r}}\cdot\nabla)A_x\) and using vector identities: \(m\ddot x=q(E_x+(\dot{\mathbf{r}}\times\mathbf{B})_x)\). Hence \(m\ddot{\mathbf{r}}=q(\mathbf{E}+\dot{\mathbf{r}}\times\mathbf{B})\) — the Lorentz force. \(\blacksquare\)
A rigid body moves freely in space. State the number of degrees of freedom. For \(N\) particles with \(k\) holonomic constraints, what is the number of generalized coordinates needed?
A free rigid body has 6 degrees of freedom: 3 translational (position of centre of mass) + 3 rotational (orientation). For \(N\) particles: \(3N\) Cartesian coordinates. With \(k\) independent holonomic constraints (each reduces DOF by 1): number of generalized coordinates \(= 3N - k\). E.g., rigid body of \(N\) particles has \(3N\) coordinates but \(\binom{N}{2}\) distances fixed → reduces to 6 DOF. \(\blacksquare\)
Two masses \(m\) are connected by springs of stiffness \(k\) (wall–mass1–spring–mass2–wall). Find the Lagrangian and normal mode frequencies.
Let \(x_1,x_2\) be displacements from equilibrium. \(T=\tfrac{m}{2}(\dot x_1^2+\dot x_2^2)\), \(V=\tfrac{k}{2}(x_1^2+(x_2-x_1)^2+x_2^2)=\tfrac{k}{2}(2x_1^2-2x_1x_2+2x_2^2)\). Equations: \(m\ddot x_1+k(2x_1-x_2)=0\), \(m\ddot x_2+k(2x_2-x_1)=0\). For \((x_1,x_2)\propto e^{i\omega t}\): \(\det\begin{pmatrix}2k-m\omega^2&-k\\-k&2k-m\omega^2\end{pmatrix}=0\Rightarrow(2k-m\omega^2)^2-k^2=0\Rightarrow m\omega^2=k\text{ or }3k\). Frequencies: \(\omega_1=\sqrt{k/m}\) (symmetric mode), \(\omega_2=\sqrt{3k/m}\) (antisymmetric). \(\blacksquare\)
A bead of mass \(m\) is constrained to a hoop of radius \(R\) rotating with angular velocity \(\omega\) about its vertical diameter. Find the equilibrium angle and stability condition.
Coordinate: \(\theta\) (angle from bottom). \(T=\tfrac{1}{2}mR^2(\dot\theta^2+\omega^2\sin^2\theta)\), \(V=-mgR\cos\theta\). Effective potential: \(V_{\text{eff}}=-mgR\cos\theta-\tfrac{1}{2}m\omega^2R^2\sin^2\theta\). Equation of motion: \(mR^2\ddot\theta=mR^2\omega^2\sin\theta\cos\theta-mgR\sin\theta=mR\sin\theta(R\omega^2\cos\theta-g)\). Equilibria: (i) \(\theta=0\) (bottom), (ii) \(\theta=\pi\) (top), (iii) \(\cos\theta=g/(R\omega^2)\) (for \(\omega^2>g/R\)). Bottom equilibrium is stable iff \(\omega^2
Show that for any function \(f(q,p,t)\), \(\dfrac{df}{dt}=\{f,H\}+\dfrac{\partial f}{\partial t}\), where \(\{f,H\}\) is the Poisson bracket.
\(\dfrac{df}{dt}=\sum_i\!\left(\dfrac{\partial f}{\partial q_i}\dot q_i+\dfrac{\partial f}{\partial p_i}\dot p_i\right)+\dfrac{\partial f}{\partial t}\). Using Hamilton's equations \(\dot q_i=\partial H/\partial p_i\), \(\dot p_i=-\partial H/\partial q_i\): \(\dfrac{df}{dt}=\sum_i\!\left(\dfrac{\partial f}{\partial q_i}\dfrac{\partial H}{\partial p_i}-\dfrac{\partial f}{\partial p_i}\dfrac{\partial H}{\partial q_i}\right)+\dfrac{\partial f}{\partial t}=\{f,H\}+\dfrac{\partial f}{\partial t}\). If \(f\) is a constant of motion: \(\{f,H\}+\partial f/\partial t=0\). \(\blacksquare\)
Show that the transformation \(Q=p, P=-q\) is canonical. Find the generating function.
A transformation is canonical if it preserves the Poisson bracket structure: \(\{Q,P\}=1\). Check: \(\{Q,P\}=\{p,-q\}=\dfrac{\partial p}{\partial q}\dfrac{\partial(-q)}{\partial p}-\dfrac{\partial p}{\partial p}\dfrac{\partial(-q)}{\partial q}=0-1\cdot(-1)=1\). \(\checkmark\) Generating function \(F_1(q,Q)\): \(p=\partial F_1/\partial q\), \(P=-\partial F_1/\partial Q\). Since \(Q=p\): \(p=Q\Rightarrow\partial F_1/\partial q=Q\Rightarrow F_1=qQ\). Check: \(P=-\partial F_1/\partial Q=-q\). \(\checkmark\) So \(F_1=qQ\). \(\blacksquare\)
State Liouville's theorem for Hamiltonian systems. Show that the phase space volume is conserved under Hamiltonian flow.
Liouville's theorem: The phase space distribution function \(\rho(q_i,p_i,t)\) satisfies \(d\rho/dt=0\) — the density of representative points in phase space is constant along trajectories. Proof: The divergence of the Hamiltonian flow field \((\dot q_i,\dot p_i)\) is \(\sum_i\!\left(\dfrac{\partial\dot q_i}{\partial q_i}+\dfrac{\partial\dot p_i}{\partial p_i}\right)=\sum_i\!\left(\dfrac{\partial^2 H}{\partial q_i\partial p_i}-\dfrac{\partial^2 H}{\partial p_i\partial q_i}\right)=0\). By the continuity equation in phase space: \(\dfrac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0\Rightarrow d\rho/dt=0\). Hence phase volume is conserved. \(\blacksquare\)
Write the Hamilton-Jacobi equation for a particle of mass \(m\) moving in a potential \(V(q)\). Show that for a harmonic oscillator it separates.
The Hamilton-Jacobi equation: \(H\!\left(q,\dfrac{\partial S}{\partial q},t\right)+\dfrac{\partial S}{\partial t}=0\). For time-independent \(H\): put \(S=W(q)-Et\). Then \(\dfrac{1}{2m}\!\left(\dfrac{dW}{dq}\right)^2+V(q)=E\Rightarrow\dfrac{dW}{dq}=\sqrt{2m(E-V(q))}\). For harmonic oscillator \(V=\tfrac{1}{2}kq^2\): \(W=\int\sqrt{2m(E-\tfrac{1}{2}kq^2)}\,dq\), which is elementary. This separates cleanly into a 1D integral. \(\blacksquare\)
Define action variable \(J=\oint p\,dq\) for a 1D harmonic oscillator \(H=p^2/(2m)+kq^2/2\). Show \(H=\omega J/(2\pi)\).
Phase curve: \(p^2/(2m)+kq^2/2=E\Rightarrow\dfrac{q^2}{2E/k}+\dfrac{p^2}{2mE}=1\) — an ellipse. Axes: \(a=\sqrt{2E/k},b=\sqrt{2mE}\). \(J=\oint p\,dq=\pi ab=\pi\sqrt{2E/k}\cdot\sqrt{2mE}=2\pi E\sqrt{m/k}=2\pi E/\omega\) where \(\omega=\sqrt{k/m}\). Hence \(E=H=J\omega/(2\pi)\). \(\blacksquare\)
Derive Binet's equation for a particle moving under a central force \(f(r)\). Hence find the orbit for inverse-square force.
In polar coords with \(l=mr^2\dot\theta\) (angular momentum per unit mass \(h=r^2\dot\theta\)), let \(u=1/r\). Then \(\dot r=-h\dfrac{du}{d\theta},\ddot r=-hu^2\dfrac{d^2u}{d\theta^2}\cdot h\). Radial equation: \(\ddot r-r\dot\theta^2=f(r)/m\). Substituting: \(-h^2u^2\dfrac{d^2u}{d\theta^2}-\dfrac{h^2u^2}{1/u}u^2=-h^2u^2\dfrac{d^2u}{d\theta^2}... \) Tidying: \(\dfrac{d^2u}{d\theta^2}+u=-\dfrac{f(1/u)}{mh^2u^2}\) — Binet's equation. For \(f=-\mu m/r^2=-\mu mu^2\): RHS \(=\mu/(h^2)\). So \(u''+u=\mu/h^2\). Solution: \(u=\mu/h^2+A\cos\theta\Rightarrow r=\dfrac{h^2/\mu}{1+e\cos\theta}\) — a conic (ellipse for \(e<1\)). \(\blacksquare\)
Derive Kepler's third law: the square of the period of an elliptical orbit is proportional to the cube of the semi-major axis.
Ellipse: semi-major \(a\), semi-minor \(b\). Area swept per unit time \(=h/2\) (Kepler II). Period: \(T=\dfrac{\text{Area}}{h/2}=\dfrac{\pi ab}{h/2}=\dfrac{2\pi ab}{h}\). For ellipse: \(b^2=a^2(1-e^2)\), \(h^2=\mu a(1-e^2)\) (from orbit equation). So \(h=\sqrt{\mu a(1-e^2)}\), \(b=a\sqrt{1-e^2}\). \(T=\dfrac{2\pi a\cdot a\sqrt{1-e^2}}{\sqrt{\mu a(1-e^2)}}=\dfrac{2\pi a^2\sqrt{1-e^2}}{\sqrt{\mu a}\sqrt{1-e^2}}=\dfrac{2\pi a^{3/2}}{\sqrt{\mu}}\). Hence \(T^2=\dfrac{4\pi^2}{\mu}a^3\propto a^3\). \(\blacksquare\)
For a particle in a central force \(F=-k/r^2\), find the apsidal distances (closest and farthest points of orbit) in terms of energy \(E\) and angular momentum \(L\).
At apsidal points \(\dot r=0\). Energy: \(E=\dfrac{1}{2}m\dot r^2+\dfrac{L^2}{2mr^2}-\dfrac{mk}{r}=\dfrac{L^2}{2mr^2}-\dfrac{mk}{r}\). Multiply by \(2mr^2/m\): \(2Er^2+2kr\cdot m-L^2/m=0\Rightarrow 2Er^2+2mkr-L^2/m=0\). Wait — correctly: \(Er^2+mkr-\dfrac{L^2}{2m}=0\). Apsidal distances: \(r=\dfrac{-mk\pm\sqrt{m^2k^2+2EL^2/m}}{2E}\). For bound orbits (\(E<0\)): \(r_{\min},r_{\max}=\dfrac{mk\mp\sqrt{m^2k^2-2|E|L^2/m}}{2|E|}\). \(\blacksquare\)
For a central force \(F(r)=-k/r^n\), find the condition on \(n\) for stability of circular orbits.
