Question 6 of 6: Rod Released by a Snapped Cable — Rigid-Body Kinetics at the Instant of Release
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 6: Rod Released by a Snapped Cable — Rigid-Body Kinetics at the Instant
of Release (20 marks)
Check: orientation of the spring and cable. The figure shows both supports running VERTICALLY — a coil spring hanging straight down from A to a fixed base, and a cable running straight up from B to a fixed ceiling — on either end of a HORIZONTAL 2 m rod. This is also the only configuration that makes the problem well posed without a stated spring constant.
Given. Horizontal rod AB, mass 1 kg, length 2 m; a vertical spring
supports end A from below, a vertical cable supports end B from above; the system is in
static equilibrium until the cable at B suddenly snaps.
Given data
Quantity
Symbol
Value
Rod mass
$m$
1 kg
Rod length
$L$
2 m
Moment of inertia about G (uniform rod)
$I_G$
$\tfrac{1}{12}mL^2$
Figure 6 -- rod AB: vertical spring at A, vertical cable at B (about to be cut).
Find. The angular acceleration $\alpha$ of rod AB immediately after
the cable at B snaps.
Approach. Find the spring force from the PRE-snap static equilibrium
(it survives the first instant unchanged, since a real spring’s force is a
continuous function of its own deformation), then apply the planar rigid-body equations
of motion with that force plus gravity as the only two forces acting.
Pre-snap static equilibrium. Taking moments about A and summing forces
vertically, with the weight $mg$ acting at the centre $G$ (1 m from each end):
$$\sum M_A = 0: \quad T_B\,L - mg\left(\frac{L}{2}\right) = 0
\;\Longrightarrow\; \boxed{T_B = \frac{mg}{2} = 4.905 \text{ N}}$$
$$\sum F_y = 0: \quad F_A + T_B - mg = 0
\;\Longrightarrow\; \boxed{F_A = \frac{mg}{2} = 4.905 \text{ N}}$$
The instant after the snap. The cable tension $T_B$ can (and does)
drop to zero discontinuously, but the spring force $F_A$ cannot — it depends only on
the spring’s own compression, which has not yet had time to change, so
$F_A = mg/2$ still, directed upward. Both remaining forces (spring, gravity) are vertical,
so $\sum F_x = 0 \Rightarrow a_{Gx} = 0$.
Vertical translation of the centre of mass.
$$\sum F_y = ma_{Gy}: \quad F_A - mg = ma_{Gy}
\;\Longrightarrow\; \boxed{a_{Gy} = \frac{F_A}{m} - g = \frac{mg/2}{m} - g
= -\frac{g}{2} = -4.905 \text{ m/s}^2}$$
Rotation about the centre of mass. Only $F_A$, applied at A ($L/2$
to the $-x$ side of $G$), produces a moment about $G$ (gravity acts at $G$ itself):
$$\sum M_G = I_G\alpha: \quad -F_A\left(\frac{L}{2}\right) = \left(\frac{1}{12}mL^2\right)\alpha$$
$$\boxed{\alpha = \frac{-F_A(L/2)}{\tfrac{1}{12}mL^2}
= \frac{-(4.905)(1)}{\tfrac{1}{12}(1)(4)} = -14.715 \text{ rad/s}^2}$$
The negative sign (with $+z$ out of the page, A on the $-x$ side of G) means the rod
rotates so that end B swings down and end A swings up.
Kinematic cross-check. Using $\vec a_P = \vec a_G + \vec\alpha\times\vec r_{P/G}$
(valid since $\omega = 0$ at this instant, the system having just left static
equilibrium):
$$a_B = a_{Gy} + \alpha\left(\frac{L}{2}\right) = -4.905 + (-14.715)(1) = -19.62 \text{ m/s}^2$$
$$a_A = a_{Gy} - \alpha\left(\frac{L}{2}\right) = -4.905 - (-14.715)(1) = +9.81 \text{ m/s}^2$$
End B, now completely unsupported, falls at exactly $2g$ — a clean sanity check
— while end A is thrown upward at $g$ as the rod snaps into rotation about a point
between A and G.
