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UPSC 2017 Maths Optional Paper 2 Q6c — Step-by-Step Solution

20 marks · Section B

Lagrange's equations · Mechanics & Fluid Dynamics · asked 9× in 13 yrs · Read the full method →

Question

Two uniform rods $AB,AC$, each of mass $m$ and length $2a$, are smoothly hinged together at $A$ and move on a horizontal plane. At time $t$, the mass centre of the rods is at the point $(\xi,\eta)$ referred to fixed perpendicular axes $Ox,Oy$ in the plane, and the rods make angles $\theta\pm\phi$ with $Ox$. Prove that the kinetic energy of the system is

Also derive Lagrange’s equations of motion for the system if an external force with components $[X,Y]$ along the axes acts at $A$.

Technique

König decomposition (CM translation + spin) for each rod; express the hinge $A$ through the system CM via $(\ast)$ using $\cos(\theta\pm\phi)$ sum formulas; Lagrangian mechanics with generalized forces $Q_q=\mathbf F\cdot\partial\mathbf r_A/\partial q$.

Solution

Setup. Each rod has length $2a$, so its own centre of mass is at distance $a$ from the hinge $A$, and its moment of inertia about its own centre is $I=\dfrac{m(2a)^2}{12}=\dfrac{ma^2}{3}$. Let $A=(x_A,y_A)$.

• Rod $AB$ lies along direction $\theta+\phi$; its centre $G_1=A+a(\cos(\theta+\phi),\sin(\theta+\phi))$, angular velocity $\dot\theta+\dot\phi$.
• Rod $AC$ lies along direction $\theta-\phi$; its centre $G_2=A+a(\cos(\theta-\phi),\sin(\theta-\phi))$, angular velocity $\dot\theta-\dot\phi$.

Step 1 — Locate the system mass centre and the hinge

Equal masses, so the system centre of mass is the midpoint of $G_1,G_2$:

Using $\cos(\theta\pm\phi)$ sum formulas, $\cos(\theta+\phi)+\cos(\theta-\phi)=2\cos\theta\cos\phi$ and similarly for sine:

Hence the hinge is

Step 2 — Kinetic energy of each rod

By König’s theorem, each rod contributes $\tfrac12 m\,v_{G_i}^2+\tfrac12 I\,\omega_i^2$:

Differentiating $G_1,G_2$ (with $A$ expressed through $(\ast)$) and substituting $I=\tfrac{ma^2}{3}$ and simplifying with the identities $\cos(\theta+\phi)+\cos(\theta-\phi)=2\cos\theta\cos\phi$, etc., all cross terms in $\dot\theta\dot\phi$, $\dot\xi\dot\theta,\ldots$ cancel, leaving the clean diagonal form.

Step 3 — Result for the kinetic energy

The structure is transparent: $m(\dot\xi^2+\dot\eta^2)$ is the translation of the total mass $2m$ written as $\tfrac12(2m)(\dot\xi^2+\dot\eta^2)$; the $\sin^2\phi$ and $\cos^2\phi$ split comes from the two rods opening symmetrically about the line $\theta$ — the $\theta$-motion swings them transversely (weight $\sin^2\phi$) while the $\phi$-motion is the scissor opening (weight $\cos^2\phi$); the $\tfrac13$ in each is the rod’s own spin inertia.

Step 4 — Generalized forces from $[X,Y]$ at $A$

Generalized coordinates: $(\xi,\eta,\theta,\phi)$. The force $(X,Y)$ acts at $A=(x_A,y_A)$, so the generalized force conjugate to a coordinate $q$ is $Q_q=X\dfrac{\partial x_A}{\partial q}+Y\dfrac{\partial y_A}{\partial q}$. From $(\ast)$:

Step 5 — Lagrange’s equations $\dfrac{d}{dt}\dfrac{\partial T}{\partial\dot q}-\dfrac{\partial T}{\partial q}=Q_q$

Answer