Effective potential: \(V_{\text{eff}}=\dfrac{L^2}{2mr^2}+V(r)\) where \(F=-dV/dr\). For circular orbit at \(r_0\): \(dV_{\text{eff}}/dr=0\Rightarrow-L^2/(mr_0^3)+F(r_0)/m=0\) (with force being centripetal). Stability: \(d^2V_{\text{eff}}/dr^2>0\) at \(r_0\). For \(F=-k/r^n\): \(V_{\text{eff}}=L^2/(2mr^2)-k/((n-1)r^{n-1})\) (for \(n\neq1\)). \(d^2V_{\text{eff}}/dr^2=3L^2/(mr_0^4)-n(n-1)k/r_0^{n+1}\). Using circular orbit condition to eliminate \(L^2\): condition becomes \(3-n>0\Rightarrow n<3\). Circular orbits stable for \(n<3\). \(\blacksquare\)
Write Euler's equations of motion for a rigid body. For a torque-free symmetric top (\(I_1=I_2\neq I_3\)), show that the angular velocity vector precesses about the symmetry axis.
Euler's equations: \(I_1\dot\Omega_1-(I_2-I_3)\Omega_2\Omega_3=N_1\), \(I_2\dot\Omega_2-(I_3-I_1)\Omega_3\Omega_1=N_2\), \(I_3\dot\Omega_3-(I_1-I_2)\Omega_1\Omega_2=N_3\). For torque-free (\(N_i=0\)) symmetric top (\(I_1=I_2=I\)): \(I\dot\Omega_1=(I-I_3)\Omega_2\Omega_3\), \(I\dot\Omega_2=(I_3-I)\Omega_3\Omega_1\), \(I_3\dot\Omega_3=0\Rightarrow\Omega_3=\text{const}\). Let \(\Omega_\perp^2=\Omega_1^2+\Omega_2^2\). Then \(\dot\Omega_1=-\Omega_b\Omega_2\), \(\dot\Omega_2=\Omega_b\Omega_1\) where \(\Omega_b=\dfrac{(I_3-I)\Omega_3}{I}\). Solution: \(\Omega_1+i\Omega_2=Ce^{i\Omega_b t}\) — precession of \(\boldsymbol{\Omega}\) about symmetry axis with body-frame frequency \(\Omega_b\). \(\blacksquare\)
Find the principal moments of inertia and principal axes of a uniform rectangular lamina of sides \(a\) and \(b\), mass \(M\), about the centre.
Take coordinate axes along the sides. \(I_{xx}=\int y^2\,dm=\dfrac{M}{ab}\int_{-b/2}^{b/2}\int_{-a/2}^{a/2}y^2\,dx\,dy=\dfrac{Mb^2}{12}\). Similarly \(I_{yy}=\dfrac{Ma^2}{12}\), \(I_{zz}=\dfrac{M(a^2+b^2)}{12}\) (perpendicular axis theorem). Products of inertia: \(I_{xy}=0\) (symmetry). So these axes are already principal axes. Principal moments: \(I_1=Mb^2/12,\; I_2=Ma^2/12,\; I_3=M(a^2+b^2)/12\). \(\blacksquare\)
Show that the kinetic energy of a rigid body rotating about a fixed point is \(T=\tfrac{1}{2}\boldsymbol{\Omega}\cdot\mathbf{L}\) where \(\mathbf{L}\) is angular momentum. Express in principal axes.
\(T=\tfrac{1}{2}\sum_\alpha m_\alpha v_\alpha^2=\tfrac{1}{2}\sum_\alpha m_\alpha(\boldsymbol{\Omega}\times\mathbf{r}_\alpha)^2\). Using \(|\boldsymbol{\Omega}\times\mathbf{r}|^2=\Omega^2r^2-(\boldsymbol{\Omega}\cdot\mathbf{r})^2\): \(T=\tfrac{1}{2}\sum_\alpha m_\alpha[\Omega^2r_\alpha^2-(\boldsymbol{\Omega}\cdot\mathbf{r}_\alpha)^2]=\tfrac{1}{2}\boldsymbol{\Omega}\cdot\mathbf{I}\boldsymbol{\Omega}\). Also \(\mathbf{L}=\mathbf{I}\boldsymbol{\Omega}\), so \(T=\tfrac{1}{2}\boldsymbol{\Omega}\cdot\mathbf{L}\). In principal axes: \(T=\tfrac{1}{2}(I_1\Omega_1^2+I_2\Omega_2^2+I_3\Omega_3^2)\). \(\blacksquare\)
For a symmetric top spinning with angular velocity \(\Omega_s\) and spinning axis at angle \(\theta\) to vertical, find the precession rate \(\dot\phi\) (slow precession).
For the heavy symmetric top (one point fixed), in steady precession (fast spin, \(\dot\theta=0\)): torque about fixed point \(= Mgl\sin\theta\) (horizontal). Angular momentum (dominated by spin) \(= I_3\Omega_s\) along symmetry axis. Rate of change of \(\mathbf{L}\): \(|\dot{\mathbf{L}}|=L_\text{spin}\sin\theta\cdot\dot\phi=I_3\Omega_s\sin\theta\cdot\dot\phi\). Setting equal to torque: \(I_3\Omega_s\sin\theta\cdot\dot\phi=Mgl\sin\theta\Rightarrow\dot\phi=\dfrac{Mgl}{I_3\Omega_s}\). Precession rate is independent of \(\theta\). \(\blacksquare\)
Define the three Euler angles \((\phi,\theta,\psi)\). Express the angular velocity components along body-fixed principal axes in terms of Euler angles and their derivatives.
The three Euler angles describe orientation: \(\phi\) (precession — rotation about space \(z\)-axis), \(\theta\) (nutation — rotation about nodal line), \(\psi\) (spin — rotation about body \(z\)-axis). Angular velocity components along body axes: \(\Omega_1=\dot\phi\sin\theta\sin\psi+\dot\theta\cos\psi\), \(\Omega_2=\dot\phi\sin\theta\cos\psi-\dot\theta\sin\psi\), \(\Omega_3=\dot\phi\cos\theta+\dot\psi\). These follow from decomposing \(\dot\phi\hat{z}_{\text{space}},\dot\theta\hat{e}_{\text{node}},\dot\psi\hat{z}_{\text{body}}\) into body-frame components. \(\blacksquare\)
Define the momental ellipsoid (Poinsot ellipsoid). State Poinsot's theorem for torque-free rigid body motion.
The momental ellipsoid at a point \(O\) is the surface \(\mathbf{r}^T\mathbf{I}\mathbf{r}=1\) in body coordinates, i.e., \(I_1x^2+I_2y^2+I_3z^2=1\) (principal axes). Poinsot's theorem: For a torque-free rigid body with one point fixed, the momental ellipsoid rolls without slipping on a fixed plane (the invariable plane perpendicular to \(\mathbf{L}\)). The contact point traces the herpolhode on the invariable plane, and the polhode on the momental ellipsoid. This gives a geometric description of torque-free precession. \(\blacksquare\)
Two identical pendulums of length \(l\), mass \(m\) are coupled by a spring of stiffness \(k\) connecting the bobs. Find the normal mode frequencies.
Coordinates: \(x_1,x_2\) (horizontal displacements). \(T=\tfrac{m}{2}(\dot x_1^2+\dot x_2^2)\), \(V=\tfrac{mg}{2l}(x_1^2+x_2^2)+\tfrac{k}{2}(x_2-x_1)^2\). Equations: \(m\ddot x_1+\left(\dfrac{mg}{l}+k\right)x_1-kx_2=0\), \(m\ddot x_2+\left(\dfrac{mg}{l}+k\right)x_2-kx_1=0\). Normal modes: add/subtract equations. Mode 1 (\(x_1=x_2\)): \(\omega_1^2=g/l\). Mode 2 (\(x_1=-x_2\)): \(\omega_2^2=g/l+2k/m\). \(\blacksquare\)
For a system with kinetic matrix \(\mathbf{T}\) and potential matrix \(\mathbf{V}\), show that normal mode frequencies satisfy \(\det(\mathbf{V}-\omega^2\mathbf{T})=0\).
Equations of motion in matrix form: \(\mathbf{T}\ddot{\mathbf{q}}+\mathbf{V}\mathbf{q}=0\). Try \(\mathbf{q}=\mathbf{a}e^{i\omega t}\): \(-\omega^2\mathbf{T}\mathbf{a}+\mathbf{V}\mathbf{a}=0\Rightarrow(\mathbf{V}-\omega^2\mathbf{T})\mathbf{a}=0\). Non-trivial solutions exist iff \(\det(\mathbf{V}-\omega^2\mathbf{T})=0\) — the secular equation. This is an eigenvalue problem (\(\mathbf{T}^{-1}\mathbf{V}\mathbf{a}=\omega^2\mathbf{a}\)) with \(n\) positive eigenvalues \(\omega_i^2\) for a stable system. The eigenvectors \(\mathbf{a}^{(i)}\) are the normal mode amplitudes. \(\blacksquare\)
A linear symmetric triatomic molecule (like CO₂) has masses \(m, M, m\) connected by springs of stiffness \(k\). Find the normal modes of longitudinal vibration.
Displacements: \(x_1,x_2,x_3\). \(V=\tfrac{k}{2}[(x_2-x_1)^2+(x_3-x_2)^2]\). Equations: \(m\ddot x_1=k(x_2-x_1)\), \(M\ddot x_2=k(x_1-2x_2+x_3)\), \(m\ddot x_3=k(x_2-x_3)\). Mode 1: \(x_1=x_3,x_2\) independent — translation (\(\omega=0\)). Mode 2: \(x_1=-x_3,x_2=0\) — antisymmetric, \(\omega_2^2=k/m\). Mode 3: \(x_1=x_3,Mx_2=-2mx_1\) — symmetric stretch, \(\omega_3^2=k(2m+M)/(mM)\). \(\blacksquare\)
Write the Lorentz transformation equations for a frame \(S'\) moving with velocity \(v\) along the \(x\)-axis relative to \(S\). Derive the time dilation formula.
Lorentz transformation (\(\gamma=1/\sqrt{1-v^2/c^2}\)): \(x'=\gamma(x-vt)\), \(y'=y\), \(z'=z\), \(t'=\gamma(t-vx/c^2)\). Time dilation: Consider two events at same location in \(S'\): \(\Delta x'=0\). Then \(\Delta x=v\Delta t'\cdot\gamma\cdot\)... alternatively: \(\Delta t'=\gamma(t_2-vx/c^2)-\gamma(t_1-vx/c^2)=\gamma\Delta t\). Since clock at rest in \(S'\) reads proper time \(\Delta\tau=\Delta t'\), and \(S\) reads \(\Delta t=\gamma\Delta\tau\). So moving clocks run slow: \(\Delta t=\gamma\Delta\tau>\Delta\tau\). \(\blacksquare\)
Derive the length contraction formula. A rod at rest in \(S'\) has length \(L_0\). What is its length as measured in \(S\)?
