UPSC Mathematics Optional · Paper I · Complete Formulae
Parallelogram Law: \(R^2=P^2+Q^2+2PQ\cos\theta\), \(\tan\alpha=\dfrac{Q\sin\theta}{P+Q\cos\theta}\).
Lami's Theorem: If three concurrent forces are in equilibrium, \(\dfrac{P}{\sin\alpha}=\dfrac{Q}{\sin\beta}=\dfrac{R}{\sin\gamma}\) where \(\alpha,\beta,\gamma\) are angles opposite to \(P,Q,R\).
\(\sum F_x=0\), \(\sum F_y=0\), \(\sum M_O=0\) (moments about any point \(O\)).
For three forces in equilibrium: they must be concurrent or all parallel.
Moment of force \(F\) about point \(O\) = \(F\times\perp\text{ distance}\). Couple = two equal, opposite, non-collinear forces. Moment of couple = \(F\times d\) (constant, independent of pivot).
Two parallel forces \(P\) and \(Q\) (same sense): Resultant \(R=P+Q\), divides \(PQ\) internally in ratio \(Q:P\).
Opposite sense: \(R=P-Q\), divides externally in ratio \(Q:P\).
\(\bar x=\dfrac{\sum m_i x_i}{\sum m_i}\), \(\bar y=\dfrac{\sum m_i y_i}{\sum m_i}\).
Continuous: \(\bar x=\dfrac{\int x\,dm}{\int dm}\).
A system is in equilibrium iff the total virtual work done by all active forces in any virtual displacement is zero: \(\delta W = \sum F_i\cdot\delta r_i = 0\).
Powerful for systems with constraints — no need to find reactions.
Equation: \(y=c\cosh\dfrac{x}{c}\) (vertex at \((0,c)\) above lowest point).
Span \(2l\), sag \(d\), parameter \(c\): \(l=c\sinh(l/c)\approx\) ... (solve numerically or by approximation).
For nearly taut chain (\(d\ll l\)): \(c\approx\dfrac{l^2}{2d}\), \(T_0=\dfrac{wl^2}{2d}\).
Use \(v\dfrac{dv}{dx}=\dfrac{d^2x}{dt^2}=f\) (acceleration). \(v=\dfrac{dx}{dt}\). Energy: \(\frac{1}{2}mv^2+V=E\).
\(\ddot{x}+\omega^2 x=0\). Solution: \(x=a\cos(\omega t+\phi)\).
\(\ddot{x}+2k\dot{x}+\omega^2 x=F_0\cos pt\).
Initial velocity \(u\), angle \(\alpha\) to horizontal. \(x=u\cos\alpha\cdot t\), \(y=u\sin\alpha\cdot t-\frac{1}{2}gt^2\).
In polar coords with force \(F\) towards pole:
Angular momentum conservation (Areal velocity): \(h=r^2\dot\theta=\text{const}\). (Kepler II: equal areas in equal times.)
Pedal equation of orbit: \(F = \dfrac{h^2}{p^3}\dfrac{dp}{dr}\cdot\dfrac{1}{r}\cdot\dfrac{p}{r}\)... use \(u=1/r\):
\[F = \frac{h^2 u^2}{m}\left(u+\frac{d^2u}{d\theta^2}\right)\] (Binet's equation)
For inverse-square law \(F=\mu/r^2\): orbit is a conic. Ellipse if \(E<0\), parabola if \(E=0\), hyperbola if \(E>0\).
Orbit \(\dfrac{l}{r}=1+e\cos\theta\) (focus-directrix). Semi-latus rectum \(l=h^2/\mu\). Eccentricity \(e=\sqrt{1+2Eh^2/\mu^2}\).
Velocity: \(v^2=\mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right)\) (vis-viva equation).
Parallel axis: \(I = I_G + Md^2\) where \(I_G\) is MOI about axis through C.G., \(d\) = distance between axes.
Perpendicular axis (lamina only): \(I_z=I_x+I_y\) where \(z\perp\) plane of lamina.
\(\tau = I\alpha\) where \(\tau\)=torque, \(I\)=MOI about axis, \(\alpha=\ddot\theta\).
KE of rotation \(=\dfrac{1}{2}I\omega^2\). Angular momentum \(L=I\omega\).
Conservation of angular momentum: if net torque = 0, \(I\omega=\text{const}\).
Period: \(T=2\pi\sqrt{\dfrac{I}{Mgh}}\) where \(I\)=MOI about pivot, \(h\)=distance from pivot to C.G.
Equivalent simple pendulum length: \(l=\dfrac{I}{Mh}=\dfrac{k^2}{h}+h\) where \(k\)=radius of gyration about C.G.
Minimum \(T\) when \(h=k\) (pivot at radius of gyration from C.G.). Then \(l_{min}=2k\).