Question 1 of 6: Sphere Rolling on a Free-to-Roll Wedge — Constraint Classification and the Lagrange-Multiplier Force

Question 1: Sphere Rolling on a Free-to-Roll Wedge — Constraint Classification and the Lagrange-Multiplier Force (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. Sphere $m=2$ kg, diameter 1 m (radius $r=0.5$ m), resting at the top of a wedge $M=5$ kg with incline angle $\theta=30^\circ$; the wedge sits on frictionless wheels (massless, for the purposes of the dynamics) so it is free to translate horizontally; the contact between sphere and incline is also frictionless; motion stays in the vertical X-Y plane; both bodies start at rest.

Find. (a) holonomic or non-holonomic classification of the contact constraint; (b) the number of generalized coordinates needed for the sphere's motion; (c) the normal contact (constraint) force between sphere and wedge, via Lagrange’s equations.

Approach. Reduce the system to its two independent, unconstrained generalized coordinates to classify the constraint and count DOF, then reintroduce the contact condition explicitly as a third coordinate tied down by a Lagrange multiplier so the normal force appears directly as an output of the equations of motion.

1. Part (a) — classify the constraint. Because there is no friction anywhere (sphere–wedge contact, wedge–wheel, wheel–ground), no tangential force can act on the sphere at the contact point. For a sphere touching a flat surface, the line from the contact point to the sphere's centre is exactly the surface normal, so the only contact force — the normal reaction $N$ — passes through the sphere's mass centre and exerts zero torque about it. The sphere therefore never spins ($\dot\phi_{\text{spin}}\equiv 0$, since it starts from rest); it simply translates while staying in contact with the incline. The contact condition itself is the ordinary algebraic (geometric) requirement that the sphere's centre remain a fixed perpendicular distance $r$ from the incline surface — a relation among the positions alone, with no non-integrable velocity term. That is the definition of a holonomic constraint (and it does not depend explicitly on time, so it is scleronomic too). This is a genuinely different situation from a sphere rolling without slipping (friction sufficient to prevent slip), where the no-slip condition ties a spin coordinate to a translation coordinate — still holonomic here as well, because the motion is confined to one fixed plane (no heading change is possible), but that spin coordinate simply does not exist in this frictionless problem. Answer: holonomic.
2. Part (b) — count the degrees of freedom. Because the sphere does not spin, its configuration is fully fixed once two translational quantities are known: the wedge's own horizontal position $s$ (it is itself free to translate) and the sphere's position $\xi$ measured down the incline relative to the wedge. Conversely, given the sphere's own absolute planar position $(X,Y)$, both $s$ and $\xi$ can be recovered uniquely (the map is invertible), so tracking the sphere alone carries exactly as much information as tracking the whole system. Answer: 2 degrees of freedom — e.g. $(s,\xi)$. (The quoted moment of inertia $2mr^2/5$ is a deliberate distractor here: with zero torque about the sphere's centre, its rotational state never enters the dynamics.)
3. Part (c) — Lagrangian with a multiplier for the constraint force. Take three coordinates: wedge position $s$, incline-tangential position $\xi$, and an incline-normal offset $\eta$ of the sphere's centre from its nominal standoff (the physical contact condition is $\eta\equiv 0$). With the down-slope direction $\hat t=(-\cos\theta,-\sin\theta)$ and outward normal $\hat n=(-\sin\theta,\cos\theta)$, the sphere's absolute position is $$\vec r_{\text{sph}} = \big(s-\xi\cos\theta-\eta\sin\theta,\ -\xi\sin\theta+\eta\cos\theta\big).$$ The Lagrangian $L=\tfrac12 M\dot s^2+\tfrac12 m\big(\dot X^2+\dot Y^2\big) - mgY$ gives, via $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=\lambda\,\frac{\partial g}{\partial q_i},\qquad g=\eta=0,$$ three equations. Setting $\eta=\dot\eta=\ddot\eta=0$ afterward (the constraint holds at every instant, not just this one): $$(M+m)\ddot s = m\cos\theta\,\ddot\xi \quad\text{(s is cyclic — horizontal momentum of the pair is conserved)}$$ $$\ddot\xi - \ddot s\cos\theta = g\sin\theta \quad\text{(equation of motion along the incline)}$$ $$\lambda = mg\cos\theta - m\ddot s\sin\theta \quad\text{(the normal-direction equation; }\lambda\text{ is the constraint force }N\text{)}$$ Solving the first two together, $$\ddot\xi = \frac{(M+m)g\sin\theta}{M+m\sin^2\theta}, \qquad \ddot s = \frac{mg\sin\theta\cos\theta}{M+m\sin^2\theta},$$ and substituting into the $\lambda$ equation collapses to the closed form $$\boxed{N=\lambda=\dfrac{Mmg\cos\theta}{M+m\sin^2\theta}}.$$
4. Evaluate. With $M=5$, $m=2$, $\theta=30^\circ$, $g=9.81\ \text{m/s}^2$: $$\ddot\xi = \frac{7(9.81)(0.5)}{5+2(0.25)} = 6.243\ \text{m/s}^2,\qquad \ddot s = \frac{2(9.81)(0.5)(0.866)}{5.5}=1.545\ \text{m/s}^2,$$ $$N = \frac{5(2)(9.81)(0.866)}{5.5} = 15.45\ \text{N}.$$ Sanity check: letting $M\to\infty$ (a fixed wedge) recovers the familiar block-on-a-fixed-incline result $N\to mg\cos\theta = 16.99$ N; a wedge free to recoil away reduces the normal force slightly, as expected.