← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

Hamilton's equations · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

Using Hamilton’s equation, find the acceleration for a sphere rolling down a rough inclined plane, if $x$ be the distance of the point of contact of the sphere from a fixed point on the plane.

Technique

Build $\mathcal L$ with the rolling constraint folded in ($\dot\phi=\dot x/r$), pass to $H=T+V$ in terms of $(x,p)$, then apply Hamilton’s equations $\dot x=\partial_p H$, $\dot p=-\partial_x H$ and combine.

Solution

Setup. Uniform solid sphere, mass $M$, radius $r$, moment of inertia $I=\tfrac25 Mr^2$, rolling without slipping down a plane inclined at angle $\theta$ to the horizontal. Let $x$ be the distance of the contact point along the plane from a fixed point (down-plane positive). Rolling: the spin $\phi$ satisfies $x=r\phi$, so $\dot\phi=\dot x/r$. There is a single generalised coordinate $x$.

Step 1 — Lagrangian (to build the Hamiltonian)

Kinetic energy (translation + rotation, with $\dot\phi=\dot x/r$):

Potential energy (height drops by $x\sin\theta$):

So $\mathcal L=T-V=\tfrac{7}{10}M\dot x^2+Mgx\sin\theta.$

Step 2 — Generalised momentum and Hamiltonian

The Hamiltonian $H=p\dot x-\mathcal L$ (a natural system, so $H=T+V$ expressed in $p$):

Step 3 — Hamilton’s equations

Step 4 — Acceleration

From $\dot x=\dfrac{5p}{7M}$, differentiate: $\ddot x=\dfrac{5\dot p}{7M}=\dfrac{5}{7M}\,(Mg\sin\theta)$.

Answer