Question 1 of 6: Rolling Disk on an Incline — Constraint Classification and the Lagrange-Multiplier Force
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-A1 Classical Mechanics, National Exams
December 2013 — a three-hour closed-book examination; one of two
approved calculator models (Casio or Sharp) is permitted, plus a single 8.5″
× 11″ aid sheet. The cover page states that five (5) questions constitute a
complete exam paper and that only the first five as they appear in the answer book are
marked, and that each question is of equal value — so each of the six printed
questions is worth 20 marks against a 100-mark paper. The candidate is invited to submit a
clear statement of any assumptions made where a question is open to interpretation; this
paper needs that licence in Questions 3 and 6, both flagged below in a Check
box. The cover page also warns that most questions require an essay-format answer, where
clarity and organisation carry marks. All six questions are worked here, because the set
is a study resource rather than a timed attempt.
Reference texts. H. Goldstein, C. Poole and J. Safko, Classical
Mechanics, 3rd ed. (Lagrange multipliers and constraint forces, holonomic versus
nonholonomic constraints, Hamiltonian mechanics, cyclic coordinates and conservation
laws); R. C. Hibbeler, Engineering Mechanics: Dynamics, 14th ed. (rolling without
slipping, relative-motion analysis using rotating axes, dependent-motion pulley analysis,
impulse and momentum for a system of particles, planar kinetics of a rigid body).
Question 1: Rolling Disk on an Incline — Constraint Classification and the
Lagrange-Multiplier Force (20 marks)
Given. A uniform solid disk released from rest at the top of a fixed
incline and rolling without slipping while its plane of motion stays fixed in the X-Y
plane (it does not swerve or tip out of plane).
Given data
Quantity
Symbol
Value
Disk mass
$m$
2 kg
Disk diameter
$2R$
1 m (so $R = 0.5$ m)
Incline angle
$\phi$
30°
Moment of inertia (uniform disk)
$I$
$\tfrac{1}{2}mR^2$
Figure 1 -- disk released from rest at the top of a 30 deg incline.
Find. (a) the type of rolling constraint; (b) the number of degrees
of freedom; (c) the constraint (friction) force, via Lagrange’s equations with an
undetermined multiplier.
Approach. Classify the rolling constraint by checking whether it is
integrable, count degrees of freedom from the reduced coordinate set, then reintroduce the
constraint explicitly with a Lagrange multiplier so the friction force appears as an
output of the equations of motion rather than disappearing into a minimal coordinate.
Part (a) — classify the constraint. Let $x$ be the disk
centre’s distance down the slope and $\theta$ its rotation angle. Rolling without
slipping requires the contact point’s velocity to vanish, $\dot{x} = R\dot{\theta}$,
which integrates directly to
$$x - R\theta = \text{const.}$$
Because the disk is restricted to a single fixed plane (it cannot swerve to a new heading
the way a coin rolling freely on a table can), this rolling condition is a genuine,
time-independent algebraic relation between the coordinates themselves — not just
their rates — so it is holonomic, and since it does not depend
explicitly on time it is also scleronomic. (A disk or sphere rolling
without slipping on a general 2-D surface, free to change heading, gives a genuinely
nonholonomic constraint; confining the motion to the X-Y plane is exactly what removes
that possibility here.)
Part (b) — count the degrees of freedom. An unconstrained rigid
lamina moving in a plane has 3 DOF: $(x_c, y_c, \theta)$. Two independent constraints act
here: (i) the disk’s centre stays on the fixed incline surface, which ties $y_c$ to
$x_c$, and (ii) the holonomic rolling condition from part (a), which ties $\theta$ to
$x_c$. Each removes one DOF:
$$\text{DOF} = 3 - 2 = \boxed{1}$$
A single generalized coordinate — the distance $x$ travelled down the slope —
fully describes the motion.
Part (c) — set up the augmented Lagrangian. To recover the
constraint (friction) force, keep $x$ and $\theta$ as two separate coordinates
and enforce the rolling condition with a multiplier $\lambda$ rather than substituting it
away. With $g(x,\theta) = x - R\theta = 0$,
$$L = \tfrac{1}{2}m\dot{x}^2 + \tfrac{1}{2}I\dot{\theta}^2 + mgx\sin\phi,
\qquad
\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i}
= \lambda \frac{\partial g}{\partial q_i}$$
which gives the two equations
$$m\ddot{x} - mg\sin\phi = \lambda, \qquad I\ddot{\theta} = -\lambda R$$
Solve the pair with the rolling condition $\ddot{x} = R\ddot{\theta}$.
Substituting $\ddot{\theta} = \ddot{x}/R$ into the second equation gives
$\lambda = -I\ddot{x}/R^2$; equating this to the first equation,
$$m\ddot{x} - mg\sin\phi = -\frac{I}{R^2}\ddot{x}
\;\Longrightarrow\;
\ddot{x}\left(m + \frac{I}{R^2}\right) = mg\sin\phi$$
With $I = \tfrac{1}{2}mR^2$, the bracket is $\tfrac{3}{2}m$, so
$$\boxed{\ddot{x} = \frac{2}{3}g\sin\phi
= \frac{2}{3}(9.81)(\sin 30^\circ) = 3.27 \text{ m/s}^2}$$
Extract the constraint force. Back-substituting into
$\lambda = m\ddot{x} - mg\sin\phi$,
$$\lambda = (2)(3.27) - (2)(9.81)(0.5) = 6.54 - 9.81 = -3.27 \text{ N}$$
The multiplier is the generalized force conjugate to $x$ in the constraint equation; the
physical friction force at the contact point is $f = -\lambda$, directed up the slope
(opposing the tendency to slip):
$$\boxed{f = \frac{1}{3}mg\sin\phi = \frac{1}{3}(2)(9.81)(0.5) = 3.27 \text{ N, up the slope}}$$
This is exactly the static friction a Newton-Euler free-body analysis would also
find — the multiplier method reproduces it directly from the augmented
Lagrangian, without ever drawing a free-body diagram.
