Topic: Engineering Mechanics
A boat is travelling at a speed of $40~m/s$ due east as shown in the figure below. An observer is stationed 100 m south of the line of travel. Determine the angular velocity of the boat, relative to the observer, when in the position $(\theta=30^{\circ})$ shown in the figure.
- A. $0.3~rad/sec$
- B. $0.4~rad/sec$
- C. $0.6~rad/sec$
- D. $0.8~rad/sec$
Answer: A. $0.3~rad/sec$
Explanation: Let the horizontal distance travelled by the boat be $x$. From the geometry of the figure, $x = 100 \tan\theta$.
Differentiating with respect to time $t$ gives the velocity of the boat: $v = \frac{dx}{dt} = 100 \sec^2\theta \cdot \frac{d\theta}{dt}$.
Here, angular velocity $\omega = \frac{d\theta}{dt}$.
Substituting the given values: $40 = 100 \sec^2(30^{\circ}) \cdot \omega$.
$\sec(30^{\circ}) = \frac{2}{\sqrt{3}}$, so $\sec^2(30^{\circ}) = \frac{4}{3}$.
$40 = 100 \times \frac{4}{3} \times \omega \Rightarrow 40 = \frac{400}{3} \omega$.
$\omega = \frac{40 \times 3}{400} = 0.3~rad/sec$.