Question 2.SP.32: An automobile is moving south with an absolute velocity of 4......
An automobile is moving south with an absolute velocity of 40 km/h. An observer at O is stationed 50 m to the east of the line of travel. When the automobile is directly west of the observer, what is the angular velocity relative to the observer? After the automobile moves 50 m south, what is its angular velocity relative to the observer at O?
As indicated in Fig. 2-23, the velocity v_{A / O} of the automobile at A relative to O is 40 \mathrm{~km} / \mathrm{h} or 11.11 \mathrm{~m} / \mathrm{s}. However, the linear velocity of A relative to O (in this case it is perpendicular to OA ) is the product of the distance OA and the angular velocity of A relative to O. Then
\begin{array}{c}v_{A / O}=O A \times \omega_{A / O} \\11.11 \mathrm{~m} / \mathrm{s}=50 \mathrm{~m} \times \omega_{A / O} \qquad \therefore \omega_{A / O}=\underline{0.222 \mathrm{rad} / \mathrm{s}}\end{array}
For the next part of the problem note that the absolute velocity v_{B} of the vehicle at B is still south 40 \mathrm{~km} / \mathrm{h} \left(11.11 \mathrm{~m} / \mathrm{s}\right. ). The component v_{B / O} (velocity of B relative to O ) is perpendicular to the arm BO; hence, v_{B / O}= 11.11 \cos 45^{\circ}=7.85 \mathrm{~m} / \mathrm{s}. Then
v_{B / O}=O B \times \omega_{B / O} \quad 7.85=\left(\frac{50}{\cos 45^{\circ}}\right)\left(\omega_{B / O}\right) \qquad \therefore \omega_{B / O}=\underline{0.111 \mathrm{rad} / \mathrm{s}}
