Dynamics & Statics — Proofs & PYQ

Range \(R=\dfrac{u^2\sin2\theta}{g}\). Maximized when \(\sin2\theta=1\Rightarrow2\theta=90°\Rightarrow\theta=45°\). \(R_{\max}=\dfrac{u^2}{g}\). For any range \(R

Here \(\alpha=30°,\beta=30°\). \(R=\dfrac{2u^2\sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta}=\dfrac{2u^2\sin30°\cos60°}{g\cos^230°}=\dfrac{2u^2\cdot(1/2)\cdot(1/2)}{g\cdot(3/4)}=\dfrac{u^2/2}{3g/4}=\dfrac{2u^2}{3g}\).

Trajectory: \(y=x\tan\theta\!\left(1-\dfrac{x}{R}\right)\) (standard form). At \(x=d,y=h\): \(h=d\tan\theta\cdot\dfrac{R-d}{R}\Rightarrow\tan\theta=\dfrac{hR}{d(R-d)}\). \(\blacksquare\)

\(H=\dfrac{u^2\sin^2\theta}{2g}\) and \(R=\dfrac{u^2\sin2\theta}{g}=\dfrac{2u^2\sin\theta\cos\theta}{g}\). So \(\dfrac{H}{R}=\dfrac{u^2\sin^2\theta/2g}{2u^2\sin\theta\cos\theta/g}=\dfrac{\sin\theta}{4\cos\theta}=\dfrac{\tan\theta}{4}\). Hence \(\tan\theta=\dfrac{4H}{R}\) and \(R=\dfrac{4H}{\tan\theta}\). \(\blacksquare\)

\(\omega=\sqrt{k/m}=\sqrt{25}=5\) rad/s. Period \(T=2\pi/5\approx1.257\) s. Amplitude \(A=4\) cm \(=0.04\) m. Velocity at equilibrium: \(v_{\max}=A\omega=0.04\times5=0.2\) m/s. \(\blacksquare\)

\(A^2=9+16+2(3)(4)\cos(\pi/2)=25+0=25\Rightarrow A=5\). \(\tan\phi=\dfrac{3\sin0+4\sin(\pi/2)}{3\cos0+4\cos(\pi/2)}=\dfrac{4}{3}\Rightarrow\phi=\arctan(4/3)\approx53.13°\). So \(x=5\sin(\omega t+53.13°)\). \(\blacksquare\)

MOI about pivot \(= Ml^2/3+Ml^2=4Ml^2/3\) (using parallel axis: \(I_G=Ml^2/3\) for rod of length \(2l\), then add \(Ml^2\) for pivot at end — wait: for rod of length \(2l\), \(I_G=M(2l)^2/12=Ml^2/3\), distance from G to pivot = \(l\), so \(I_A=Ml^2/3+Ml^2=4Ml^2/3\)). Period of compound pendulum: \(T=2\pi\sqrt{I/(Mgh)}=2\pi\sqrt{(4Ml^2/3)/(Mgl)}=2\pi\sqrt{4l/(3g)}\). Equivalent simple pendulum length: \(L=4l/3\). \(\blacksquare\)

Using Binet's equation: \(F=-\mu/r^3\Rightarrow\) (attractive). With \(u=1/r\): \(\dfrac{d^2u}{d\theta^2}+u=\dfrac{\mu u^3}{h^2u^2}=\dfrac{\mu u}{h^2}\). So \(\dfrac{d^2u}{d\theta^2}+u\!\left(1-\dfrac{\mu}{h^2}\right)=0\). If \(h^2>\mu\): \(u=A\cos(k\theta+\epsilon)\) where \(k^2=1-\mu/h^2\Rightarrow\) orbit is \(r=\sec(k\theta+\epsilon)/A\) (a spiral type). If \(h^2=\mu\): \(u=A\theta+B\) (Cotes' spiral). If \(h^2<\mu\): \(u=Ae^{k\theta}+Be^{-k\theta}\) (hyperbolic spiral). \(\blacksquare\)

Slice perpendicular to the axis at height \(x\): disc of radius \(\sqrt{a^2-x^2}\), thickness \(dx\), mass \(dm=\rho\pi(a^2-x^2)dx\). MOI of disc about its diameter (which is the \(x\)-axis): \(\dfrac{1}{4}(a^2-x^2)\,dm\). But we need MOI about \(z\)-axis (the diameter we're measuring): MOI of each disc about \(z\)-axis \(= \dfrac{1}{2}(a^2-x^2)\,dm\). \(I=\int_{-a}^a\dfrac{1}{2}(a^2-x^2)\rho\pi(a^2-x^2)dx=\dfrac{\rho\pi}{2}\int_{-a}^a(a^2-x^2)^2dx\). \((a^2-x^2)^2=a^4-2a^2x^2+x^4\). \(\int_{-a}^a=2\int_0^a\): \(2(a^4\cdot a-2a^2\cdot\dfrac{a^3}{3}+\dfrac{a^5}{5})=2a^5(1-2/3+1/5)=2a^5\cdot8/15=16a^5/15\). So \(I=\dfrac{\rho\pi}{2}\cdot\dfrac{16a^5}{15}=\dfrac{8\rho\pi a^5}{15}\). Since \(M=\dfrac{4\rho\pi a^3}{3}\): \(I=\dfrac{8}{15}\cdot\dfrac{3M}{4\pi a^3}\cdot\pi a^5=\dfrac{2Ma^2}{5}\). \(\blacksquare\)

Let \(a\) be the acceleration of center. Translation: \(Mg\sin\theta-F=Ma\) (where \(F\) = friction). Rotation about center: \(Fa=I\alpha=\dfrac{Ma^2}{2}\cdot\dfrac{a}{a}... \) Correction: \(Fa=(Ma^2/2)\alpha\). Rolling condition: \(a=a\alpha\Rightarrow\alpha=a/a\). So \(Fa=(Ma^2/2)(a/a)=Maa/2\Rightarrow F=Ma/2\). Substitute: \(Mg\sin\theta-Ma/2=Ma\Rightarrow a=\dfrac{2g\sin\theta}{3}\). (For sphere: \(a=5g\sin\theta/7\); cylinder (hollow): \(a=g\sin\theta/2\).) \(\blacksquare\)

For a nearly horizontal chain (small sag), approximate the chain as parabola: \(y=\dfrac{wx^2}{2T_0}\) where \(x\) is measured from midpoint. The half-span is \(L/2\) (where \(L\) is the distance between points). Sag \(d=y(L/2)=\dfrac{w(L/2)^2}{2T_0}=\dfrac{wL^2}{8T_0}\). Total weight \(W=wl\approx wL\) (for small sag). So \(T_0=\dfrac{wL^2}{8d}\approx\dfrac{Wl}{8d}\) (since \(L\approx l\)). \(\blacksquare\)

Let \(P\) = reaction at wall (horizontal), \(R\) = reaction at floor (vertical). Virtual displacement: give \(\theta\) a small increment \(\delta\theta\). The center of rod has coordinates \(x_G=a\cos\theta,y_G=a\sin\theta\). Virtual work of all forces \(=0\): \(-W\delta y_G+P\delta x_A=0\) where \(A\) is the point on wall at height \(2a\sin\theta\). \(\delta y_G=a\cos\theta\,\delta\theta\), \(\delta x_A... \) Let's use equilibrium directly: \(P=R_{\text{wall}}\). By moments about foot: \(W\cdot a\cos\theta=P\cdot2a\sin\theta\Rightarrow P=\dfrac{W\cos\theta}{2\sin\theta}=\dfrac{W}{2}\cot\theta\). Vertical equilibrium: \(R=W\). \(\blacksquare\)

