Question

A laser pointer is placed on a platform that rotates at a rate of 60 revolutions per minute. The beam hits a wall 10 m away, producing a dot of light that moves horizontally along the wall. Let θθ be the angle between the beam and the line through the searchlight perpendicular to the wall.

Answer 1

Answer:

$v = 20\pi sec^2\theta$

Explanation:

As we know that the angle made with the vertical is given so we have

$tan\theta = \frac{x}{10}$

so we have

$x = 10 tan\theta$

now differentiate both sides with time

$v = \frac{dx}{dt} = 10 sec^2\theta \frac{d\theta}{dt}$

$v = 10 \omega sec^2\theta$

now we know that

$\omega = 2\pi f$

$\omega = 2\pi (\frac{60}{60})$

$\omega = 2\pi$

now we have

$v = 10(2\pi) sec^2\theta$

$v = 20\pi sec^2\theta$