Question 2 of 6: Two-Particle Rod with Perpendicular-Slide Supports (Lagrange Multiplier)

Question 2: Two-Particle Rod with Perpendicular-Slide Supports (Lagrange Multiplier) (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two identical particles of mass $m$ at the ends of a massless rigid rod of length $l$; each particle sits on a frictionless flat-base ("skate") support oriented so its own velocity is constrained to be perpendicular to the rod; $\theta$ is the rod's angle from the $X$-axis, $\dot\theta$ its angular rate.

Find. (a) the holonomic (geometric) constraints; (b) the nonholonomic (velocity) constraint and its scleronomic/rheonomic classification; (c) the system kinetic energy in terms of the centre coordinates; (d) the equations of motion via a Lagrange multiplier; (e) the constraint force and its physical meaning; (f) the shape of the path the bar's centre traces.

Approach. Reduce the four raw coordinates to the centre coordinates $(x,y,\theta)$ using the rod's fixed length (holonomic), express each skate's no-slip-along-the-rod condition as a single velocity (Pfaffian) constraint, build $L=T$ (no potential on the frictionless horizontal plane), and adjoin the constraint with an undetermined multiplier $\lambda$ in the Euler–Lagrange equations.

Part (a) — holonomic constraints. With $x=(x_1+x_2)/2$, $y=(y_1+y_2)/2$ and the rod's fixed length $l$ oriented at angle $\theta$ (so $\cos\theta=(x_2-x_1)/l$, $\sin\theta=(y_2-y_1)/l$), the four raw coordinates reduce to $(x,y,\theta)$ through the purely geometric — hence holonomic — relations: $$\boxed{x_1 = x-\tfrac{l}{2}\cos\theta,\quad y_1 = y-\tfrac{l}{2}\sin\theta,\quad x_2 = x+\tfrac{l}{2}\cos\theta,\quad y_2 = y+\tfrac{l}{2}\sin\theta}$$ (equivalently: the two centroid definitions plus the fixed-length condition $(x_2-x_1)^2+(y_2-y_1)^2=l^2$). No velocities appear, so all four are holonomic; they cut the configuration space from 4 raw coordinates down to 3, $(x,y,\theta)$.

Part (b) — nonholonomic constraint. Each skate forces its particle's velocity to have zero component along the rod direction $(\cos\theta,\sin\theta)$. Differentiating Part (a) and dotting with the rod direction, the $\dot\theta$ terms cancel identically for BOTH particles, leaving the same single condition from either one: $$\boxed{\dot x\cos\theta+\dot y\sin\theta = 0}$$ — i.e. the centre's own velocity is likewise perpendicular to the rod. Its coefficients $(\cos\theta,\sin\theta,0)$ depend only on the configuration, never on $t$ explicitly, and there is no function $f(x,y,\theta)$ whose exact differential reproduces it once $\theta$ is itself a dynamical variable — so it is nonholonomic. Because time never enters explicitly, it is scleronomic, not rheonomic.

Part (c) — kinetic energy. Summing $\tfrac12 m(\dot x_i^2+\dot y_i^2)$ over both particles and substituting Part (a), the cross terms cancel exactly: $$\boxed{T = m(\dot x^2+\dot y^2) + \tfrac14 m l^2\dot\theta^2}$$ — translational kinetic energy of the centre (total mass $2m$) plus rotational kinetic energy about the centre, with $I_{cm}=2m(l/2)^2=ml^2/2$.

Part (d) — Lagrange's equations with multiplier. There is no potential energy (frictionless horizontal plane), so $L=T$. Writing the Part (b) constraint in Pfaffian form $a_x\dot x+a_y\dot y+a_\theta\dot\theta=0$ with $(a_x,a_y,a_\theta)=(\cos\theta,\sin\theta,0)$ and adjoining it with a multiplier $\lambda$ in $\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=\lambda a_i$ gives: $$\boxed{2m\ddot x=\lambda\cos\theta,\qquad 2m\ddot y=\lambda\sin\theta,\qquad \tfrac12 ml^2\ddot\theta=0}$$ together with the constraint $\dot x\cos\theta+\dot y\sin\theta=0$. The $\theta$-equation carries no multiplier term at all (since $a_\theta=0$), so it immediately gives $\ddot\theta=0$, i.e. $\dot\theta=\Omega=\text{const}$: the bar's own spin rate is unaffected by the constraint force.

Part (e) — constraint force. The generalized constraint force is $(\lambda\cos\theta,\ \lambda\sin\theta,\ 0)$: a force of magnitude $|\lambda|$ acting on the CENTRE, directed along the rod — exactly what is needed to keep enforcing $\dot x\cos\theta+\dot y\sin\theta=0$ at every instant (the two skates individually only react perpendicular to themselves, but their combined effect on the centroid is a net force along the rod's own axis). Differentiating the constraint once more and substituting the Part (d) equations of motion gives: $$\boxed{\lambda = -2m\,\Omega\,v_\perp},\qquad v_\perp \equiv \dot y\cos\theta-\dot x\sin\theta$$ where $v_\perp$ is the centre's speed along the one direction the skates DO allow. Physically, $\lambda$ is a Coriolis-type reaction: it vanishes unless the bar is BOTH spinning ($\Omega\neq0$) AND translating ($v_\perp\neq0$) at once, in which case the skates must supply a steady lateral push to keep bending the centre's path into a circle (Part f).

Part (f) — motion path. Since $\ddot\theta=0$ and $\dot v_\perp=0$ as well, both $\Omega$ and $v_\perp$ are constants of the motion, and $\dot x=-v_\perp\sin\theta$, $\dot y=v_\perp\cos\theta$ are exactly the velocity components of uniform circular motion. So:

• If $\Omega=0$: the bar translates in a straight line perpendicular to its own (fixed) orientation — pure sideways sliding, no rotation.
• If $\Omega\neq0$: the centre traces a circle of radius $R=v_\perp/\Omega$, with the rod's orientation turning in lock-step with the angular position around that circle — the same structure as a two-wheeled cart or ice-skate with a fixed steering/edge angle.