Question 2 of 4

Question 2 (15 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. In each case the athlete's whole-body angular velocity is the vector sum of rotation about two of the three orthogonal body axes (anteroposterior/AP, mediolateral/ML, longitudinal). Because instantaneous angular velocity is a true vector, two simultaneous single-axis rotations add exactly like any other orthogonal vector pair (Pythagorean magnitude, arctangent angle), even though the corresponding finite rotations would not commute.

Find. a),b) the resultant angular velocity and its angle from each given component; c) the two orthogonal components, given the resultant and (from the figure) its angle to the ML axis.

[Figure not reproduced: Figure 2 (redrawn) — orthogonal angular-velocity components (blue = long/longitudinal axis, red = AP or ML axis as labelled) and resultant (green), right-hand-rule sense, for each athlete. See the official exam paper.]

Approach. Each panel is a right-triangle vector sum: resultant magnitude $\omega_{res}=\sqrt{\omega_1^2+\omega_2^2}$, angle from the first-named component $\theta=\tan^{-1}(\omega_2/\omega_1)$; part c) reverses this, decomposing a known resultant using the angle its axis makes with the ML axis in the photograph (measured ≈ 30°).

1. a) High jumper — frontal-plane (AP-axis) 180°/s + long-axis 120°/s. By the right-hand rule, the frontal-plane rotation (head left) is a vector along the AP axis, and the spin (right shoulder up) is a vector along the long (longitudinal) axis; the two are already orthogonal, so $$\omega_{res} = \sqrt{180^2+120^2} = \boxed{216.3\ ^\circ/\text{s}}, \qquad \theta = \tan^{-1}\!\left(\frac{120}{180}\right) = \boxed{33.7^\circ}\ \text{from the AP-axis component}$$
2. b) Ski racer — forward somersault (ML axis) 180°/s + long-axis twist 360°/s. $$\omega_{res} = \sqrt{180^2+360^2} = \boxed{402.5\ ^\circ/\text{s}}, \qquad \theta = \tan^{-1}\!\left(\frac{360}{180}\right) = \boxed{63.4^\circ}\ \text{from the ML-axis component}$$
3. c) Aerial skier — decompose the 220°/s resultant. The dotted resultant axis in the photograph lies in the ML–AP plane, measured at ≈30° from the horizontal (ML) crosshair toward the vertical (AP) crosshair: $$\omega_{ML}\,(\text{backward somersault}) = 220\cos30^\circ = \boxed{190.5\ ^\circ/\text{s}}$$ $$\omega_{AP}\,(\text{left-shoulder-left cartwheel}) = 220\sin30^\circ = \boxed{110.0\ ^\circ/\text{s}}$$