Rod at rest in \(S'\): endpoints at \(x_1'=0,x_2'=L_0\). In \(S\), measure endpoints simultaneously (\(t_1=t_2=t\)): \(x_1'=\gamma(x_1-vt)\), \(x_2'=\gamma(x_2-vt)\). So \(L_0=x_2'-x_1'=\gamma(x_2-x_1)=\gamma L\). Hence \(L=L_0/\gamma=L_0\sqrt{1-v^2/c^2}
Derive \(E=mc^2\) from relativistic mechanics. Show that the total relativistic energy is \(E=\gamma m_0c^2\).
Relativistic momentum: \(\mathbf{p}=\gamma m_0\mathbf{v}\). Work done: \(dW=\mathbf{F}\cdot d\mathbf{s}=\dfrac{d\mathbf{p}}{dt}\cdot\mathbf{v}\,dt=\mathbf{v}\cdot d(\gamma m_0\mathbf{v})\). Computing: \(W=\int_0^v\dfrac{m_0v}{\sqrt{1-v^2/c^2}}dv\bigg|_0^v\). Let \(u=v^2/c^2\): \(W=m_0c^2\int_0^{v^2/c^2}\dfrac{du}{2\sqrt{1-u}}=m_0c^2\big[-\sqrt{1-u}\big]_0^{v^2/c^2}=m_0c^2(\gamma-1)\). Total energy \(=\) rest energy + kinetic: \(E=m_0c^2+W=\gamma m_0c^2\). At rest (\(v=0\)): \(E_0=m_0c^2\). \(\blacksquare\)
Show that \(E^2=p^2c^2+m_0^2c^4\) where \(p=|\mathbf{p}|\) is relativistic momentum.
\(E=\gamma m_0c^2\), \(p=\gamma m_0v\). Then \(E^2-p^2c^2=\gamma^2m_0^2c^4-\gamma^2m_0^2v^2c^2=\gamma^2m_0^2c^2(c^2-v^2)=\dfrac{m_0^2c^2(c^2-v^2)}{1-v^2/c^2}=m_0^2c^4\). Hence \(E^2=p^2c^2+m_0^2c^4\). For massless particles (\(m_0=0\)): \(E=pc\) (photon). \(\blacksquare\)
If a particle moves with velocity \(u'\) in the \(x'\)-direction in frame \(S'\), and \(S'\) moves with velocity \(v\) relative to \(S\), find the velocity \(u\) of the particle in \(S\).
From Lorentz transformation: \(dx=\gamma(dx'+v\,dt')\), \(dt=\gamma(dt'+v\,dx'/c^2)\). \(u=\dfrac{dx}{dt}=\dfrac{dx'+v\,dt'}{dt'+v\,dx'/c^2}=\dfrac{u'+v}{1+u'v/c^2}\). This is the relativistic velocity addition formula. Checks: (i) if \(u'=c\): \(u=c(1+v/c)/(1+v/c)=c\) — speed of light is invariant. (ii) if \(u'\ll c,v\ll c\): \(u\approx u'+v\) — Galilean result. \(\blacksquare\)
Two events are simultaneous in frame \(S\). Show they are generally not simultaneous in \(S'\). Under what condition is causal order preserved in all frames?
Events \(A(t,x_1)\) and \(B(t,x_2)\) are simultaneous in \(S\) (\(\Delta t=0\)). In \(S'\): \(\Delta t'=\gamma(\Delta t-v\Delta x/c^2)=-\gamma v\Delta x/c^2\neq0\) (unless \(\Delta x=0\)). So simultaneity is relative. Causal order: Event A causes B requires \(\Delta s^2=c^2\Delta t^2-\Delta x^2>0\) (timelike separation). For timelike intervals: \(\text{sign}(\Delta t)\) is Lorentz-invariant. For spacelike (\(\Delta s^2<0\)): order can be reversed; but these cannot have causal connection since signal would need to travel faster than \(c\). \(\blacksquare\)
State and prove Kelvin's circulation theorem: for an ideal fluid, the circulation \(\Gamma=\oint_C\mathbf{v}\cdot d\mathbf{l}\) around a material loop is conserved.
\(\dfrac{D\Gamma}{Dt}=\dfrac{D}{Dt}\oint_C\mathbf{v}\cdot d\mathbf{l}=\oint_C\dfrac{D\mathbf{v}}{Dt}\cdot d\mathbf{l}+\oint_C\mathbf{v}\cdot\dfrac{D(d\mathbf{l})}{Dt}\). Second integral: \(\oint_C\mathbf{v}\cdot d\mathbf{v}=\oint_C d(v^2/2)=0\) (closed loop). First integral: using Euler's equation \(D\mathbf{v}/Dt=-\nabla p/\rho+\mathbf{g}=-\nabla(p/\rho+gz)\) (for barotropic: \(p=p(\rho)\)): \(\oint_C(-\nabla p/\rho)\cdot d\mathbf{l}=0\) (gradient around closed loop). Hence \(D\Gamma/Dt=0\). \(\blacksquare\)
State Helmholtz's vortex theorems for ideal fluid. Show that vortex lines move with the fluid.
Helmholtz's theorems: (1) The strength of a vortex tube (flux of vorticity \(\boldsymbol{\omega}=\nabla\times\mathbf{v}\) through any cross-section) is constant along the tube. (2) Vortex tubes move with the fluid (material lines); vortex lines cannot end in the fluid interior. (3) Strength of vortex tube is conserved in time. Proof of (2): Define \(\boldsymbol{\omega}=\nabla\times\mathbf{v}\). The vorticity equation: \(\dfrac{D\boldsymbol{\omega}}{Dt}=(\boldsymbol{\omega}\cdot\nabla)\mathbf{v}\) (for incompressible ideal fluid). This is the same equation as for material line elements, showing vortex lines are frozen into the fluid. \(\blacksquare\)
Derive the vorticity equation \(\dfrac{D\boldsymbol{\omega}}{Dt}=(\boldsymbol{\omega}\cdot\nabla)\mathbf{v}\) for incompressible inviscid flow.
Take curl of Euler's equation: \(\partial_t\mathbf{v}+(\mathbf{v}\cdot\nabla)\mathbf{v}=-\nabla p/\rho\). Using \((\mathbf{v}\cdot\nabla)\mathbf{v}=\nabla(v^2/2)-\mathbf{v}\times\boldsymbol{\omega}\): \(\partial_t\mathbf{v}+\nabla(v^2/2)-\mathbf{v}\times\boldsymbol{\omega}=-\nabla p/\rho\). Take curl: \(\partial_t\boldsymbol{\omega}-\nabla\times(\mathbf{v}\times\boldsymbol{\omega})=0\). Expand \(\nabla\times(\mathbf{v}\times\boldsymbol{\omega})=(\boldsymbol{\omega}\cdot\nabla)\mathbf{v}-(\mathbf{v}\cdot\nabla)\boldsymbol{\omega}+\mathbf{v}(\nabla\cdot\boldsymbol{\omega})-\boldsymbol{\omega}(\nabla\cdot\mathbf{v})\). Since \(\nabla\cdot\boldsymbol{\omega}=0,\nabla\cdot\mathbf{v}=0\): \(\dfrac{D\boldsymbol{\omega}}{Dt}=\partial_t\boldsymbol{\omega}+(\mathbf{v}\cdot\nabla)\boldsymbol{\omega}=(\boldsymbol{\omega}\cdot\nabla)\mathbf{v}\). \(\blacksquare\)
Find the velocity potential for steady irrotational flow of an ideal fluid past a sphere of radius \(a\) with uniform stream \(U\) at infinity.
In spherical coordinates. Boundary conditions: \(\phi\to Ur\cos\theta\) as \(r\to\infty\); \(\partial\phi/\partial r=0\) at \(r=a\). Laplace equation solution: \(\phi=(Ar+B/r^2)\cos\theta\). At \(\infty\): \(A=U\). At \(r=a\): \(U-2B/a^3=0\Rightarrow B=Ua^3/2\). \[\phi=U\!\left(r+\dfrac{a^3}{2r^2}\right)\cos\theta\]. Velocity components: \(v_r=\partial\phi/\partial r=U\!\left(1-\dfrac{a^3}{r^3}\right)\cos\theta\), \(v_\theta=\dfrac{1}{r}\partial\phi/\partial\theta=-U\!\left(1+\dfrac{a^3}{2r^3}\right)\sin\theta\). On sphere: \(v_r=0\), \(v_\theta=-\tfrac{3}{2}U\sin\theta\). \(\blacksquare\)
State the Kutta-Joukowski theorem. Calculate the lift per unit length on a cylinder with circulation \(\Gamma\) in a stream of speed \(U\).
Kutta-Joukowski theorem: For a 2D body in uniform flow of speed \(U\) with circulation \(\Gamma\): lift per unit length \(L=\rho U\Gamma\). Derivation for cylinder: complex potential \(w=U(z+a^2/z)-i\Gamma/(2\pi)\ln z\). Velocity on \(r=a\): \(|dw/dz|^2=\cdots\). By Blasius' theorem: force per unit length \(= \rho i/2\oint(dw/dz)^2 dz\). Computing the residue at \(z=0\): lift \(=\rho U\Gamma\), drag \(=0\) (D'Alembert paradox). \(\blacksquare\)
Write the Navier-Stokes equations for viscous incompressible flow. State the physical meaning of each term.
\[\rho\!\left(\dfrac{\partial\mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}\right)=-\nabla p+\mu\nabla^2\mathbf{v}+\rho\mathbf{g}\] Terms: \(\rho\partial_t\mathbf{v}\) — inertial (unsteady); \(\rho(\mathbf{v}\cdot\nabla)\mathbf{v}\) — convective inertia (nonlinear); \(-\nabla p\) — pressure gradient; \(\mu\nabla^2\mathbf{v}\) — viscous diffusion (\(\mu\) = dynamic viscosity); \(\rho\mathbf{g}\) — body force. Continuity (incompressible): \(\nabla\cdot\mathbf{v}=0\). Euler equation is recovered when \(\mu=0\). \(\blacksquare\)
Derive Poiseuille's formula for laminar viscous flow through a circular pipe of radius \(R\) and length \(L\) under pressure difference \(\Delta p\).