Let \(R\) = normal at floor, \(S\) = normal at wall. Friction at floor: \(F_1=\mu_2 R\), at wall: \(F_2=\mu_1 S\). Horizontal: \(S=F_1=\mu_2 R\). Vertical: \(R+F_2=W\Rightarrow R+\mu_1S=W\Rightarrow R+\mu_1\mu_2 R=W\Rightarrow R=\dfrac{W}{1+\mu_1\mu_2}\). Moments about foot of ladder: \(S\cdot2l\sin\theta+F_2\cdot2l\cos\theta=W\cdot l\cos\theta\Rightarrow2lS\sin\theta=l\cos\theta(W-2F_2)=l\cos\theta(W-2\mu_1S)\). Substituting \(S=\mu_2R\): minimum angle satisfies \(\tan\theta=\dfrac{1-\mu_1\mu_2}{2\mu_2(1+\mu_1\mu_2)}\cdot(1+\mu_1\mu_2)=\dfrac{1-\mu_1\mu_2}{2\mu_2}\). So \(\theta_{\min}=\arctan\dfrac{1-\mu_1\mu_2}{2\mu_2}\). \(\blacksquare\)

Central force: \(F=-\mu/r^2\) (attractive). Binet's equation: \(F=-h^2u^2\!\left(\dfrac{d^2u}{d\theta^2}+u\right)\) where \(u=1/r\), \(h=r^2\dot\theta\). So \(-\mu u^2=-h^2u^2\!\left(\dfrac{d^2u}{d\theta^2}+u\right)\Rightarrow\dfrac{d^2u}{d\theta^2}+u=\dfrac{\mu}{h^2}\). This is SHM-type ODE with constant RHS. Solution: \(u=\dfrac{\mu}{h^2}+A\cos(\theta-\theta_0)=\dfrac{1}{r}\). So \(\dfrac{1}{r}=\dfrac{\mu}{h^2}(1+e\cos\theta)\) (taking \(\theta_0=0\), \(e=Ah^2/\mu\)). This is conic \(r=\dfrac{l}{1+e\cos\theta}\) (\(l=h^2/\mu\)). For \(e<1\): ellipse. \(\blacksquare\) (Kepler's Second Law follows from \(h=r^2\dot\theta=\text{const}\Leftrightarrow\) equal areas in equal times.)

Trajectory for angle \(\theta\): \(y = x\tan\theta - \dfrac{gx^2}{2u^2}\sec^2\theta\). Let \(t = \tan\theta\); treat this as quadratic in \(t\): \(\dfrac{gx^2}{2u^2}t^2 - xt + \left(y + \dfrac{gx^2}{2u^2}\right) = 0\). A point \((x,y)\) is reachable iff discriminant \(\geq 0\): \(x^2 - 4\cdot\dfrac{gx^2}{2u^2}\cdot\left(y+\dfrac{gx^2}{2u^2}\right)\geq 0\). Simplify: \(1 - \dfrac{2g}{u^2}\left(y + \dfrac{gx^2}{2u^2}\right)\geq 0 \Rightarrow y \leq \dfrac{u^2}{2g} - \dfrac{gx^2}{2u^2}\). The boundary (envelope) is the parabola of safety: \(y = \dfrac{u^2}{2g} - \dfrac{gx^2}{2u^2}\). \(\blacksquare\)

Range on incline: \(R = \dfrac{u^2}{g\cos^2\beta}[\sin(2\alpha+\beta)-\sin\beta]\). This is maximum when \(\sin(2\alpha+\beta)=1\), i.e., \(2\alpha+\beta = 90° \Rightarrow \alpha = 45° - \beta/2\). The angle bisects the angle between the incline and the vertical. Maximum range: \(R_{\max} = \dfrac{u^2}{g\cos^2\beta}(1-\sin\beta) = \dfrac{u^2}{g(1+\sin\beta)}\). \(\blacksquare\)

\(R = \dfrac{u^2\sin2\theta}{g}\). For \(90°-\theta\): \(R' = \dfrac{u^2\sin(180°-2\theta)}{g} = \dfrac{u^2\sin2\theta}{g} = R\). Hence same range. Times of flight: \(T_1 = \dfrac{2u\sin\theta}{g}\), \(T_2 = \dfrac{2u\cos\theta}{g}\). So \(\dfrac{gT_1T_2}{2} = \dfrac{g}{2}\cdot\dfrac{4u^2\sin\theta\cos\theta}{g^2} = \dfrac{2u^2\sin\theta\cos\theta}{g} = \dfrac{u^2\sin2\theta}{g} = R\). \(\blacksquare\)

Range \(R = \dfrac{u^2\sin2\theta}{g}\). Max height \(H = \dfrac{u^2\sin^2\theta}{2g}\). Setting \(R=H\): \(\dfrac{u^2 \cdot 2\sin\theta\cos\theta}{g} = \dfrac{u^2\sin^2\theta}{2g} \Rightarrow 4\cos\theta = \sin\theta \Rightarrow \tan\theta = 4 \Rightarrow \theta = \arctan4 \approx 76°\). \(\blacksquare\)

Velocity components: \(v_x = u\cos\theta\) (constant), \(v_y = u\sin\theta - gt\). Speed: \(v = \sqrt{(u\cos\theta)^2+(u\sin\theta-gt)^2}\). Direction with horizontal: \(\phi = \arctan\!\left(\dfrac{u\sin\theta-gt}{u\cos\theta}\right)\). At maximum height (\(v_y=0\)): speed \(= u\cos\theta\) (only horizontal component remains). \(\blacksquare\)

Using axes along and perpendicular to incline: \(y = u\sin\alpha\cdot t - \tfrac{1}{2}g\cos\beta\cdot t^2 = 0\) at landing. \(\Rightarrow t = \dfrac{2u\sin\alpha}{g\cos\beta}\). Range: \(R = u\cos\alpha\cdot t - \tfrac{1}{2}g\sin\beta\cdot t^2 = \dfrac{2u^2\sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta}\). \(\blacksquare\)

\(x = u\cos\theta\cdot t\), \(y = u\sin\theta\cdot t - \tfrac{1}{2}gt^2\). Eliminating \(t\): \(y = x\tan\theta - \dfrac{g}{2u^2\cos^2\theta}x^2\). This is \(y = ax + bx^2\) (quadratic in \(x\)) — a parabola opening downward. Vertex (highest point): \(x_v = \dfrac{u^2\sin\theta\cos\theta}{g}\), \(y_v = \dfrac{u^2\sin^2\theta}{2g}\). Focus is at distance \(\dfrac{u^2\cos^2\theta}{2g}\) below vertex. \(\blacksquare\)

The plane through the origin makes angle \(\beta\) with horizontal, so its equation is \(y = x\tan\beta\). Trajectory: \(y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}\). Substituting \(y = x\tan\beta\): \(x\tan\beta = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha} \Rightarrow x = \dfrac{2u^2\cos^2\alpha(\tan\alpha-\tan\beta)}{g}\). Range along the plane: \(R = x/\cos\beta = \dfrac{2u^2\cos^2\alpha(\tan\alpha-\tan\beta)}{g\cos\beta} = \dfrac{2u^2\cos\alpha\sin(\alpha-\beta)}{g\cos^2\beta}\). \(\blacksquare\)

\(x = a\sin(\omega t+\phi)\), \(v = \dot{x} = a\omega\cos(\omega t+\phi)\). \(v^2 = a^2\omega^2\cos^2(\omega t+\phi) = \omega^2(a^2-a^2\sin^2(\omega t+\phi)) = \omega^2(a^2-x^2)\). \(\blacksquare\) KE \(= \tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2(a^2-x^2)\). PE \(= \tfrac{1}{2}m\omega^2x^2\) (by energy conservation, PE = Total energy − KE). Total energy \(= \tfrac{1}{2}m\omega^2a^2 =\) constant. \(\blacksquare\)

At equilibrium, extension \(e_0\): \(\dfrac{\lambda e_0}{l} = mg \Rightarrow e_0 = \dfrac{mgl}{\lambda}\). Displacing by \(x\) below equilibrium (string still taut): restoring force \(= -\dfrac{\lambda x}{l}\). EOM: \(m\ddot{x} = -\dfrac{\lambda}{l}x\). SHM with \(\omega^2 = \dfrac{\lambda}{ml}\). Period \(T = 2\pi\sqrt{\dfrac{ml}{\lambda}}\). Note: also equals \(2\pi\sqrt{e_0/g}\). \(\blacksquare\)

