Question

A racing car is travelling along a striaght track at a constant velocity of $40\text{m/s}$. A fixed TV camera is recording the event as shown in figure. In order to keep the car in view, the angular velocity of camera in the position shown should be

• A. $3\text{rad/s}$
• B. $2\text{rad/s}$
• C. $4\text{rad/s}$
• D. $1\text{rad/s}$

Answer 1

The correct option is D $1\text{rad/s}$

By the diagram,


Relation between linear and angular speed is,

$\omega =\frac{v_{\perp }}{r}=\frac{40\cos {30}^{\circ }}{r}$

Here $\Rightarrow r=\frac{h}{\cos {30}^{\circ }}=\frac{30}{\cos {30}^{\circ }}$

$\Rightarrow \omega =\frac{v_{\perp }}{r}=\frac{40}{30}{\cos }^2{30}^{\circ }=\frac{4}{3}\times \frac{3}{4}=1\text{rad/s}$

$(\cos {30}^{\circ }=\frac{\sqrt{3}}{2})$

Hence option D is the correct answer.