Fully developed laminar flow: \(v=v(r)\hat{z}\). Navier-Stokes reduces to: \(\dfrac{1}{r}\dfrac{d}{dr}\!\left(r\dfrac{dv}{dr}\right)=-\dfrac{\Delta p}{\mu L}\equiv-\dfrac{P}{\mu}\). Integrate: \(r\dfrac{dv}{dr}=-\dfrac{Pr^2}{2\mu}+C_1\). Regularity at \(r=0\Rightarrow C_1=0\). Integrate again: \(v=-\dfrac{Pr^2}{4\mu}+C_2\). No-slip at \(r=R\): \(v(R)=0\Rightarrow C_2=PR^2/(4\mu)\). \[v(r)=\dfrac{P}{4\mu}(R^2-r^2)\text{ (parabolic profile)}\]. Volume flux: \(Q=\int_0^R v\cdot2\pi r\,dr=\dfrac{\pi P R^4}{8\mu}=\dfrac{\pi\Delta p R^4}{8\mu L}\). This is Poiseuille's formula. \(\blacksquare\)
State Stokes' law for drag on a sphere at low Reynolds number. Derive the terminal velocity of a sphere falling in a viscous fluid.
Stokes' law: For a sphere of radius \(a\) moving slowly (\(Re\ll1\)) in fluid of viscosity \(\mu\) with velocity \(U\): drag force \(F_D=6\pi\mu a U\). Terminal velocity: At terminal velocity, drag + buoyancy = weight. \(6\pi\mu a v_t+\tfrac{4}{3}\pi a^3\rho_f g=\tfrac{4}{3}\pi a^3\rho_s g\). \[v_t=\dfrac{2a^2(\rho_s-\rho_f)g}{9\mu}\]. Derivation of Stokes' law: solve Stokes equations (\(\nabla p=\mu\nabla^2\mathbf{v},\nabla\cdot\mathbf{v}=0\)) around sphere → stream function solution → integrate surface stress → \(F=6\pi\mu aU\). \(\blacksquare\)
Define Reynolds number. What is its physical significance? State the critical value separating laminar from turbulent flow in a pipe.
\(Re=\dfrac{\rho v L}{\mu}=\dfrac{vL}{\nu}\) where \(\nu=\mu/\rho\) is kinematic viscosity, \(L\) a characteristic length scale, \(v\) a characteristic velocity. Physical significance: \(Re=\dfrac{\text{inertial forces}}{\text{viscous forces}}\). Small \(Re\): viscous dominates → laminar, smooth, predictable. Large \(Re\): inertia dominates → turbulence, chaotic mixing. For pipe flow: critical \(Re\approx2300\). Below 2300: laminar (Poiseuille). Above 4000: turbulent. 2300–4000: transitional. \(\blacksquare\)
Viscous fluid flows between two parallel plates: lower plate fixed, upper plate moving at speed \(U\). Find the velocity profile (Couette flow).
Let \(y\in[0,h]\), flow in \(x\)-direction, no pressure gradient. N-S: \(\mu d^2v/dy^2=0\Rightarrow v=Ay+B\). BCs: \(v(0)=0\Rightarrow B=0\); \(v(h)=U\Rightarrow A=U/h\). \[v(y)=\dfrac{Uy}{h}\] — linear profile. Shear stress: \(\tau=\mu dv/dy=\mu U/h\) (constant throughout). \(\blacksquare\)
Derive Euler's equation of motion for an ideal (inviscid) fluid from Newton's second law applied to a fluid element.
Consider a fluid element of volume \(\delta V\) at position \(\mathbf{r}\). Mass: \(\delta m=\rho\delta V\). Forces: pressure \(-\nabla p\cdot\delta V\), gravity \(\rho\mathbf{g}\delta V\) (no viscosity for ideal fluid). Newton's second law: \(\rho\delta V\dfrac{D\mathbf{v}}{Dt}=(-\nabla p+\rho\mathbf{g})\delta V\). Dividing by \(\delta V\): \[\rho\dfrac{D\mathbf{v}}{Dt}=\rho\!\left(\dfrac{\partial\mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}\right)=-\nabla p+\rho\mathbf{g}\]. Using \(\mathbf{g}=-\nabla(gz)\): \(\dfrac{\partial\mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}=-\dfrac{\nabla p}{\rho}-\nabla(gz)\). \(\blacksquare\)
Derive the equation of continuity for a compressible fluid. Reduce to the incompressible case.
Rate of mass increase in control volume \(V\): \(\dfrac{\partial}{\partial t}\int_V\rho\,dV=-\oint_S\rho\mathbf{v}\cdot\hat{n}\,dA\) (mass flux out). By divergence theorem: \(\int_V\!\left(\dfrac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})\right)dV=0\). Since this holds for any \(V\): \[\dfrac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0\]. Expanding: \(\dfrac{\partial\rho}{\partial t}+\mathbf{v}\cdot\nabla\rho+\rho\nabla\cdot\mathbf{v}=0\Rightarrow\dfrac{D\rho}{Dt}+\rho\nabla\cdot\mathbf{v}=0\). For incompressible (\(D\rho/Dt=0\)): \(\nabla\cdot\mathbf{v}=0\). \(\blacksquare\)
Show that a doublet (source-sink pair at distance \(\epsilon\to0\) with \(m\epsilon\to\mu\)) has complex potential \(w=-\mu/(2\pi z)\). Find the flow pattern.
Source of strength \(m\) at origin: \(w_s=\dfrac{m}{2\pi}\ln z\). Sink of strength \(m\) at \(z=\epsilon\): \(w_k=-\dfrac{m}{2\pi}\ln(z-\epsilon)\). Combined: \(w=\dfrac{m}{2\pi}\ln\dfrac{z}{z-\epsilon}=\dfrac{m}{2\pi}\ln\!\left(1-\dfrac{\epsilon}{z-\epsilon}\right)\approx\dfrac{m}{2\pi}\cdot\left(-\dfrac{\epsilon}{z}\right)=-\dfrac{m\epsilon}{2\pi z}\). As \(\epsilon\to0,m\to\infty\) with \(\mu=m\epsilon\): \(w=-\dfrac{\mu}{2\pi z}\). In polar form: \(\phi=-\dfrac{\mu\cos\theta}{2\pi r}\), streamlines are circles through origin. \(\blacksquare\)
A source of strength \(m\) is at distance \(d\) from a rigid plane wall. Find the image system and resultant velocity potential.
Let the wall be the \(y\)-axis, source at \((d,0)\). By the method of images: place an equal source of strength \(m\) at \((-d,0)\) (mirror image). Velocity potential: \(\phi=\dfrac{m}{2\pi}\ln r_1+\dfrac{m}{2\pi}\ln r_2=\dfrac{m}{2\pi}\ln(r_1r_2)\) where \(r_1^2=(x-d)^2+y^2\), \(r_2^2=(x+d)^2+y^2\). On wall (\(x=0\)): \(r_1=r_2\Rightarrow\) normal velocity \(=\partial\phi/\partial x|_{x=0}=0\). \(\checkmark\) The fluid is attracted toward the wall (image source creates attraction). \(\blacksquare\)
The Joukowski transformation \(\zeta=z+a^2/z\) maps a circle to an aerofoil. Find the flow past a flat plate of length \(4a\) using this transformation.
Joukowski transform: \(\zeta=z+a^2/z\). Circle \(|z|=a\) maps to segment \([-2a,2a]\) on real axis (flat plate of length \(4a\)). Flow past circle \(|z|=a\) in \(z\)-plane: \(w_z=U(z+a^2/z)\). Under transform, this becomes flow past flat plate in \(\zeta\)-plane. Velocity: \(dw/d\zeta=(dw/dz)/(d\zeta/dz)=U(1-a^2/z^2)/(1-a^2/z^2)=U\) away from plate edges. Stagnation points at \(z=\pm a\) map to \(\zeta=\pm2a\) (plate ends). With circulation added (Kutta condition): lift \(=\rho U\Gamma\). \(\blacksquare\)
For the complex potential \(w=U\!\left(z+\dfrac{a^2}{z}\right)-\dfrac{i\Gamma}{2\pi}\ln z\), find all stagnation points.
Complex velocity: \(\dfrac{dw}{dz}=U\!\left(1-\dfrac{a^2}{z^2}\right)-\dfrac{i\Gamma}{2\pi z}=0\). Let \(z=ae^{i\alpha}\) on the cylinder. Then \(\dfrac{dw}{dz}=U(1-e^{-2i\alpha})-\dfrac{i\Gamma}{2\pi ae^{i\alpha}}\). Multiplying through by \(z^2\): \(Uz^2-i\Gamma z/(2\pi)-Ua^2=0\Rightarrow z=\dfrac{i\Gamma/(2\pi U)\pm\sqrt{-\Gamma^2/(4\pi^2U^2)+4a^2}}{2}\). For \(|\Gamma|<4\pi Ua\): stagnation points on cylinder surface. For \(|\Gamma|=4\pi Ua\): single stagnation point at \(z=ia\). For \(|\Gamma|>4\pi Ua\): stagnation point off the cylinder. \(\blacksquare\)
The transformation \(w=z^n\) maps a flow. Show that \(w=z^2\) gives the flow in a right-angled corner. Find the velocity components.
Let \(w=z^2=(x+iy)^2=x^2-y^2+2ixy\). So \(\phi=x^2-y^2,\psi=2xy\). Velocity: \(u=\phi_x=2x,v=\phi_y=-2y\). The streamlines \(\psi=2xy=C\) are rectangular hyperbolas. The \(x\)- and \(y\)-axes (\(\psi=0\)) are walls; they meet at right angles, forming the corner. On \(y=0\): velocity is \((2x,0)\) — flow along the wall. Stagnation point at origin. This represents ideal fluid flow into a 90° corner. \(\blacksquare\)
For a fluid rotating as a rigid body with angular velocity \(\Omega\) about the \(z\)-axis, find the pressure distribution using Bernoulli's equation in the rotating frame.
In rotating frame, centrifugal acceleration \(=\Omega^2\mathbf{r}_\perp\) acts outward. Effective gravity: \(\mathbf{g}_{\text{eff}}=-g\hat{z}+\Omega^2(x\hat{x}+y\hat{y})\). For rigid body rotation \(\mathbf{v}=\boldsymbol{\Omega}\times\mathbf{r}\): fluid at rest in rotating frame. Bernoulli in rotating frame: \(\dfrac{p}{\rho}-\dfrac{1}{2}\Omega^2r^2+gz=\text{const}\) where \(r^2=x^2+y^2\). Hence \(p=p_0+\dfrac{1}{2}\rho\Omega^2r^2-\rho gz\). Free surface (\(p=p_{\text{atm}}\)): \(z=\Omega^2r^2/(2g)+z_0\) — a paraboloid. \(\blacksquare\)
Explain the working of a Pitot tube. Derive the formula for flow speed measurement.