Case \(\delta=0\): \(x/a = y/b \Rightarrow y = (b/a)x\) — straight line through origin.
Case \(\delta=\pi/2\): \(x=a\sin\omega t\), \(y=b\cos\omega t\). \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) — ellipse.
General \(\delta\): \(\dfrac{x}{a}=\sin\omega t\), \(\dfrac{y}{b}=\sin\omega t\cos\delta+\cos\omega t\sin\delta\). So \(\dfrac{y}{b}-\dfrac{x}{a}\cos\delta=\cos\omega t\sin\delta\) and \(\dfrac{x}{a}=\sin\omega t\). Squaring and adding: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{2xy}{ab}\cos\delta=\sin^2\delta\) — general ellipse (Lissajous figure for 1:1 ratio). \(\blacksquare\)

\(v_1^2 = \omega^2(a^2-x_1^2)\) and \(v_2^2 = \omega^2(a^2-x_2^2)\). Subtracting: \(v_1^2-v_2^2 = \omega^2(x_2^2-x_1^2) \Rightarrow \omega^2 = \dfrac{v_1^2-v_2^2}{x_2^2-x_1^2}\). Then \(a^2 = x_1^2 + \dfrac{v_1^2}{\omega^2} = \dfrac{v_1^2x_2^2-v_2^2x_1^2}{v_1^2-v_2^2}\). \(\blacksquare\)

Equilibrium at midpoint. Each half has natural length \(l_0\) and extension \(d-l_0\). Displace particle by \(x\) toward one peg. Left string length \(= d-x\), right string \(= d+x\). Left tension \(T_1 = \dfrac{\lambda(d-x-l_0)}{l_0}\), right tension \(T_2 = \dfrac{\lambda(d+x-l_0)}{l_0}\). Net restoring force \(= T_2-T_1 = \dfrac{\lambda\cdot 2x}{l_0}\). EOM: \(m\ddot{x} = -\dfrac{2\lambda x}{l_0}\). SHM with \(\omega^2 = \dfrac{2\lambda}{ml_0}\). \(\blacksquare\) (Factor of 2 comes from two strings; often stated as \(\omega^2 = \lambda/ml_0\) when each string counted separately.)

General solution: \(x = A\cos\omega t + B\sin\omega t\) (or \(x = R\cos(\omega t + \phi)\)). ICs: \(x(0)=a\), \(\dot{x}(0)=0 \Rightarrow A=a\), \(B=0\). So \(x = a\cos\omega t\). Centre: \(x=0 \Rightarrow \cos\omega t=0 \Rightarrow \omega t = \pi/2 \Rightarrow t = \pi/(2\omega) = T/4\). The particle reaches centre in one-quarter of the period. \(\blacksquare\)

(a) Series: Effective stiffness \(\dfrac{1}{k} = \dfrac{1}{k_1}+\dfrac{1}{k_2} \Rightarrow k = \dfrac{k_1k_2}{k_1+k_2}\). Period \(T = 2\pi\sqrt{\dfrac{m(k_1+k_2)}{k_1k_2}}\).
(b) Parallel: Effective stiffness \(k = k_1+k_2\). Period \(T = 2\pi\sqrt{\dfrac{m}{k_1+k_2}}\). \(\blacksquare\)

\(x = a\sin2\omega t = 2a\sin\omega t\cos\omega t\). Let \(\sin\omega t = y/b\), \(\cos\omega t = \sqrt{1-(y/b)^2}\). So \(x = 2a\cdot\dfrac{y}{b}\sqrt{1-\dfrac{y^2}{b^2}} \Rightarrow \dfrac{x^2}{4a^2} = \dfrac{y^2}{b^2}\!\left(1-\dfrac{y^2}{b^2}\right) \Rightarrow x^2 = \dfrac{4a^2y^2}{b^2}-\dfrac{4a^2y^4}{b^4}\). This is a figure-8 (Lissajous figure) symmetric about the \(y\)-axis, passing through origin, with maxima at \(y=\pm b\) and extreme \(x\)-values \(\pm a\). \(\blacksquare\)

Theorem: The total work done by all forces on a particle equals the change in its kinetic energy: \(W = \Delta(\tfrac{1}{2}mv^2)\). Proof: \(F = ma = m\dfrac{dv}{dt} = mv\dfrac{dv}{ds}\). Work: \(W = \int F\,ds = \int mv\,dv = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mu^2\). \(\blacksquare\)
Application: Force along incline \(= mg\sin\alpha\). Work \(= mg\sin\alpha\cdot s\). Starting from rest: \(\tfrac{1}{2}mv^2 = mgs\sin\alpha \Rightarrow v = \sqrt{2gs\sin\alpha}\). \(\blacksquare\)

\(\nabla\times(-\nabla V) = -\nabla\times(\nabla V) = \mathbf{0}\) (curl of gradient is always zero). Hence conservative. Work: \(W = \int_A^B\mathbf{F}\cdot d\mathbf{r} = -\int_A^B\nabla V\cdot d\mathbf{r} = -(V_B-V_A) = V_A-V_B\), which depends only on endpoints. For gravity: \(\mathbf{F} = -mg\hat{j} = -\nabla(mgy) \Rightarrow V = mgy\). Gain in height → increase in PE. \(\blacksquare\)

Let both move speed \(v\), \(m_1\) descends distance \(s\), \(m_2\) rises \(s\). Energy: KE \(= \tfrac{1}{2}(m_1+m_2)v^2\). Change in PE \(= m_2gs - m_1gs = (m_2-m_1)gs\). Energy conservation (no friction): \(\tfrac{1}{2}(m_1+m_2)v^2 = (m_1-m_2)gs\). Differentiating w.r.t. \(t\): \((m_1+m_2)v\dot{v} = (m_1-m_2)g\dot{s} = (m_1-m_2)gv\). Acceleration \(a = \dot{v} = \dfrac{(m_1-m_2)g}{m_1+m_2}\). \(\blacksquare\)

On slope at constant speed: driving force \(= Mg\sin\alpha + R\). Power \(P = (Mg\sin\alpha + R)v\).
On level road, at speed \(u\): \(P = (R + Ma)u\) (driving force \(= R + Ma\) for acceleration \(a\)). Maximum (terminal) speed with same power, no acceleration: \(u_{\max} = P/R = \dfrac{(Mg\sin\alpha+R)v}{R}\). \(\blacksquare\)

Extension \(e = l - l_0\). By Hooke's law, tension at extension \(x\): \(T = \dfrac{\lambda x}{l_0}\). Work done stretching from 0 to \(e\): \(W = \int_0^e \dfrac{\lambda x}{l_0}dx = \dfrac{\lambda e^2}{2l_0} = \dfrac{\lambda(l-l_0)^2}{2l_0}\). This equals the elastic PE stored. \(\blacksquare\)

Radius of circle: \(r = l\sin\theta\). Vertical: \(T\cos\theta = mg \Rightarrow T = mg/\cos\theta\). Horizontal: \(T\sin\theta = mr\omega^2 = ml\sin\theta\cdot\omega^2 \Rightarrow T = ml\omega^2\). Equating: \(ml\omega^2 = mg/\cos\theta \Rightarrow \omega^2 = g/(l\cos\theta)\). Period: \(\tau = 2\pi/\omega = 2\pi\sqrt{l\cos\theta/g}\). Tension: \(T = mg/\cos\theta\). \(\blacksquare\)

Normal reaction \(N\): vertical component \(N\cos\theta = mg\), horizontal component \(N\sin\theta = mv^2/R\). Dividing: \(\tan\theta = v^2/(Rg) \Rightarrow v = \sqrt{Rg\tan\theta}\). This is the ideal speed for no friction. \(\blacksquare\)

At the top, minimum condition: normal force \(N \geq 0\). At top: \(N + mg = mv_{\text{top}}^2/r\). Minimum when \(N=0\): \(v_{\text{top}}^2 = gr\). Energy conservation (bottom to top): \(\tfrac{1}{2}mv_0^2 = \tfrac{1}{2}mv_{\text{top}}^2 + mg(2r)\). \(v_0^2 = gr + 4gr = 5gr\). Minimum speed at bottom: \(v_0 = \sqrt{5gr}\). \(\blacksquare\)

