UPSC 2016 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

A hoop with radius $r$ is rolling, without slipping, down an inclined plane of length $l$ and with angle of inclination $\phi$ . Assign appropriate generalized coordinates to the system. Determine the constraints, if any. Write down the Lagrangian equations for the system. Hence or otherwise determine the velocity of the hoop at the bottom of the inclined plane.

Technique

One holonomic rolling constraint $s=r\psi$ ⇒ single DOF $s$ ; Lagrangian $L=M\dot s^2-Mg(l-s)\sin\phi$ (hoop’s rotational KE equals translational KE since $I=Mr^2$ ); Euler–Lagrange gives $a=\tfrac12 g\sin\phi$ ; kinematics/energy give $v=\sqrt{gl\sin\phi}$ .

Solution

Setup. A hoop (thin ring) of mass $M$ , radius $r$ , moment of inertia about its centre $I=Mr^2$ (all mass at the rim). It rolls without slipping down a plane inclined at angle $\phi$ .

Step 1 — Generalized coordinates and constraint

Natural coordinates: $s$ = distance the centre has travelled down the incline, and $\psi$ = angle the hoop has rotated. Rolling without slipping links them:

\boxed{\text{constraint: } s=r\psi\ \ (\dot s=r\dot\psi).}

This is a holonomic constraint, reducing the system to one degree of freedom; take the generalized coordinate to be $s$ .

Step 2 — Lagrangian

Kinetic energy = translation of the centre + rotation about the centre:

T=\tfrac12 M\dot s^2+\tfrac12 I\dot\psi^2=\tfrac12 M\dot s^2+\tfrac12 (Mr^2)\Big(\frac{\dot s}{r}\Big)^2=\tfrac12 M\dot s^2+\tfrac12 M\dot s^2=M\dot s^2.

(For a hoop the rotational KE equals the translational KE.) Taking the bottom of the incline as datum, with the hoop having descended a height $s\sin\phi$ below the top, the potential energy is $V=Mg(l-s)\sin\phi$ (decreasing as $s$ grows). The Lagrangian:

L=T-V=M\dot s^2-Mg(l-s)\sin\phi.

Step 3 — Lagrange’s equation

\frac{d}{dt}\frac{\partial L}{\partial\dot s}-\frac{\partial L}{\partial s}=0,\qquad \frac{\partial L}{\partial\dot s}=2M\dot s,\quad \frac{\partial L}{\partial s}=Mg\sin\phi.

2M\ddot s-Mg\sin\phi=0\ \Longrightarrow\ \boxed{\ddot s=\tfrac12 g\sin\phi.}

So the hoop accelerates at half the value $g\sin\phi$ of a frictionless slide — the other half of the gravity component drives the rotation.

Step 4 — Velocity at the bottom

Constant acceleration $a=\tfrac12 g\sin\phi$ over distance $l$ from rest ( $v^2=2al$ ):

v^2=2\Big(\tfrac12 g\sin\phi\Big)l=gl\sin\phi.

Answer

\boxed{\;v=\sqrt{g\,l\sin\phi}.\;}