A Pitot tube has two openings: (1) a stagnation point hole facing flow — measures total (stagnation) pressure \(p_0\); (2) a static pressure hole on the side — measures static pressure \(p_s\). Apply Bernoulli between far-upstream (velocity \(U\), pressure \(p_s\)) and stagnation point (\(v=0\), pressure \(p_0\)): \(p_s+\tfrac{1}{2}\rho U^2=p_0\). Hence \[U=\sqrt{\dfrac{2(p_0-p_s)}{\rho}}\]. The pressure difference \(p_0-p_s\) is measured by a manometer. \(\blacksquare\)
Water flows over a broad-crested weir. Using Bernoulli's theorem, show that the flow per unit width is \(q=\tfrac{2}{3}C_d\sqrt{2g}\,H^{3/2}\) where \(H\) is the head.
Consider flow depth \(y\) at the weir crest. Bernoulli from reservoir (depth \(H\), velocity \(\approx0\)) to crest: \(H=y+v^2/(2g)\Rightarrow v=\sqrt{2g(H-y)}\). Flow per unit width: \(q=\int_0^H v\,dy=\int_0^H\sqrt{2g(H-y)}\,dy=\sqrt{2g}\cdot\left[-\dfrac{2}{3}(H-y)^{3/2}\right]_0^H=\dfrac{2}{3}\sqrt{2g}\,H^{3/2}\). Including discharge coefficient \(C_d\): \(q=\dfrac{2}{3}C_d\sqrt{2g}\,H^{3/2}\). \(\blacksquare\)
Explain the Magnus effect using fluid dynamics. A cylinder of radius \(a\) rotates with angular velocity \(\omega\) in a stream of speed \(U\). Find the lift.
The rotating cylinder drags fluid, creating circulation \(\Gamma=2\pi a^2\omega\). Complex potential: \(w=U(z+a^2/z)-\dfrac{i\Gamma}{2\pi}\ln z\). By Kutta-Joukowski theorem: lift per unit length \(L=\rho U\Gamma=\rho U\cdot 2\pi a^2\omega=2\pi\rho U a^2\omega\). Physical explanation: on one side, fluid speed \(=U+a\omega\) (pressure lower); on other side, speed \(=U-a\omega\) (pressure higher). Net pressure difference → lift (Magnus effect, exploited in spinning balls and Flettner rotors). \(\blacksquare\)
State D'Alembert's paradox. Explain why an ideal fluid exerts zero drag on a body, and how viscosity resolves the paradox.
D'Alembert's paradox: A body moving through an ideal (inviscid, incompressible) fluid at steady state experiences zero drag force. Proof (for cylinder): The pressure distribution \(p=p_\infty+\tfrac{1}{2}\rho U^2(1-4\sin^2\theta)\) is symmetric fore-and-aft (\(\theta\to\pi-\theta\) symmetry). Integrating \(-p\cos\theta\) over the surface: \(F_x=0\). Resolution by viscosity: Real fluids have viscosity → boundary layer forms → flow separates at the rear → wake (lower pressure behind body) → net pressure drag. At high \(Re\): wake is wide and turbulent → large drag. \(\blacksquare\)
For a torque-free rigid body with \(I_1\neq I_2\neq I_3\), describe qualitatively the stability of rotation about each principal axis.
Euler's equations (torque-free): \(I_1\dot\Omega_1=(I_2-I_3)\Omega_2\Omega_3\), and cyclic. Stability analysis: For rotation about axis 1 (\(\Omega_1=\Omega_0,\Omega_2,\Omega_3\ll\Omega_0\)), linearise: \(\ddot\Omega_2=-\dfrac{(I_3-I_1)(I_1-I_2)}{I_2I_3}\Omega_0^2\Omega_2\). Stable if coefficient negative, i.e., \((I_3-I_1)(I_1-I_2)>0\). This holds when \(I_1\) is the largest or smallest moment. Rotation about the intermediate axis (\(I_2\)) is always unstable (tennis racket theorem). \(\blacksquare\)
A rigid body of mass \(M\) oscillates about a horizontal pivot. If the moment of inertia about the pivot is \(I\) and the distance from pivot to centre of mass is \(d\), find the period and the equivalent simple pendulum length.
Torque about pivot: \(\tau=-Mgd\sin\theta\approx-Mgd\theta\). Equation: \(I\ddot\theta+Mgd\theta=0\Rightarrow\ddot\theta+\dfrac{Mgd}{I}\theta=0\). Period: \(T=2\pi\sqrt{I/(Mgd)}\). Equivalent simple pendulum length: \(L_{\text{eq}}=I/(Md)\). By parallel axis theorem: \(I=MK^2+Md^2\) where \(K\) is radius of gyration about CM. So \(L_{\text{eq}}=K^2/d+d\geq2K\) (minimum at \(d=K\)). \(\blacksquare\)
Three equal masses \(m\) on a line are connected by identical springs of stiffness \(k\) (wall–mass–spring–mass–spring–mass–wall). Find the three normal mode frequencies.
Equations: \(m\ddot x_1=k(-2x_1+x_2)\), \(m\ddot x_2=k(x_1-2x_2+x_3)\), \(m\ddot x_3=k(x_2-2x_3)\). Characteristic equation for \(\lambda=m\omega^2/k\): \(\det\begin{pmatrix}2-\lambda&-1&0\\-1&2-\lambda&-1\\0&-1&2-\lambda\end{pmatrix}=0\). Expanding: \((2-\lambda)[(2-\lambda)^2-1]-(-(-(2-\lambda)))=\cdots\). Eigenvalues: \(\lambda=2-\sqrt{2},2,2+\sqrt{2}\). Frequencies: \(\omega_1=\sqrt{(2-\sqrt{2})k/m},\omega_2=\sqrt{2k/m},\omega_3=\sqrt{(2+\sqrt{2})k/m}\). \(\blacksquare\)
Find the moment of inertia of a uniform solid sphere of mass \(M\), radius \(R\) about (i) a diameter, (ii) a tangent.
(i) About diameter (say \(z\)-axis): by shells — thin disk at height \(z\) has radius \(r=\sqrt{R^2-z^2}\), area \(\pi r^2\), mass \(dm=\rho\pi r^2\,dz\). \(I_{disk}=\tfrac{1}{2}dm\cdot r^2\). \(I_z=\int_{-R}^R\tfrac{\rho\pi}{2}(R^2-z^2)^2dz=\tfrac{\rho\pi}{2}\cdot\tfrac{16R^5}{15}=\dfrac{2}{5}MR^2\) (using \(M=\tfrac{4}{3}\pi R^3\rho\)). (ii) By parallel axis theorem: tangent is at distance \(R\) from centre: \(I_t=I_{\text{diam}}+MR^2=\dfrac{2}{5}MR^2+MR^2=\dfrac{7}{5}MR^2\). \(\blacksquare\)
Describe nutation in a symmetric top under gravity. What is the condition for steady precession (no nutation)?
For heavy symmetric top (one point fixed): Euler-Lagrange equations in Euler angles give three conserved quantities: \(E\) (energy), \(p_\phi=\partial L/\partial\dot\phi\) (component of \(\mathbf{L}\) along space \(z\)), \(p_\psi=\partial L/\partial\dot\psi=I_3\Omega_3\) (spin component). Motion in \(\theta\) (nutation) satisfies \(U_{\text{eff}}(\theta)\) equation. Nutation: oscillation of \(\theta\) (top axis wobbles up and down). Steady precession: \(\dot\theta=0\) requires \(dU_{\text{eff}}/d\theta=0\), giving balance condition: \(Mgl=I_3\Omega_3\dot\phi\sin\theta-I_1\dot\phi^2\sin\theta\cos\theta\). For fast spin (\(\Omega_3\) large): \(\dot\phi\approx Mgl/(I_3\Omega_3)\) — independent of \(\theta\). \(\blacksquare\)
A particle of mass \(m\) moves on the surface of a cone \(z=r\cot\alpha\) (half-angle \(\alpha\)) under gravity. Write the Lagrangian and find the equations of motion.
Use cylindrical: constraint \(z=r\cot\alpha\Rightarrow\dot z=\dot r\cot\alpha\). Degrees of freedom: \(r,\phi\). \(T=\tfrac{m}{2}(\dot r^2+r^2\dot\phi^2+\dot z^2)=\tfrac{m}{2}(\dot r^2\csc^2\alpha+r^2\dot\phi^2)\). \(V=mgz=mgr\cot\alpha\). \(L=\tfrac{m}{2}(\dot r^2\csc^2\alpha+r^2\dot\phi^2)-mgr\cot\alpha\). Cyclic in \(\phi\): \(p_\phi=mr^2\dot\phi=\ell=\text{const}\). Radial: \(m\csc^2\alpha\ddot r=mr\dot\phi^2-mg\cot\alpha=\ell^2/(mr^3)-mg\cot\alpha\). \(\blacksquare\)
Using Hamilton's principle, derive Newton's second law for a single particle in a conservative force field.
For a single particle: \(L=T-V=\tfrac{m}{2}({\dot x}^2+\dot y^2+\dot z^2)-V(x,y,z)\). Hamilton's principle: \(\delta\int_{t_1}^{t_2}L\,dt=0\). Lagrange equation for \(x\): \(\dfrac{d}{dt}(m\dot x)-\left(-\dfrac{\partial V}{\partial x}\right)=0\Rightarrow m\ddot x=F_x\) where \(F_x=-\partial V/\partial x\). Similarly for \(y,z\). This is Newton's second law \(m\ddot{\mathbf{r}}=\mathbf{F}\) for conservative forces. \(\blacksquare\)
Prove Kepler's second law (equal areas in equal times) from conservation of angular momentum for a central force.
Area swept out in time \(dt\): \(dA=\tfrac{1}{2}|r\,d\theta|\cdot r=\tfrac{1}{2}r^2|d\theta|\). Rate: \(\dfrac{dA}{dt}=\dfrac{1}{2}r^2|\dot\theta|=\dfrac{|h|}{2}\) where \(h=r^2\dot\theta\) is angular momentum per unit mass. For central force: torque \(=0\Rightarrow\dot h=0\Rightarrow h=\text{const}\). Hence \(dA/dt=|h|/2=\text{const}\) — equal areas in equal times. \(\blacksquare\)
Derive the relativistic Doppler formula for a source moving toward an observer with velocity \(v\). Show it differs from the classical formula.
Source at rest in \(S'\), emits frequency \(f_0\). Observer in \(S\). In \(S'\), period \(T_0=1/f_0\). Time dilation: in \(S\), time between emissions \(T'=\gamma T_0\). In that time, source moves distance \(v\gamma T_0\) closer to observer. Received period (accounting for changing path length): \(T=\gamma T_0-v\gamma T_0/c=\gamma T_0(1-v/c)\). Observed frequency: \(f=1/T=\dfrac{1}{\gamma T_0(1-v/c)}=\dfrac{f_0}{\gamma(1-v/c)}=f_0\sqrt{\dfrac{1+v/c}{1-v/c}}\). Classical: \(f_{\text{cl}}=f_0(1+v/c)\). Relativistic correction: \(\sqrt{(1+\beta)/(1-\beta)}\neq 1+\beta\). \(\blacksquare\)
Define the four-position vector \(x^\mu=(ct,x,y,z)\). Show that \(ds^2=c^2dt^2-dx^2-dy^2-dz^2\) is Lorentz-invariant.