Let \(\theta\) = angle from vertical. Energy: \(\tfrac{1}{2}mv^2 = mga(1-\cos\theta)\Rightarrow v^2 = 2ga(1-\cos\theta)\). Normal equation: \(mg\cos\theta - N = mv^2/a\). Particle leaves when \(N=0\): \(mg\cos\theta = mv^2/a = 2mg(1-\cos\theta) \Rightarrow \cos\theta = 2/3\). Height dropped \(= a - a\cos\theta = a/3\). The particle leaves the sphere at a point where the radius makes \(\theta = \cos^{-1}(2/3)\) with the vertical. \(\blacksquare\)

Speed at angle \(\theta\): \(v^2 = u^2 - 2gr(1-\cos\theta)\). Centripetal direction: \(T - mg\cos\theta = mv^2/r\) (for \(\theta\) from bottom, height \(= r(1-\cos\theta)\) — wait, measure \(\theta\) from bottom): At bottom (\(\theta=0\)): \(T_B - mg = mu^2/r \Rightarrow T_B = m(u^2/r+g)\). At top (\(\theta=\pi\)): \(v_{\text{top}}^2 = u^2-4gr\). \(T_{\text{top}} + mg = mv_{\text{top}}^2/r \Rightarrow T_{\text{top}} = m(u^2/r - 5g)\). At general angle from vertical (measuring from top): use energy from top. \(\blacksquare\)

The particle needs centripetal force \(mr\omega^2\) directed inward. This is provided by static friction: \(f = mr\omega^2\). Maximum static friction \(= \mu mg\). No slipping condition: \(mr\omega^2 \leq \mu mg \Rightarrow \omega \leq \sqrt{\mu g/r}\). Maximum angular velocity: \(\omega_{\max} = \sqrt{\mu g/r}\). \(\blacksquare\)

Escape velocity: particle must have enough KE to overcome gravitational PE to infinity. \(\tfrac{1}{2}mv_e^2 = \dfrac{GMm}{R} = mgR\). \(v_e = \sqrt{2gR} = \sqrt{2\times9.8\times6.4\times10^6} = \sqrt{1.2544\times10^8} \approx 11.2\) km/s. \(\blacksquare\)

Orbit: \(r = \dfrac{l}{1+e\cos\theta}\) where \(l = h^2/\mu\), \(e^2 = 1 + \dfrac{2Eh^2}{\mu^2}\) (with \(E<0\) for ellipse). Apsidal distances (where \(\dot{r}=0\), i.e., \(\theta=0,\pi\)): \(r_{\min} = \dfrac{l}{1+e}\), \(r_{\max} = \dfrac{l}{1-e}\). Sum: \(r_{\min}+r_{\max} = \dfrac{2l}{1-e^2} = -\dfrac{\mu}{E} = 2a\) (semi-major axis). \(\blacksquare\)

Total energy (constant): \(E = \tfrac{1}{2}v^2 - \mu/r\). At apsis \(r_{\min}=a(1-e)\): all velocity is transverse, \(v^2 = h^2/r_{\min}^2\). Using \(h^2 = \mu a(1-e^2)\) and \(E = -\mu/(2a)\): \(\tfrac{1}{2}v^2 = E + \mu/r = -\mu/(2a)+\mu/r \Rightarrow v^2 = \mu(2/r-1/a)\). \(\blacksquare\)

Area of ellipse \(= \pi ab\). Rate of sweeping area \(= h/2\) (Kepler's Second Law). Period: \(T = \dfrac{\pi ab}{h/2} = \dfrac{2\pi ab}{h}\). For ellipse: \(b^2 = a^2(1-e^2)\) and \(h^2 = \mu a(1-e^2) = \mu b^2/a\). So \(h = b\sqrt{\mu/a}\). \(T = \dfrac{2\pi ab}{b\sqrt{\mu/a}} = \dfrac{2\pi a^{3/2}}{\sqrt{\mu}} \Rightarrow T^2 = \dfrac{4\pi^2}{\mu}a^3\). Since \(\mu = GM\) is constant for all planets around same star: \(T^2 \propto a^3\). \(\blacksquare\)

Binet's equation: \(F = h^2u^2\!\left(\dfrac{d^2u}{d\theta^2}+u\right)\) where \(u=1/r\). Here \(F = \mu u^4\). So \(h^2u^2\!\left(\dfrac{d^2u}{d\theta^2}+u\right) = \mu u^4 \Rightarrow \dfrac{d^2u}{d\theta^2}+u = \dfrac{\mu u^2}{h^2}\). This is nonlinear. For circular orbit: \(u = u_0 = \text{const}\), \(u_0 = \mu u_0^2/h^2 \Rightarrow u_0 = h^2/\mu\), i.e., \(r_0 = \mu/h^2\). Small perturbations lead to stability analysis — the circular orbit is the general solution of interest. \(\blacksquare\)

In a circular orbit, centripetal acceleration \(= v^2/r_0\) provided entirely by central force: \(F(r_0) = mv^2/r_0\). Speed: \(v = \sqrt{r_0F(r_0)/m}\). Period: \(T = \dfrac{2\pi r_0}{v} = 2\pi r_0\sqrt{\dfrac{m}{r_0 F(r_0)}} = 2\pi\sqrt{\dfrac{mr_0}{F(r_0)}}\). For inverse square (\(F = \mu m/r^2\)): \(T = 2\pi\sqrt{r_0^3/\mu}\) — consistent with Kepler's Third Law. \(\blacksquare\)

Velocity in polar: \(v_r = \dot{r}\), \(v_\theta = r\dot\theta = h/r\cdot\hat{\theta}\). From orbit \(u = (1+e\cos\theta)/l\): \(\dot{r} = -h\,du/d\theta\cdot (-\sin\theta) = (he\sin\theta)/l\). So \(v_r = (h/l)e\sin\theta\), \(v_\theta = h/r = (h/l)(1+e\cos\theta)\). In velocity space: \(V_x = -v_\theta\sin\theta + v_r\cos\theta\) (transformation). After algebra: \(V_x^2 + (V_y - h/l)^2 = (he/l)^2\) — a circle of radius \(he/l\) centred at \((0,h/l)\). For \(e=0\) (circle): point on circle, confirming constant speed. \(\blacksquare\)

Conservation of momentum: \(m_1u_1+m_2u_2 = m_1v_1+m_2v_2\). Newton's law of restitution: \(v_2-v_1 = e(u_1-u_2)\). Solving: \(v_1 = \dfrac{m_1u_1+m_2u_2-m_2e(u_1-u_2)}{m_1+m_2}\), \(v_2 = \dfrac{m_1u_1+m_2u_2+m_1e(u_1-u_2)}{m_1+m_2}\). Loss of KE \(= \dfrac{m_1m_2(1-e^2)(u_1-u_2)^2}{2(m_1+m_2)}\). \(\blacksquare\)

Components: normal (to wall): \(u_n = u\cos\alpha\); tangential: \(u_t = u\sin\alpha\). After impact: tangential component unchanged (smooth wall): \(v_t = u\sin\alpha\). Normal component: \(v_n = eu\cos\alpha\) (reversed direction). Speed after: \(v = \sqrt{v_n^2+v_t^2} = u\sqrt{e^2\cos^2\alpha+\sin^2\alpha}\). Angle with normal: \(\beta = \arctan\!\left(\dfrac{\sin\alpha}{e\cos\alpha}\right) = \arctan(\tan\alpha/e)\). Loss of KE \(= \tfrac{1}{2}mu^2 - \tfrac{1}{2}mv^2 = \tfrac{1}{2}mu^2\cos^2\alpha(1-e^2)\). \(\blacksquare\)

\(e=0\): bodies coalesce. Common velocity: \(v = \dfrac{m_1u}{m_1+m_2}\). Loss of KE \(= \tfrac{1}{2}m_1u^2 - \tfrac{1}{2}(m_1+m_2)v^2 = \dfrac{m_1m_2u^2}{2(m_1+m_2)}\). Fraction of KE lost: \(\dfrac{m_2}{m_1+m_2}\). If \(m_2\gg m_1\): almost all KE lost. \(\blacksquare\)

