Question 2 of 4: Angular Velocity Vector Composition
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, May 2017 — 04-Bio-A4, Biomechanics. 3-hour open-book exam;
FOUR (4) questions of equal value constitute a complete paper, so all four are
answered in full below.
Reference texts: Winter, Biomechanics and Motor Control of
Human Movement; Zatsiorsky, Kinematics of Human Motion; Nordin &
Frankel, Basic Biomechanics of the Musculoskeletal System.
Given. Rotation "in the frontal plane" (a cartwheel-type tumble) has
its angular-velocity vector along the anteroposterior (AP) axis; "spinning"/"twisting"
about the long axis has its vector along the longitudinal axis; "somersaulting"
(sagittal-plane tumbling) has its vector along the medio-lateral (ML) axis. These three
body axes are mutually orthogonal, so each part reduces to combining two perpendicular
angular-velocity vectors (or, in (c), splitting a known resultant into two perpendicular
unknowns).
Find. The resultant angular velocity and its angle to each component
(a, b); the two unknown orthogonal components, with the resultant angle confirmed (c).
Approach. Because both component rotations share the same
instantaneous axis intersection (the body's centre of mass) about mutually perpendicular
anatomical axes, they add like any pair of perpendicular vectors: Pythagorean magnitude,
inverse-tangent angle.
(a) High jumper. $\omega_{AP}=90\ °/\text{s}$ (frontal-plane
rotation), $\omega_{long}=27\ °/\text{s}$ (long-axis spin), orthogonal by construction.
$$\omega_R=\sqrt{90^2+27^2}=\boxed{93.96\ °/\text{s}}$$
Angle from the 90°/s component: $\theta=\tan^{-1}(27/90)=\boxed{16.7°}$ (equivalently
73.3° from the 27°/s component).
(b) Ski racer. $\omega_{long}=36\ °/\text{s}$ (twist),
$\omega_{ML}=10\ °/\text{s}$ (somersault).
$$\omega_R=\sqrt{36^2+10^2}=\boxed{37.36\ °/\text{s}}$$
Angle from the 36°/s component: $\theta=\tan^{-1}(10/36)=\boxed{15.5°}$ (74.5°
from the 10°/s component).
(c) Aerial skier. Here the resultant $\omega_R=360\ °/\text{s}$ is
given and the two orthogonal components (AP-axis, ML-axis) are unknown. The source figure
shows the dashed resultant axis sitting at approximately 45° to both orthogonal axes;
adopting that reading (see check note),
$$\omega_{AP}=\omega_{ML}=\omega_R\cos 45° = \boxed{254.6\ °/\text{s}}\ \text{each}$$
and, by construction with equal components, the resultant sits at exactly 45° from
both the AP-axis and ML-axis components.
Orthogonal
angular-velocity components (black) and resultant (orange) for parts (a), (b), (c).
Final results — Question 2
Part
Component 1
Component 2
Resultant ω
Angle to comp. 1
(a) High jumper
90°/s (frontal/AP)
27°/s (long-axis)
93.96°/s
16.7°
(b) Ski racer
36°/s (long-axis)
10°/s (somersault/ML)
37.36°/s
15.5°
(c) Aerial skier
254.6°/s (AP)
254.6°/s (ML)
360.0°/s (given)
45.0°
Check: part (c) states only that the resultant axis is "dotted"
in the source figure and does not print a numeric angle; the figure's dashed line is
drawn at approximately 45° to both orthogonal axes, which we adopt exactly (equal
AP/ML components). A different adopted angle would change both component magnitudes but
not the method above.