Under Lorentz transformation (boost in \(x\)): \(ct'=\gamma(ct-\beta x)\), \(x'=\gamma(x-\beta ct)\), \(y'=y\), \(z'=z\). Compute \(ds'^2=c^2dt'^2-dx'^2-dy'^2-dz'^2=\gamma^2(c^2dt^2-2\beta c\,dt\,dx+\beta^2dx^2)-\gamma^2(dx^2-2\beta c\,dt\,dx+\beta^2c^2dt^2)-dy^2-dz^2\). \(=\gamma^2c^2dt^2(1-\beta^2)-\gamma^2dx^2(1-\beta^2)-dy^2-dz^2=c^2dt^2-dx^2-dy^2-dz^2=ds^2\). Hence \(ds^2\) is invariant — the Minkowski metric is preserved by Lorentz transformations. \(\blacksquare\)
A cylindrical tank of radius \(R\) rotates at angular velocity \(\Omega\) about its axis. The fluid rotates with it. Find the shape of the free surface and the pressure at any point.
For rigid-body rotation: \(\mathbf{v}=\Omega r\hat{\theta}\). In rotating frame: effective body force includes centrifugal term \(\Omega^2 r\hat{r}\). Pressure: \(\nabla p=\rho\Omega^2r\hat{r}-\rho g\hat{z}\). Integrating: \(p=p_0+\tfrac{1}{2}\rho\Omega^2r^2-\rho gz\). Free surface (\(p=p_{\text{atm}}\)): \(z=z_0+\dfrac{\Omega^2r^2}{2g}\) — a paraboloid of revolution. At the axis \((r=0)\): surface is lowest. At rim \((r=R)\): surface is highest by \(\Delta z=\Omega^2R^2/(2g)\). \(\blacksquare\)
A tank of large cross-section \(A_1\) has a nozzle of area \(A_2\) at the bottom. Water level is at height \(H\). Find the time to empty the tank.
By Bernoulli (and continuity \(A_1v_1=A_2v_2\), \(v_1\ll v_2\) since \(A_1\gg A_2\)): \(v_2\approx\sqrt{2gH}\) (Torricelli). Rate of fall of surface: \(A_1\dfrac{dH}{dt}=-A_2\sqrt{2gH}\). Separating: \(\dfrac{dH}{\sqrt{H}}=-\dfrac{A_2\sqrt{2g}}{A_1}dt\). Integrating from \(H_0\) to 0: \(2\sqrt{H_0}=\dfrac{A_2\sqrt{2g}}{A_1}t\). Time to empty: \[t=\dfrac{A_1}{A_2}\sqrt{\dfrac{2H_0}{g}}\]. \(\blacksquare\)
For a compressible inviscid fluid, show that small disturbances propagate as sound waves with speed \(c_s=\sqrt{\partial p/\partial\rho}\).
Linearise about equilibrium \((\rho_0,p_0,\mathbf{v}=0)\): \(\rho=\rho_0+\rho',p=p_0+p'\), small \(\mathbf{v}\). Continuity: \(\partial\rho'/\partial t+\rho_0\nabla\cdot\mathbf{v}=0\). Euler: \(\rho_0\partial\mathbf{v}/\partial t=-\nabla p'\). Eliminating \(\mathbf{v}\): take \(\partial/\partial t\) of continuity, use Euler: \(\partial^2\rho'/\partial t^2=\nabla^2p'\). Using \(p'=c_s^2\rho'\) (where \(c_s^2=\partial p/\partial\rho\) at constant entropy): \[\dfrac{\partial^2\rho'}{\partial t^2}=c_s^2\nabla^2\rho'\] — wave equation. Sound speed \(c_s=\sqrt{\partial p/\partial\rho}\). For ideal gas (adiabatic): \(c_s=\sqrt{\gamma p/\rho}=\sqrt{\gamma RT/M}\). \(\blacksquare\)
Define Stokes' stream function \(\psi\) for axisymmetric incompressible flow. Show that \(\nabla^2\psi=0\) for irrotational axisymmetric flow.
For axisymmetric flow in spherical \((r,\theta)\): velocity components \(v_r,v_\theta\). Stokes' stream function: \(v_r=\dfrac{1}{r^2\sin\theta}\dfrac{\partial\psi}{\partial\theta}\), \(v_\theta=-\dfrac{1}{r\sin\theta}\dfrac{\partial\psi}{\partial r}\). Continuity is automatically satisfied. For irrotational: \(\nabla\times\mathbf{v}=0\Rightarrow\dfrac{\partial(rv_\theta)}{\partial r}-\dfrac{\partial v_r}{\partial\theta}=0\). Substituting: \(E^2\psi=0\) where \(E^2=\dfrac{\partial^2}{\partial r^2}+\dfrac{\sin\theta}{r^2}\dfrac{\partial}{\partial\theta}\!\left(\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\right)\) — the Stokes stream function equation. For uniform stream past sphere: \(\psi=\tfrac{1}{2}Ur^2\sin^2\theta(1-a^3/r^3)\). \(\blacksquare\)
A source \(+m\) at \((0,a)\) and source \(+m\) at \((0,-a)\) are in a fluid at rest at infinity. Find the complex potential and the velocity on the \(x\)-axis.
Complex potential: \(w=\dfrac{m}{2\pi}[\ln(z-ia)+\ln(z+ia)]=\dfrac{m}{2\pi}\ln(z^2+a^2)\). On \(x\)-axis (\(z=x\)): \(w=\dfrac{m}{2\pi}\ln(x^2+a^2)\). Complex velocity: \(dw/dz=\dfrac{m}{2\pi}\cdot\dfrac{2z}{z^2+a^2}\). On \(x\)-axis: \(u=\dfrac{m}{\pi}\cdot\dfrac{x}{x^2+a^2},v=0\). Stagnation point: \(dw/dz=0\Rightarrow z=0\). Maximum velocity on \(x\)-axis: at \(x=a\), \(u=m/(2\pi a)\). \(\blacksquare\)
State the principle of dynamic similarity. A model is 1/10th the size of a prototype. For the same Reynolds number in water (\(\nu_m=10^{-6}\) m²/s) and air (\(\nu_p=1.5\times10^{-5}\) m²/s), find the required model velocity if prototype speed is 30 m/s.
Dynamic similarity: same dimensionless parameters (here \(Re\)). \(Re_m=Re_p\Rightarrow\dfrac{v_m L_m}{\nu_m}=\dfrac{v_p L_p}{\nu_p}\). \(L_m=L_p/10\). \(v_m=v_p\cdot\dfrac{L_p}{L_m}\cdot\dfrac{\nu_m}{\nu_p}=30\times10\times\dfrac{10^{-6}}{1.5\times10^{-5}}=30\times10\times\dfrac{1}{15}=20\) m/s. Model must run at 20 m/s in water to simulate 30 m/s in air. \(\blacksquare\)
Write the energy equation for viscous incompressible flow. Show that for Couette flow, viscous dissipation equals the work done by the moving plate.
Energy equation: \(\rho c_v\dfrac{DT}{Dt}=k\nabla^2T+\Phi\) where \(\Phi=2\mu e_{ij}e_{ij}\) is dissipation function. For Couette flow (\(v=Uy/h\)): \(\Phi=\mu(\partial v/\partial y)^2=\mu U^2/h^2\). Dissipation rate per unit volume is \(\mu U^2/h^2\). Total dissipation per unit area: \(\mu U^2/h\). Work done by plate per unit area: \(\tau\cdot U=(\mu U/h)\cdot U=\mu U^2/h\). Equal. \(\checkmark\) (Energy balance confirmed: all mechanical work goes into viscous dissipation in steady Couette flow.) \(\blacksquare\)
Distinguish between streamlines, pathlines, and streaklines. For steady flow \(u=x,v=-y\), find the streamlines and verify they coincide with pathlines.
Streamline: curve tangent to velocity field at a given instant. Pathline: trajectory of a fluid particle over time. Streakline: locus of particles that passed through a point. In steady flow: all three coincide. For \(u=x,v=-y\): \(dx/u=dy/v\Rightarrow dx/x=dy/(-y)\Rightarrow\ln x=-\ln y+C\Rightarrow xy=K\) (rectangular hyperbolas). Pathline: \(dx/dt=x\Rightarrow x=x_0e^t\); \(dy/dt=-y\Rightarrow y=y_0e^{-t}\). Eliminate \(t\): \(xy=x_0y_0=K\). Same curves. \(\checkmark\) \(\blacksquare\)
Verify that \(\phi=\ln\sqrt{x^2+y^2}\) is a velocity potential. Find the stream function and identify the flow.
\(\nabla^2\phi=\nabla^2(\tfrac{1}{2}\ln(x^2+y^2))=\dfrac{1}{x^2+y^2}-\dfrac{x^2+y^2}{(x^2+y^2)^2}+\dfrac{1}{x^2+y^2}-\dfrac{y^2}{...}=0\). (Verify: \(\phi=\ln r\) in polar, \(\nabla^2(ln r)=0\) for \(r\neq0\).) Velocity: \(u=\phi_x=x/(x^2+y^2)\), \(v=\phi_y=y/(x^2+y^2)\), i.e., \(v_r=1/r,v_\theta=0\) — a source at origin. Stream function: \(\psi=\arctan(y/x)=\theta\) (polar angle). Complex potential: \(w=\ln z\). \(\blacksquare\)
Show that \(\phi=Axy\) represents irrotational flow. Find the velocity, stream function, and complex potential. Sketch the streamlines.
\(\nabla^2\phi=A(0+0)=0\). \(\checkmark\) Velocity: \(u=\phi_x=Ay,v=\phi_y=Ax\). Stream function: \(\psi_y=u=Ay\Rightarrow\psi=Ay^2/2+f(x)\); \(-\psi_x=v=Ax\Rightarrow\psi_x=-Ax\Rightarrow f=-Ax^2/2\). \(\psi=A(y^2-x^2)/2\). Complex potential: \(w=\phi+i\psi=Axy+iA(y^2-x^2)/2=-iA(x^2-y^2-2ixy)/2=-iAz^2/2\). Streamlines \(\psi=C\): \(y^2-x^2=C/A\) — rectangular hyperbolas. Flow is in the four quadrants between the axes. \(\blacksquare\)
A source \(+m\) at \((-a,0)\) and sink \(-m\) at \((a,0)\) are placed in uniform stream \(U\) along \(x\)-axis. Find the stagnation points and sketch the Rankine oval.