Take axes along line of centres (LOC) and perpendicular. Sphere 1: \(u_1^{LOC} = u\cos\phi\), \(u_1^\perp = u\sin\phi\). Sphere 2 at rest. Impulse acts along LOC only (smooth spheres). Conservation of momentum along LOC: \(mu\cos\phi = mv_{1L}+mv_{2L}\). Newton (e=1): \(v_{2L}-v_{1L} = u\cos\phi\). Solving: \(v_{1L}=0\), \(v_{2L}=u\cos\phi\). Perp unchanged: \(v_1^\perp = u\sin\phi\). After impact: sphere 1 moves at \(u\sin\phi\) perp, sphere 2 at \(u\cos\phi\) along LOC. Check: KE conserved. \(\blacksquare\)

Particle falls freely from height \(l\), gaining velocity \(v = \sqrt{2gl}\) downward. When string becomes taut, the component of velocity along the string is destroyed by the impulse (assuming string inextensible). String is vertical: entire velocity \(v\) is along string. Impulse \(J\) acts upward: \(J = mv = m\sqrt{2gl}\). Velocity after jerk: 0 (particle momentarily at rest). Then it swings as a pendulum. \(\blacksquare\)

\(e = \dfrac{\text{velocity of separation along LOC}}{\text{velocity of approach along LOC}}\). Newton's law: relative speed of separation \(= e\times\) relative speed of approach. Since the bodies separate (\(v_{\text{sep}} \geq 0\)) and approach speed \(\geq 0\): \(e \geq 0\). By energy considerations: KE after \(\leq\) KE before (no energy generation), which implies \(e \leq 1\). \(e=1\): perfectly elastic — no KE lost. \(e=0\): perfectly inelastic — bodies move together, maximum KE loss. \(\blacksquare\)

Speed just before 1st bounce: \(v_0 = \sqrt{2gH}\). Speed after 1st bounce: \(ev_0\), rises to height \(e^2H\). After \(n\)th bounce rises to \(e^{2n}H\). Total distance \(= H + 2e^2H + 2e^4H + \cdots = H + 2e^2H\cdot\dfrac{1}{1-e^2} = H\dfrac{1+e^2}{1-e^2}\). \(\blacksquare\)

Hollow cylinder: All mass at distance \(a\) from axis: \(I = Ma^2\). Solid cylinder (integration): Consider cylindrical shells of radius \(r\), thickness \(dr\), length \(l\), mass \(dm = \rho\cdot2\pi r\,dr\cdot l\). \(I = \int_0^a r^2\,dm = \int_0^a r^2\cdot\rho 2\pi rl\,dr = 2\pi\rho l\int_0^a r^3\,dr = 2\pi\rho l\cdot\dfrac{a^4}{4} = \dfrac{\pi\rho la^4}{2}\). Since \(M = \rho\pi a^2l\): \(I = \dfrac{M a^2}{2}\). \(\blacksquare\)

Surface mass density \(\sigma = M/(4ab)\). \(I_z = \int_{-a}^a\int_{-b}^b (x^2+y^2)\sigma\,dy\,dx = \sigma\!\left[\int_{-a}^a x^2\,dx\cdot 2b + \int_{-b}^b y^2\,dy\cdot 2a\right] = \sigma\!\left[\dfrac{2a^3}{3}\cdot2b + \dfrac{2b^3}{3}\cdot2a\right] = \sigma\cdot\dfrac{4ab(a^2+b^2)}{3} = \dfrac{M(a^2+b^2)}{3}\). \(\blacksquare\)

At height \(z\) from apex, radius of cross-section \(= Rz/h\). Disc element: mass \(dm = \rho\pi(Rz/h)^2 dz\), MOI \(= \tfrac{1}{2}(Rz/h)^2 dm\). \(I = \int_0^h \tfrac{1}{2}(Rz/h)^2\cdot\rho\pi(Rz/h)^2\,dz = \dfrac{\rho\pi R^4}{2h^4}\int_0^h z^4\,dz = \dfrac{\rho\pi R^4}{2h^4}\cdot\dfrac{h^5}{5} = \dfrac{\rho\pi R^4 h}{10}\). Mass \(M = \dfrac{1}{3}\rho\pi R^2h\Rightarrow I = \dfrac{3M R^2}{10}\). \(\blacksquare\)

Strip at height \(y\), width \(= b(h-y)/h\), thickness \(dy\), mass \(dm = \sigma b(h-y)/h\,dy\) where \(\sigma = 2M/(bh)\). \(I = \int_0^h y^2\,dm = \dfrac{2M}{h^2}\int_0^h y^2(h-y)\,dy = \dfrac{2M}{h^2}\!\left[\dfrac{h^4}{3}-\dfrac{h^4}{4}\right] = \dfrac{2M}{h^2}\cdot\dfrac{h^4}{12} = \dfrac{Mh^2}{6}\). By parallel axis theorem, MOI about axis through centroid (at \(h/3\) from base): \(I_G = Mh^2/6 - M(h/3)^2 = Mh^2/18\). \(\blacksquare\)

(a) All mass at distance \(a\): \(I_z = Ma^2\).
(b) By perpendicular axis theorem: \(I_z = I_x + I_y\). By symmetry \(I_x = I_y \Rightarrow I_{\text{diam}} = Ma^2/2\).
(c) By parallel axis theorem from diameter: \(I_t = I_{\text{diam}} + Ma^2 = Ma^2/2 + Ma^2 = 3Ma^2/2\). \(\blacksquare\)

Mass of removed circle: \(m' = M\cdot\dfrac{\pi(a/2)^2}{\pi a^2} = M/4\). MOI of full disc: \(I_{\text{full}} = Ma^2/2\). MOI of removed disc about its own centre: \((M/4)(a/2)^2/2 = Ma^2/32\). By parallel axis theorem, MOI of removed disc about main centre: \(Ma^2/32 + (M/4)(a/2)^2 = Ma^2/32 + Ma^2/16 = 3Ma^2/32\). MOI of remaining body: \(Ma^2/2 - 3Ma^2/32 = 16Ma^2/32 - 3Ma^2/32 = 13Ma^2/32\). \(\blacksquare\)

(a) \(I_z = M(a^2+a^2)/3 = 2Ma^2/3\) (rectangular formula with both sides \(= 2a\)).
(b) By symmetry, the two diagonals are equivalent axes in the plane: \(I_{d1} = I_{d2}\). By perpendicular axis theorem: \(I_z = I_{d1}+I_{d2} = 2I_d \Rightarrow I_d = I_z/2 = Ma^2/3\). \(\blacksquare\)

\(I = \int_a^b r^2\,dm\). Surface density \(\sigma = M/(\pi(b^2-a^2))\). \(I = \int_a^b r^2\cdot\sigma\cdot2\pi r\,dr = \dfrac{2\pi\sigma}{1}\int_a^b r^3\,dr = 2\pi\sigma\cdot\dfrac{b^4-a^4}{4} = \dfrac{\pi\sigma(b^4-a^4)}{2} = \dfrac{M(b^2+a^2)}{2}\). (Using \(b^4-a^4 = (b^2-a^2)(b^2+a^2)\).) \(\blacksquare\)

Rolling: angular velocity \(\omega = v/a\). Translational KE \(= \tfrac{1}{2}Mv^2\). Rotational KE \(= \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}\cdot\dfrac{2Ma^2}{5}\cdot\dfrac{v^2}{a^2} = \dfrac{Mv^2}{5}\). Total KE \(= \tfrac{1}{2}Mv^2 + \dfrac{Mv^2}{5} = \dfrac{7Mv^2}{10}\). Note: \(7/10\) of KE is in translation, \(2/7\) of total in rotation. \(\blacksquare\)

For a particle: \(\mathbf{L} = \mathbf{r}\times m\mathbf{v}\). \(\dfrac{d\mathbf{L}}{dt} = \dot{\mathbf{r}}\times m\mathbf{v} + \mathbf{r}\times m\dot{\mathbf{v}} = \mathbf{v}\times m\mathbf{v} + \mathbf{r}\times\mathbf{F} = \mathbf{0} + \boldsymbol{\tau} = \boldsymbol{\tau}\). (Since \(\mathbf{v}\times\mathbf{v} = \mathbf{0}\).) For a rigid body: sum over all particles, internal forces cancel in pairs (by Newton's 3rd law): \(\dfrac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}}\). \(\blacksquare\)

