Question

A spot light S rotates in a horizontal plane with a constant angular velocity of 0.1 rad/s. The spot of light P moves along the wall at a distance 3m. What is the velocity of the spot P when $\theta = 45^\circ$?

• A. 0.6 m/s
• B. 0.5 m/s
• C. 0.4 m/s
• D. 0.3 m/s

Answer 1

Answer: (A)

0.6 m/s

Answer 2

Let S be the position of the spotlight and O be the point on the wall closest to S, such that SO is perpendicular to the wall. The distance SO is given as 3m. The angle $\theta$ is the angle between the line SO and the line SP, where P is the spot of light on the wall.

From the geometry, we have a right-angled triangle SOP, with the right angle at O. The distance OP, denoted by $x$, can be expressed using trigonometry: $\tan \theta = \frac{x}{SO}$ Given $SO = 3$ m, we have: $x = 3 \tan \theta$

The angular velocity of the spotlight is $\omega = \frac{d\theta}{dt} = 0.1$ rad/s. We need to find the velocity of the spot P, which is $v_P = \frac{dx}{dt}$. Differentiating the expression for $x$ with respect to time $t$ using the chain rule: $v_P = \frac{dx}{dt} = \frac{d}{dt}(3 \tan \theta)$ $v_P = 3 \frac{d}{d\theta}(\tan \theta) \frac{d\theta}{dt}$ Since $\frac{d}{d\theta}(\tan \theta) = \sec^2 \theta$, we get: $v_P = 3 \sec^2 \theta \frac{d\theta}{dt}$

We are given $\theta = 45^\circ$ and $\frac{d\theta}{dt} = 0.1$ rad/s. First, calculate $\sec^2 \theta$ for $\theta = 45^\circ$: $\cos 45^\circ = \frac{1}{\sqrt{2}}$ $\sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2}$ $\sec^2 45^\circ = (\sqrt{2})^2 = 2$

Now, substitute the values into the expression for $v_P$: $v_P = 3 \times (2) \times 0.1$ $v_P = 6 \times 0.1$ $v_P = 0.6 \text{ m/s}$