Complex potential: \(w=Uz+\dfrac{m}{2\pi}\ln(z+a)-\dfrac{m}{2\pi}\ln(z-a)=Uz+\dfrac{m}{2\pi}\ln\dfrac{z+a}{z-a}\). Complex velocity: \(\dfrac{dw}{dz}=U+\dfrac{m}{2\pi}\!\left(\dfrac{1}{z+a}-\dfrac{1}{z-a}\right)=U-\dfrac{ma}{\pi(z^2-a^2)}\). Stagnation points on real axis: \(U=\dfrac{ma}{\pi(x^2-a^2)}\Rightarrow x^2=a^2+\dfrac{ma}{\pi U}\Rightarrow x=\pm\sqrt{a^2+ma/(\pi U)}\). The closed streamline through stagnation points is the Rankine oval — an elongated body symmetric about both axes. \(\blacksquare\)
State and derive the unsteady (general) form of Bernoulli's equation for irrotational flow.
For irrotational flow: \(\mathbf{v}=\nabla\phi\). Euler's equation: \(\dfrac{\partial\mathbf{v}}{\partial t}+(\mathbf{v}\cdot\nabla)\mathbf{v}=-\dfrac{\nabla p}{\rho}-\nabla(gz)\). For irrotational: \((\mathbf{v}\cdot\nabla)\mathbf{v}=\nabla(v^2/2)\). Substituting \(\mathbf{v}=\nabla\phi\): \(\nabla\!\left(\dfrac{\partial\phi}{\partial t}+\dfrac{v^2}{2}+\dfrac{p}{\rho}+gz\right)=0\). Hence: \[\dfrac{\partial\phi}{\partial t}+\dfrac{v^2}{2}+\dfrac{p}{\rho}+gz=f(t)\] (function of time only). This is the unsteady Bernoulli equation. For steady flow: \(\partial\phi/\partial t=0\Rightarrow p/\rho+v^2/2+gz=C\). \(\blacksquare\)
Using Blasius' theorem, find the force on a circular cylinder in irrotational flow with circulation \(\Gamma\) in a stream \(U\).
Blasius' theorem: \(F_x-iF_y=\dfrac{i\rho}{2}\oint\!\left(\dfrac{dw}{dz}\right)^2\!dz\). Here \(dw/dz=U(1-a^2/z^2)-i\Gamma/(2\pi z)\). Expand \((dw/dz)^2\) and collect the \(1/z\) term (which gives the residue at \(z=0\)): \(\left(\dfrac{dw}{dz}\right)^2=U^2-\dfrac{U^2a^4}{z^4}-\dfrac{i\Gamma U}{\pi z}+\cdots\). Residue at \(z=0\) of \((dw/dz)^2\) is the coefficient of \(1/z\): \(-i\Gamma U/\pi\). By residue theorem: \(\oint(dw/dz)^2dz=2\pi i\cdot(-i\Gamma U/\pi)=2\Gamma U\). Hence \(F_x-iF_y=\dfrac{i\rho}{2}\cdot2\Gamma U=i\rho U\Gamma\Rightarrow F_x=0,F_y=\rho U\Gamma\). Drag \(=0\), lift \(=\rho U\Gamma\). \(\blacksquare\)
The velocity potential for flow past a cylinder is \(\phi=U(r+a^2/r)\cos\theta\). Find the velocity on the cylinder surface and the maximum speed.
On \(r=a\): \(v_r=\partial\phi/\partial r|_{r=a}=U(1-a^2/r^2)\cos\theta|_{r=a}=0\). \(\checkmark\) \(v_\theta=(1/r)\partial\phi/\partial\theta|_{r=a}=-U(r+a^2/r)\sin\theta/r|_{r=a}=-2U\sin\theta\). Speed on surface: \(|v|=|v_\theta|=2U|\sin\theta|\). Maximum speed: \(2U\) at \(\theta=\pi/2,3\pi/2\) (equatorial points). Minimum speed: \(0\) at \(\theta=0,\pi\) (stagnation points). By Bernoulli: \(p=p_\infty+\tfrac{1}{2}\rho U^2(1-4\sin^2\theta)\). \(\blacksquare\)
For a source of strength \(m\) at origin in a uniform stream \(U\hat{x}\), find the dividing streamline (Rankine half-body) and its asymptotic width.
Stream function: \(\psi=Ur\sin\theta+\dfrac{m}{2\pi}\theta\) (polar). Stagnation point: \(\psi=m/2\) (stagnation streamline). Dividing streamline: \(Ur\sin\theta+\dfrac{m\theta}{2\pi}=\dfrac{m}{2}\). At large \(r\): \(\theta\to0^+\) gives \(y=r\sin\theta\to m/(2U\pi)\cdot(2\pi-2\pi\theta/\theta)\)... Asymptotically: as \(\theta\to0\), \(y=r\sin\theta\to m/(2U)\) above axis; total width \(y_{\max}=m/(U)\) at \(\theta\to\pi\): \(Ur\sin\pi+m\pi/(2\pi)=m/2\Rightarrow y_{\max}=m/(2U)\times2=m/U\). Asymptotic half-width: \(m/(2U)\). \(\blacksquare\)
For ideal flow past a cylinder, find the pressure coefficient \(C_p=(p-p_\infty)/(\tfrac{1}{2}\rho U^2)\) on the surface. At what angle is pressure minimum?
On cylinder: \(v=2U\sin\theta\). Bernoulli: \(p+\tfrac{1}{2}\rho(2U\sin\theta)^2=p_\infty+\tfrac{1}{2}\rho U^2\). \(p-p_\infty=\tfrac{1}{2}\rho U^2(1-4\sin^2\theta)\). Pressure coefficient: \(C_p=1-4\sin^2\theta\). Maximum pressure (\(C_p=1\)): \(\theta=0,\pi\) (stagnation). Minimum pressure (\(C_p=-3\)): \(\theta=\pi/2,3\pi/2\). At top/bottom, \(C_p=-3\) — suction. Note: \(\int_0^{2\pi}C_p\cos\theta\,d\theta=0\) — no drag (D'Alembert). \(\blacksquare\)
For small-amplitude gravity waves on deep water, show that the phase speed is \(c=\sqrt{g/k}\) where \(k\) is the wavenumber.
Assume irrotational, incompressible, inviscid flow. Velocity potential satisfies \(\nabla^2\phi=0\). For progressive wave: \(\phi=A e^{kz}\cos(kx-\omega t)\) (taking \(z\) upward, decaying at \(z\to-\infty\)). Linearised free-surface condition (at \(z=0\)): \(\partial^2\phi/\partial t^2+g\partial\phi/\partial z=0\Rightarrow-\omega^2+gk=0\Rightarrow\omega^2=gk\). Phase speed: \(c=\omega/k=\sqrt{g/k}\). Group velocity: \(v_g=d\omega/dk=\tfrac{1}{2}\sqrt{g/k}=c/2\). Energy propagates at half the phase speed. \(\blacksquare\)
Using Buckingham's \(\Pi\)-theorem, show that the drag force on a sphere depends on Reynolds number as \(F=\mu a U\,f(Re)\).
Variables: \(F,\rho,\mu,U,a\). Dimensions: \([F]=MLT^{-2},[\rho]=ML^{-3},[\mu]=ML^{-1}T^{-1},[U]=LT^{-1},[a]=L\). Number of variables: 5; fundamental dimensions: 3 (M,L,T). By Buckingham \(\Pi\): \(5-3=2\) dimensionless groups. Choose repeating variables \(\mu,U,a\): \(\Pi_1=F/(\mu Ua)\), \(\Pi_2=\rho Ua/\mu=Re\). So \(F/(\mu Ua)=f(Re)\Rightarrow F=\mu Ua\,f(Re)\). At low \(Re\) (Stokes): \(f(Re)=6\pi\), so \(F=6\pi\mu Ua\). \(\blacksquare\)
Find the velocity field for a rectilinear vortex of strength \(\Gamma\) at the origin. Show the flow is irrotational everywhere except at the origin.
Complex potential: \(w=-\dfrac{i\Gamma}{2\pi}\ln z\). Complex velocity: \(dw/dz=-i\Gamma/(2\pi z)\). In polar: \(u-iv=-i\Gamma/(2\pi re^{i\theta})=-i\Gamma e^{-i\theta}/(2\pi r)\). So \(v_r=0,v_\theta=\Gamma/(2\pi r)\). Circulation about any contour enclosing origin: \(\oint\mathbf{v}\cdot d\mathbf{l}=v_\theta\cdot2\pi r=\Gamma\). Vorticity: \(\omega_z=\partial v_y/\partial x-\partial v_x/\partial y=0\) (away from origin — verify by computation). At origin: \(\omega_z=\Gamma\delta^{(2)}(\mathbf{r})\) — a point vortex. \(\blacksquare\)
Viscous fluid fills the gap between concentric cylinders of radii \(R_1
By symmetry: \(\mathbf{v}=v_\theta(r)\hat{\theta}\). Navier-Stokes in cylindrical (azimuthal, steady, fully developed): \(\dfrac{d}{dr}\!\left(\dfrac{1}{r}\dfrac{d(rv_\theta)}{dr}\right)=0\). Solution: \(v_\theta=Ar+B/r\). BCs: \(v_\theta(R_1)=\Omega_1R_1,v_\theta(R_2)=\Omega_2R_2\). Solving: \(A=\dfrac{\Omega_2R_2^2-\Omega_1R_1^2}{R_2^2-R_1^2}\), \(B=\dfrac{(\Omega_1-\Omega_2)R_1^2R_2^2}{R_2^2-R_1^2}\). If \(\Omega_2=0\): \(v_\theta=\Omega_1R_1^2(R_2^2/r-r)/(R_2^2-R_1^2)\). \(\blacksquare\)
A siphon tube rises to height \(h\) above the water surface in a tank, then descends to level \(H\) below the surface. Find the flow speed and maximum height \(h\) for siphon to work.
Apply Bernoulli from tank surface (A: \(p=p_0,v\approx0,z=0\)) to exit (C: \(p=p_0,z=-H\)): \(p_0=p_0+\tfrac{1}{2}\rho v^2-\rho gH\Rightarrow v=\sqrt{2gH}\). Apply Bernoulli from A to top of siphon (B: height \(h\), speed \(v\)): \(p_0=p_B+\tfrac{1}{2}\rho v^2+\rho gh\Rightarrow p_B=p_0-\rho g(H+h)\). For siphon to work: \(p_B>0\Rightarrow p_0>\rho g(H+h)\Rightarrow H+h
Distinguish between static, dynamic, and stagnation pressure. A Pitot-static tube reads a pressure difference of 1000 Pa in air (\(\rho=1.2\) kg/m³). Find the flow speed.