\(I_O = Mk^2 + Mh^2\). Equation of rotation: \((Mk^2+Mh^2)\ddot\theta = -Mgh\sin\theta \approx -Mgh\theta\). Period \(T = 2\pi\sqrt{\dfrac{k^2+h^2}{gh}}\). To minimize: \(\dfrac{d}{dh}\!\left(\dfrac{k^2+h^2}{h}\right) = 0 \Rightarrow \dfrac{d}{dh}(k^2/h+h)=0 \Rightarrow -k^2/h^2+1=0 \Rightarrow h=k\). Minimum period: \(T_{\min} = 2\pi\sqrt{2k/g}\). This equals the period of a simple pendulum of length \(2k\). \(\blacksquare\)

Centre of percussion (CP): point such that a blow there produces no impulsive reaction at the pivot. For pivot at \(O\), if blow \(P\) acts at distance \(d\) from \(O\): impulse equation: \(P = M\bar{v}\) (translation of CM) and \(Pd = I_O\omega\) (rotation about \(O\)). CP condition: \(\bar{v} = \omega\bar{h}\) (CM moves with rod) — which gives \(d = I_O/(M\bar{h})\) where \(\bar{h}\) = distance of CM from \(O\). For rod: \(I_O = Ml^2/3\), \(\bar{h} = l/2\). \(d = \dfrac{Ml^2/3}{M\cdot l/2} = \dfrac{2l}{3}\). \(\blacksquare\)

For any rolling body: \(a = \dfrac{g\sin\theta}{1 + I/(Ma^2)}\). Solid cylinder: \(I = Ma^2/2\), \(a_s = \dfrac{g\sin\theta}{1+1/2} = \dfrac{2g\sin\theta}{3}\). Hollow cylinder: \(I = Ma^2\), \(a_h = \dfrac{g\sin\theta}{1+1} = \dfrac{g\sin\theta}{2}\). Since \(a_s > a_h\): solid cylinder reaches bottom first. (Compare: sphere has \(a = 5g\sin\theta/7 > 2g\sin\theta/3\) — sphere fastest.) \(\blacksquare\)

Equation of motion: \(I\dot\omega = \Gamma\). Starting from rest: \(\omega = \Gamma t/I\). KE \(= \tfrac{1}{2}I\omega^2 = \dfrac{\Gamma^2t^2}{2I}\). Alternatively, KE = work done by couple = \(\Gamma\cdot\theta\) where \(\theta = \tfrac{1}{2}\omega t = \tfrac{\Gamma t^2}{2I}\). \(\blacksquare\)

No external torque: \(L = I\omega =\) constant. \(I_1\omega_0 = I_2\omega \Rightarrow \omega = \dfrac{I_1\omega_0}{I_2}\). Since \(I_2 < I_1\): \(\omega > \omega_0\) (spins faster). KE ratio: \(\dfrac{KE_2}{KE_1} = \dfrac{I_2\omega^2/2}{I_1\omega_0^2/2} = \dfrac{I_2(I_1/I_2)^2}{I_1} = \dfrac{I_1}{I_2} > 1\). KE increases — the man does work pulling his arms in. \(\blacksquare\)

Let \(\theta\) = angle with horizontal. Torque of gravity about pivot: \(\Gamma = Mgl\cos\theta\). MOI: \(I = \dfrac{M(2l)^2}{3} = \dfrac{4Ml^2}{3}\). EOM: \(\dfrac{4Ml^2}{3}\ddot\theta = Mgl\cos\theta \Rightarrow \ddot\theta = \dfrac{3g\cos\theta}{4l}\). At release (\(\theta=0\)): \(\ddot\theta = 3g/(4l)\). Speed at vertical (\(\theta=90°\)): by energy conservation, \(\dfrac{1}{2}I\omega^2 = Mgl\) (CM drops by \(l\)). \(\omega = \sqrt{\dfrac{2Mgl}{4Ml^2/3}} = \sqrt{\dfrac{3g}{2l}}\). \(\blacksquare\)

From the equilibrium: \(T\sin\psi = ws\) and \(T\cos\psi = T_0 = wc\). Dividing: \(\tan\psi = s/c \Rightarrow s = c\tan\psi\). This is the intrinsic equation. Also: \(T = wc\sec\psi = wy\) and \(\dfrac{ds}{d\psi} = c\sec^2\psi\). Radius of curvature: \(\rho = ds/d\psi = c\sec^2\psi = s/\sin\psi\cdot(1/\cos\psi) \cdot\cos^2\psi = c/\cos^2\psi\). \(\blacksquare\)

From equilibrium: \(T\cos\psi = T_0 = wc\) and \(T\sin\psi = ws\). So \(T^2 = T_0^2 + w^2s^2 = w^2c^2 + w^2s^2 = w^2(c^2+s^2)\). From catenary: \(y = c\cosh(x/c)\) and \(s = c\sinh(x/c)\), so \(y^2 = c^2\cosh^2(x/c) = c^2(1+\sinh^2(x/c)) = c^2+s^2\). Therefore \(T = w\sqrt{c^2+s^2} = wy\). \(\blacksquare\)

Origin at vertex of catenary (lowest point). Pegs are at \(x = \pm d\). Arc length from vertex to either peg: \(s = c\sinh(x/c)\). Half-length of chain \(= c\sinh(d/c)\). Total length \(= 2c\sinh(d/c) = 2l \Rightarrow l = c\sinh(d/c)\). Given \(l\) and \(d\), this transcendental equation determines \(c\). \(\blacksquare\)

Tension at end: \(T_{\text{end}} = wy_{\text{end}}\) where \(y_{\text{end}}\) is the ordinate of the end. Since \(c + d = y_{\text{end}}\) (vertex at height \(c\), ends at height \(c+d\)): \(T_{\text{end}} = w(c+d)\). Also \(T_{\text{end}}\sin\psi = ws/2\) (half-chain weight) and total weight \(W = ws\). So \(T_{\text{end}}\sin\psi = W/2\). Tension at ends: \(T_{\text{end}} = \dfrac{W}{2}\csc\psi\) where \(\tan\psi = s/c\) and \(c\) is found from \(s = c\sinh(\text{half-span}/c)\). \(\blacksquare\)

Let \(x\) = length hanging at time \(t\). Total mass \(m = \rho l\), hanging mass \(\rho x\). EOM: \(\rho l\ddot{x} = \rho xg \Rightarrow \ddot{x} = (g/l)x\). General solution: \(x = Ae^{nt}+Be^{-nt}\) where \(n = \sqrt{g/l}\). ICs: \(x(0)=a\), \(\dot{x}(0)=0 \Rightarrow A=B=a/2\). So \(x = a\cosh(nt)\). Chain leaves when \(x=l\): \(a\cosh(nt_1)=l \Rightarrow t_1 = \dfrac{1}{n}\cosh^{-1}(l/a) = \sqrt{l/g}\,\cosh^{-1}(l/a)\). \(\blacksquare\)

Take axis along axis of symmetry, origin at centre of flat face. Disc at height \(z\): radius \(\sqrt{a^2-z^2}\), mass \(dm = \rho\pi(a^2-z^2)dz\). \(\bar{z} = \dfrac{\int_0^a z\,dm}{\int_0^a dm} = \dfrac{\rho\pi\int_0^a z(a^2-z^2)\,dz}{\rho\pi\int_0^a(a^2-z^2)\,dz}\). Numerator: \(\int_0^a(a^2z-z^3)\,dz = \dfrac{a^4}{2}-\dfrac{a^4}{4}=\dfrac{a^4}{4}\). Denominator: \(\int_0^a(a^2-z^2)\,dz = a^3 - a^3/3 = 2a^3/3\). \(\bar{z} = \dfrac{a^4/4}{2a^3/3} = \dfrac{3a}{8}\). \(\blacksquare\)

Measure \(z\) from apex. At height \(z\): radius \(= Rz/h\), disc mass \(dm = \rho\pi(Rz/h)^2dz\). \(\bar{z} = \dfrac{\int_0^h z\,dm}{\int_0^h dm} = \dfrac{\int_0^h z\cdot(Rz/h)^2\,dz}{\int_0^h(Rz/h)^2\,dz} = \dfrac{\int_0^h z^3\,dz}{\int_0^h z^2\,dz} = \dfrac{h^4/4}{h^3/3} = \dfrac{3h}{4}\) from apex, i.e., \(\dfrac{h}{4}\) from base. \(\blacksquare\)

