A double pendulum swings in a vertical plane under gravity,

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Lagrangian mechanics · Lagrangian mechanics

Action and Symmetry, Ch Constraints, generalised coordinates and Lagrange's equations twice over

Problem

g=9.8\,\mathrm{m\,s^{-2}}

. A light rigid rod of length

\ell_1=0.40\,\mathrm m

is pivoted at a fixed point and carries a bob of mass

m_1=0.50\,\mathrm{kg}

at its end; a second light rigid rod of length

\ell_2=0.30\,\mathrm m

is pivoted at that bob and carries a second bob of mass

m_2=0.30\,\mathrm{kg}

at its end. Let

\theta_1

and

\theta_2

be the angles the upper and lower rods make with the downward vertical, both measured the same way round, so that with

\hat{\mathbf x}

horizontal and

\hat{\mathbf y}

vertically downward,

\mathbf r_1=\ell_1(\sin\theta_1,\cos\theta_1),\qquad \mathbf r_2=\mathbf r_1+\ell_2(\sin\theta_2,\cos\theta_2).

The upper bob moves through air with negligible resistance, but the lower bob is dragged through a viscous fluid that exerts on it a force

-b\mathbf v_2

, where

\mathbf v_2=\dot{\mathbf r}_2

is its velocity in the laboratory frame and

b=0.050\,\mathrm{kg\,s^{-1}}

. Both pivots are frictionless. ((a)) Using d'Alembert's principle, derive the generalised forces

\mathcal Q_{\theta_1}

and

\mathcal Q_{\theta_2}

(gravity on both bobs together with the drag on the lower bob) as functions of

\theta_1,\theta_2,\dot\theta_1,\dot\theta_2

. ((b)) State, in one sentence, why this problem cannot be solved by writing a single

L=T-V

for the whole system and applying Hamilton's principle alone. ((c)) Evaluate

\mathcal Q_{\theta_1}

and

\mathcal Q_{\theta_2}

at the instant

\theta_1=30^\circ

\theta_2=60^\circ

\dot\theta_1=1.0\,\mathrm{rad\,s^{-1}}

\dot\theta_2=-0.50\,\mathrm{rad\,s^{-1}}