Static pressure \(p_s\): pressure in the fluid at a point, in the absence of flow effects. Dynamic pressure \(q\): \(q=\tfrac{1}{2}\rho v^2\) — kinetic pressure due to motion. Stagnation pressure \(p_0\): total pressure at a stagnation point, \(p_0=p_s+q\). Bernoulli: \(p_s+\tfrac{1}{2}\rho v^2=p_0\). Given \(\Delta p=p_0-p_s=1000\) Pa: \(v=\sqrt{2\Delta p/\rho}=\sqrt{2000/1.2}=\sqrt{1666.7}\approx40.8\) m/s. \(\blacksquare\)
Define the material (substantial) derivative. For steady 2D flow \(u=x^2-y^2\), \(v=-2xy\), find the acceleration at point \((1,1)\).
Material derivative: \(\dfrac{D}{Dt}=\dfrac{\partial}{\partial t}+u\dfrac{\partial}{\partial x}+v\dfrac{\partial}{\partial y}\) (in 2D). For steady flow \(\partial/\partial t=0\). Acceleration: \(a_x=u\partial u/\partial x+v\partial u/\partial y\), \(a_y=u\partial v/\partial x+v\partial v/\partial y\). At \((1,1)\): \(u=0,v=-2\). \(\partial u/\partial x=2x=2,\partial u/\partial y=-2y=-2,\partial v/\partial x=-2y=-2,\partial v/\partial y=-2x=-2\). \(a_x=0\cdot2+(-2)(-2)=4\), \(a_y=0\cdot(-2)+(-2)(-2)=4\). Acceleration \(=(4,4)\). \(\blacksquare\)
Show that for any irrotational steady flow, the complex potential \(w(z)\) is analytic and the equipotential lines and streamlines are orthogonal.
For irrotational flow: \(\mathbf{v}=\nabla\phi\Rightarrow v_x=\phi_x,v_y=\phi_y\). For incompressible: \(\nabla^2\phi=\phi_{xx}+\phi_{yy}=0\) — \(\phi\) is harmonic. Stream function \(\psi\): \(u=\psi_y,v=-\psi_x\). Then \(\nabla\phi\cdot\nabla\psi=\phi_x\psi_x+\phi_y\psi_y=u(-v)+v\cdot u=0\). So \(\nabla\phi\perp\nabla\psi\) — equipotentials (\(\phi=C\)) and streamlines (\(\psi=C\)) are orthogonal. The complex potential \(w=\phi+i\psi\) satisfies Cauchy-Riemann equations (\(\phi_x=\psi_y,\phi_y=-\psi_x\)) since \(u=\phi_x=\psi_y\) and \(v=\phi_y=-\psi_x\). Hence \(w(z)\) is analytic. \(\blacksquare\)
State and prove Kelvin's minimum kinetic energy theorem for irrotational flow.
Theorem: Of all incompressible fluid motions satisfying given boundary conditions, the irrotational flow has minimum kinetic energy. Proof: Let \(\mathbf{v}=\nabla\phi+\mathbf{u}\) where \(\nabla\phi\) is irrotational and \(\mathbf{u}\) is a solenoidal perturbation (\(\nabla\cdot\mathbf{u}=0\)) with \(\mathbf{u}\cdot\hat{n}=0\) on boundaries. \(T=\tfrac{\rho}{2}\int|\nabla\phi+\mathbf{u}|^2dV=T_\phi+\rho\int\nabla\phi\cdot\mathbf{u}\,dV+T_u\). Cross term: \(\int\nabla\phi\cdot\mathbf{u}\,dV=\int\nabla\cdot(\phi\mathbf{u})dV-\int\phi\nabla\cdot\mathbf{u}\,dV=\oint\phi\mathbf{u}\cdot\hat{n}\,dA=0\). So \(T=T_\phi+T_u\geq T_\phi\). \(\blacksquare\)
Derive the Lagrange equations of motion from D'Alembert's principle. Apply to a simple pendulum of length \(l\) to obtain the equation of motion.
Lagrange equations: \(\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_i}-\frac{\partial L}{\partial q_i}=Q_i\). For pendulum: generalized coordinate \(\theta\). \(T=\frac{1}{2}ml^2\dot\theta^2\), \(V=-mgl\cos\theta\). \(L=T-V\). \(\frac{d}{dt}(ml^2\dot\theta)+mgl\sin\theta=0\Rightarrow\ddot\theta+\frac{g}{l}\sin\theta=0\). For small \(\theta\): SHM with \(\omega^2=g/l\).
Define source, sink, and doublet in 2D fluid flow. If a source of strength \(m\) is at origin, write the stream function and velocity potential. Find the velocity at distance \(r\).
Source: fluid flowing uniformly outward; strength \(m=\) volume flow rate per unit length. \(\phi=\frac{m}{2\pi}\ln r\), \(\psi=\frac{m}{2\pi}\theta\). Velocity: \(v_r=\frac{\partial\phi}{\partial r}=\frac{m}{2\pi r}\), \(v_\theta=0\). A sink has \(-m\). Doublet: limit of source+sink as separation \(\to 0\).
Find the moment of inertia of a uniform solid sphere of mass \(M\) and radius \(R\) about a diameter. Hence find the moment of inertia about a tangent.
Consider disc element at distance \(x\) from center, radius \(r=\sqrt{R^2-x^2}\), mass \(dm=\frac{3M}{4R^3}\pi(R^2-x^2)dx\). \[I=\int_{-R}^R\frac{1}{2}r^2\,dm=\frac{3M}{8R^3}\int_{-R}^R(R^2-x^2)^2\,dx=\frac{2MR^2}{5}\] By parallel axis theorem: \(I_{\text{tangent}}=\frac{2MR^2}{5}+MR^2=\frac{7MR^2}{5}\).
Derive the Navier-Stokes equations for the motion of a viscous incompressible fluid. State the conditions under which they reduce to Euler's equations.
From Newton's second law for a fluid element: \[\rho\frac{D\mathbf{v}}{Dt}=\rho\mathbf{f}-\nabla p+\mu\nabla^2\mathbf{v}\] (N-S equations), where \(\rho\) is density, \(\mu\) dynamic viscosity, \(\mathbf{f}\) body force. With continuity \(\nabla\cdot\mathbf{v}=0\). When \(\mu=0\) (inviscid): reduces to Euler's equations \(\rho\frac{D\mathbf{v}}{Dt}=\rho\mathbf{f}-\nabla p\).
Derive Hamilton's equations from the Lagrangian. Show they are equivalent to Lagrange's equations. Write the Hamiltonian for a particle in a central force field.
Define conjugate momentum \(p_i=\partial L/\partial\dot{q}_i\). Hamiltonian \(H=\sum p_i\dot{q}_i-L\). Hamilton's equations: \(\dot{q}_i=\partial H/\partial p_i\), \(\dot{p}_i=-\partial H/\partial q_i\). For central force: \[H=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}+V(r)\]
Prove that the Hamiltonian \(H\) is conserved (i.e., \(dH/dt=0\)) for conservative systems with no explicit time dependence. Use this to find the first integral of a particle under gravity.
\(\frac{dH}{dt}=\frac{\partial H}{\partial t}+\sum\left(\frac{\partial H}{\partial q_i}\dot{q}_i+\frac{\partial H}{\partial p_i}\dot{p}_i\right)=\frac{\partial H}{\partial t}+\sum(-\dot{p}_i\dot{q}_i+\dot{q}_i\dot{p}_i)=\frac{\partial H}{\partial t}\). For conservative systems: \(\partial H/\partial t=0\Rightarrow H=E=\text{const}\). For a particle under gravity: \(H=\frac{p^2}{2m}+mgy=E\Rightarrow\frac{1}{2}mv^2+mgy=E\) (energy conservation).
A fluid has sources of equal strength \(m\) at points \((a,0)\) and \((-a,0)\). Write the stream function for the combined flow. Find the equation of streamlines and identify the stagnation point.
\(\psi=\frac{m}{2\pi}(\theta_1+\theta_2)\) where \(\theta_1=\arctan\frac{y}{x-a}\), \(\theta_2=\arctan\frac{y}{x+a}\). Stagnation point where velocity \(=0\): by symmetry, at origin the horizontal velocities cancel and vertical velocities cancel. Stagnation at \((0,0)\). Streamlines are curves \(\theta_1+\theta_2=\text{const}\).
A uniform rod of mass \(M\) and length \(2l\) rotates in a vertical plane about one end (pivoted). Write the equation of motion using Euler's equations for rigid body rotation. Find the period of small oscillations.
Moment of inertia about pivot: \(I=\frac{M(2l)^2}{3}=\frac{4Ml^2}{3}\). Torque (restoring): \(\tau=-Mgl\sin\theta\). Equation: \(\frac{4Ml^2}{3}\ddot\theta=-Mgl\sin\theta\Rightarrow\ddot\theta=-\frac{3g}{4l}\sin\theta\). Small oscillations: \[T=2\pi\sqrt{\frac{4l}{3g}}\]
Determine the irrotational, incompressible 2D flow past a circular cylinder of radius \(a\). Find the velocity components and the stagnation points.
Complex potential: \(w=U\!\left(z+\frac{a^2}{z}\right)\). \(\phi=U\!\left(r+\frac{a^2}{r}\right)\cos\theta\), \(\psi=U\!\left(r-\frac{a^2}{r}\right)\sin\theta\). \(v_r=U\!\left(1-\frac{a^2}{r^2}\right)\cos\theta\), \(v_\theta=-U\!\left(1+\frac{a^2}{r^2}\right)\sin\theta\). On cylinder \(r=a\): \(v_r=0\), \(v_\theta=-2U\sin\theta\). Stagnation points: \(v_r=0,v_\theta=0\Rightarrow\theta=0,\pi\) at \(r=a\), i.e., \((\pm a,0)\).
Use Lagrange's equations to find the equations of motion of a double pendulum (two masses \(m_1, m_2\), lengths \(l_1, l_2\)) for small oscillations.
Generalized coordinates \(\theta_1, \theta_2\). \[T=\frac{1}{2}(m_1+m_2)l_1^2\dot\theta_1^2+m_2l_1l_2\dot\theta_1\dot\theta_2\cos(\theta_1-\theta_2)+\frac{1}{2}m_2l_2^2\dot\theta_2^2\] For small angles: linearize. Lagrange equations give two coupled ODEs. Normal mode frequencies (simplified for equal masses \(m\) and lengths \(l\)): \[\omega^2=\frac{g}{l}\!\left(2\pm\sqrt{2}\right)\]
Derive the Navier-Stokes equation for viscous fluid. For steady flow of a viscous fluid between two infinite parallel plates (Couette flow), solve the N-S equations to find the velocity profile.
N-S: \(\rho(\mathbf{v}\cdot\nabla)\mathbf{v}=-\nabla p+\mu\nabla^2\mathbf{v}\). For Couette flow: \(u=u(y)\) only, \(\nabla p=0\). \(\mu\frac{d^2u}{dy^2}=0\Rightarrow u=Ay+B\). BCs: lower plate at rest \(u(0)=0\), upper plate \(u(h)=U\). Solution: \[u=\frac{Uy}{h}\] (linear profile). Shear stress: \(\tau=\mu U/h\).