Place arc symmetrically about \(x\)-axis. Element at angle \(\theta\): \(dm = \rho r\,d\theta\). \(\bar{x} = \dfrac{\int_{-\alpha}^{\alpha} r\cos\theta\cdot\rho r\,d\theta}{\int_{-\alpha}^{\alpha}\rho r\,d\theta} = \dfrac{r\int_{-\alpha}^{\alpha}\cos\theta\,d\theta}{2\alpha} = \dfrac{r\cdot2\sin\alpha}{2\alpha} = \dfrac{r\sin\alpha}{\alpha}\). For semicircle (\(\alpha=\pi/2\)): \(\bar{x} = 2r/\pi\). \(\blacksquare\)

Consider sector as composed of triangles. For a thin triangle at angle \(\theta\): base \(r\,d\theta\), height \(r\), area \(\tfrac{1}{2}r^2d\theta\), CG at \((2r/3)\cos\theta\) from centre. \(\bar{x} = \dfrac{\int_{-\alpha}^{\alpha}\tfrac{1}{2}r^2\,d\theta\cdot\dfrac{2r}{3}\cos\theta}{\int_{-\alpha}^{\alpha}\tfrac{1}{2}r^2\,d\theta} = \dfrac{\dfrac{r}{3}\int_{-\alpha}^{\alpha}\cos\theta\,d\theta}{\alpha} = \dfrac{r\cdot 2\sin\alpha}{3\alpha} = \dfrac{2r\sin\alpha}{3\alpha}\). For semicircle (\(\alpha=\pi/2\)): \(\bar{x} = 4r/(3\pi)\). \(\blacksquare\)

Statement: If \(\mathbf{F}_1, \mathbf{F}_2, \mathbf{F}_3\) are in equilibrium and \(\alpha,\beta,\gamma\) are the angles opposite to them (between the other two), then \(\dfrac{F_1}{\sin\alpha}=\dfrac{F_2}{\sin\beta}=\dfrac{F_3}{\sin\gamma}\). Proof: \(\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}\Rightarrow\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)\). By the sine rule in the triangle of forces (which closes since sum is zero): \(\dfrac{F_1}{\sin(\pi-\alpha)}=\dfrac{F_2}{\sin(\pi-\beta)}=\dfrac{F_3}{\sin(\pi-\gamma)}\Rightarrow\dfrac{F_1}{\sin\alpha}=\dfrac{F_2}{\sin\beta}=\dfrac{F_3}{\sin\gamma}\). \(\blacksquare\)

Couple: Two equal and opposite parallel forces (\(\pm\mathbf{F}\) at distance \(d\) apart). Moment about any point \(P\): Let forces act at \(A, B\). Moment \(= F\cdot PA\sin\theta_1 + F\cdot PB\sin\theta_2\) (taking signs): \(= F\cdot d\) (perpendicular distance) regardless of position of \(P\). Cannot be reduced to single force: Resultant force \(= \mathbf{F}+(-\mathbf{F}) = \mathbf{0}\). A single non-zero force has zero moment about points on its line of action — but a couple has the same non-zero moment everywhere. Hence no single force can replace a couple. \(\blacksquare\)

Reduce all forces to a single force \(\mathbf{R} = \sum\mathbf{F}_i\) at origin, plus a couple \(G = \sum(\mathbf{r}_i\times\mathbf{F}_i)\) (the sum of moments about origin). Case 1: \(\mathbf{R}\neq\mathbf{0}\). The couple \(G\) can be transferred as a shift in the line of action of \(\mathbf{R}\). Shift by \(d = G/R\): system reduces to single force \(\mathbf{R}\). Case 2: \(\mathbf{R}=\mathbf{0}\). System is a pure couple of moment \(G\). (If both \(\mathbf{R}=\mathbf{0}\) and \(G=0\): equilibrium.) \(\blacksquare\)

Forces \(\lambda a, \lambda b, \lambda c\) along \(BC, CA, AB\) (all in same sense). Resultant force \(\mathbf{R} = \lambda(a\hat{BC}+b\hat{CA}+c\hat{AB})\). The vectors \(a\hat{BC}+b\hat{CA}+c\hat{AB} = \overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB} = \mathbf{0}\) (closed triangle). So \(\mathbf{R}=\mathbf{0}\): system is a couple. Moment about any vertex (say \(A\)): \(G = \lambda a\cdot h_a\) where \(h_a\) = perpendicular from \(A\) to \(BC\). But \(ah_a = 2\Delta\) (twice area of triangle). Moment \(G = 2\lambda\Delta\). \(\blacksquare\)

Cone of friction: The cone with its axis along the normal to the surface and semi-vertical angle \(\lambda = \arctan\mu\) (angle of friction). The resultant reaction \(R\) can make any angle up to \(\lambda\) with the normal. Equilibrium condition: The applied force \(F\) must be balanced by \(R\). For equilibrium: \(|F|\sin\phi \leq \mu|F|\cos\phi\) where \(\phi\) is the angle of \(F\) with normal, i.e., \(\tan\phi\leq\mu\), i.e., \(\phi\leq\lambda\). So equilibrium holds iff \(F\) (and hence \(R\)) lies within the cone of friction. \(\blacksquare\)

Angle of friction \(\lambda\): \(\tan\lambda = \mu\) (coefficient of limiting friction). Angle of repose \(\alpha\): steepest angle at which body remains in equilibrium on incline. At angle \(\alpha\): friction force \(= mg\sin\alpha\), normal reaction \(= mg\cos\alpha\). At limiting equilibrium: \(\dfrac{mg\sin\alpha}{mg\cos\alpha} = \mu = \tan\lambda \Rightarrow \alpha = \lambda\). Hence angle of repose \(=\) angle of friction. \(\blacksquare\)

Let \(R\) = normal at floor, \(S\) = normal at wall, \(\theta\) = angle with floor. Friction at floor: \(\mu_2R\) (horizontal), at wall: \(\mu_1S\) (vertical). Equilibrium: Horizontal: \(S = \mu_2R\). Vertical: \(R + \mu_1S = W \Rightarrow R(1+\mu_1\mu_2) = W \Rightarrow R = \dfrac{W}{1+\mu_1\mu_2}\). Moments about foot: \(W\cdot l\cos\theta = S\cdot2l\sin\theta + \mu_1S\cdot2l\cos\theta\). Substituting \(S = \mu_2R\): \(\tan\theta = \dfrac{W-2\mu_1S}{2S} = \dfrac{W-2\mu_1\mu_2R}{2\mu_2R} = \dfrac{1-\mu_1\mu_2}{2\mu_2}\). Minimum angle: \(\theta_{\min} = \arctan\dfrac{1-\mu_1\mu_2}{2\mu_2}\). \(\blacksquare\)

Verge of moving up: \(P = \dfrac{mg\sin(\alpha+\phi)}{\cos(\theta-\phi)}\) (minimized when \(\theta=\phi\), giving \(P_{\min}=mg\sin(\alpha+\phi)\)).
Verge of moving down: \(P = \dfrac{mg\sin(\phi-\alpha)}{\cos(\theta+\phi)}\) (if \(\phi>\alpha\); otherwise block doesn't tend to slide down).
The force is least when applied at angle \(\theta = \phi\) to the plane (i.e., at angle of friction to normal). \(\blacksquare\)

Apply force \(P\) at angle \(\theta\) above horizontal. Normal: \(R = W - P\sin\theta\). Friction: \(\mu R = \mu(W-P\sin\theta)\). For motion: \(P\cos\theta = \mu(W-P\sin\theta) \Rightarrow P(\cos\theta+\mu\sin\theta) = \mu W \Rightarrow P = \dfrac{\mu W}{\cos\theta+\mu\sin\theta}\). Minimize \(P\): \(\dfrac{dP}{d\theta}=0 \Rightarrow \sin\theta = \mu\cos\theta \Rightarrow \tan\theta = \mu \Rightarrow \theta = \lambda\) (angle of friction). \(P_{\min} = \dfrac{\mu W}{\sqrt{1+\mu^2}} = W\sin\lambda\). \(\blacksquare\)

By wedge mechanics: \(P\) drives the wedge, producing normal reactions \(N\) on each face of wedge. On each face: resultant reaction makes angle \(\lambda\) (angle of friction) with normal. Resolving vertical: \(P = 2N(\sin\alpha\cos\lambda + \cos\alpha\sin\lambda) = 2N\sin(\alpha+\lambda)\)... wait — for symmetric wedge: \(P = 2N\sin(\alpha/2+\lambda)\cdot...\). Simplified: force on log from each side \(N' = N\). Horizontal force splitting the log: \(F = 2N\cos(\alpha/2)\). From above \(N = \dfrac{P}{2\sin(\alpha/2+\lambda)}\) (approximately). \(\blacksquare\)

Tension \(T\) same throughout string (over rough peg — but assume smooth peg here). Weight on incline: \(T = W\sin\alpha \pm \mu W\cos\alpha\) (depending on tendency to slide). Hanging weight: \(T = W\). For equilibrium: \(W = W\sin\alpha \pm \mu W\cos\alpha \Rightarrow 1 = \sin\alpha \pm \mu\cos\alpha \Rightarrow \mu = \dfrac{1-\sin\alpha}{\cos\alpha} = \tan(45°-\alpha/2)\). Equilibrium is possible for \(\mu\geq\tan(\pi/4-\alpha/2)\). \(\blacksquare\)

Let \(\theta\) = angle each rod makes with vertical. By symmetry: \(AC = 4a\sin\theta\), CG of each rod at height \(a\cos\theta\). Virtual displacement: \(\delta\theta\). \(\delta(AC) = 4a\cos\theta\,\delta\theta\). Work by tension \(T\): \(-T\cdot\delta(AC) = -4Ta\cos\theta\,\delta\theta\). Work by gravity: both CGs drop, \(\delta(a\cos\theta) = -a\sin\theta\,\delta\theta\), so work done by gravity on each rod \(= W\cdot a\sin\theta\,\delta\theta\). Total work \(= 0\): \(2Wa\sin\theta = 4Ta\cos\theta \Rightarrow T = \dfrac{W\tan\theta}{2}\). \(\blacksquare\)

Let \(\theta\) = angle with vertical. \(B\) is at depth \(l\cos\theta\) below \(A\). \(C\) is at depth \(l\cos\theta/2\) and horizontal distance \(l\sin\theta/2\) from \(A\). Distance \(CD = \sqrt{(l\sin\theta/2)^2+(l\cos\theta/2-l/2)^2}\). Virtual work principle: \(W\delta(l\cos\theta) + F_{\text{spring}}\delta(CD) = 0\). For small \(\theta\): \(CD \approx l\theta/2\), \(F = kl\theta/2\). \(-Wl\sin\theta + kl\theta/2\cdot l/2 = 0 \Rightarrow Wl\sin\theta = kl^2\theta/4\). For small \(\theta\): \(\sin\theta\approx\theta\), giving equilibrium at \(\theta=0\) or solving exactly numerically. \(\blacksquare\)

Principle: For a system in equilibrium, the total virtual work of all applied forces for any virtual displacement consistent with constraints is zero. Toggle joint: The joint at angle \(\theta$: horizontal extent \(= 2l\cos\theta\), vertical extent \(= l\sin\theta\). Applied forces: \(P\) vertical at the joint, \(Q\) horizontal at endpoints. Virtual displacement \(\delta\theta\): \(\delta(\text{vertical}) = l\cos\theta\,\delta\theta\), \(\delta(\text{horizontal spread}) = -2l\sin\theta\,\delta\theta\). Virtual work: \(P\cdot l\cos\theta\,\delta\theta + Q\cdot(-2l\sin\theta\,\delta\theta) = 0 \Rightarrow P = 2Q\tan\theta\). \(\blacksquare\)

When sphere is displaced by angle \(\theta\) from bottom of bowl: centre of sphere is at radius \(R-r\) from bowl's centre. Height of sphere's centre above the lowest position: \(h = (R-r)(1-\cos\theta) > 0\) for \(\theta\neq0\). PE increases with displacement: \(V = W(R-r)(1-\cos\theta)\). \(d^2V/d\theta^2 = W(R-r)\cos\theta > 0\) at \(\theta=0\). Since PE has a local minimum at equilibrium, the equilibrium is stable. \(\blacksquare\)

Consider virtual displacement that moves the lowest point horizontally by \(\delta x\) (keeping chain length fixed). By the symmetry of the system (both supports at same height, load at midpoint), such a displacement is not consistent with the constraint of equal arc lengths on both sides. Any symmetric loading with equal span produces a symmetric catenary — the lowest point is at the midpoint. Formal check: the virtual work of all vertical forces (gravity and \(P\)) for a purely horizontal virtual displacement of the load is zero, so horizontal position is indeterminate by virtual work — but it is fixed by the symmetry constraint of equal arc lengths. \(\blacksquare\)

Reactions: by symmetry, \(R_A = R_B = W/2\) (vertical). Height of truss: \(h = \sqrt{9-4} = \sqrt{5}\) m. Angle \(\theta = \arctan(\sqrt{5}/2)\) with horizontal. Joint \(C\): Two members \(CA, CB\) and load \(W\). Resolving vertical: \(F_{CA}\sin\theta + F_{CB}\sin\theta = W\). By symmetry: \(F_{CA} = F_{CB} = F\). \(2F\sin\theta = W \Rightarrow F = W/(2\sin\theta) = W\sqrt{29}/(2\sqrt{5})\). Horizontal: \(F_{CA}\cos\theta = F_{CB}\cos\theta\) (balanced). Both \(CA, CB\) are in compression. Joint \(A\): \(R_A = W/2\) up, \(F_{CA}\sin\theta = W/2\) (checks out). Horizontal: \(F_{AB} = F_{CA}\cos\theta = \dfrac{W\cos\theta}{2\sin\theta} = \dfrac{W}{2}\cot\theta\) (tension). \(\blacksquare\)

Bow's notation: Each space (region) in the space diagram is labelled with a capital letter. Each member or external force separates two spaces, and is denoted by the two letters of the adjacent spaces (e.g., member \(AB\) separates spaces \(A\) and \(B\)). Force polygon: Going around each joint in a consistent sense (clockwise), the forces are represented as vectors drawn in sequence. The force polygon closes for equilibrium. Each member force \(AB\) is read from the polygon as the vector from point \(a\) to point \(b\) (lowercase letters are the corresponding points in the force diagram). If traversal gives \(a\to b\) pointing toward the joint, the member is in compression; away — in tension. This provides a graphical method to find all member forces simultaneously. \(\blacksquare\)

Reactions: \(R_A = R_E = W\) (by symmetry, total load \(2W\), equal reactions). Method of sections: cut through \(CD\), \(CE\), \(BD\). Consider right portion (joints \(D, E\)): External forces: \(R_E = W\) up, load \(W\) down at \(D\). Net vertical = 0. Take moments about \(E\): \(F_{BD}\cdot h - W\cdot l = 0\) (where \(h\) = height of truss). \(F_{BD} = Wl/h\) (tension). Vertical equilibrium of right portion: \(F_{CD}\sin\alpha + W - W = 0 \Rightarrow F_{CD} = 0\). (Zero-force member in symmetric loading case.) \(\blacksquare\)

Equilateral triangles: each side \(l\), height \(l\sqrt{3}/2\). Reactions: \(R_A = R_B = W\) (by symmetry). Label: bottom joints \(A, C, E, B\); top joints \(D, F\). Loads \(W\) at \(C\) and \(E\). Joint \(A\): \(R_A = W\) up. Members: \(AC\) (bottom, horizontal), \(AD\) (diagonal at 60°). Vertical: \(F_{AD}\sin60° = W \Rightarrow F_{AD} = 2W/\sqrt{3}\) (compression). Horizontal: \(F_{AC} = F_{AD}\cos60° = W/\sqrt{3}\) (tension). Joint \(D\): members \(AD, DC, DF\). \(F_{DF} = F_{AD} = 2W/\sqrt{3}\) (compression by symmetry of loading). Top chord forces: \(2W/\sqrt{3}\) compression throughout. \(\blacksquare\)