Question

An object is executing uniform circular motion with an angular speed of $\pi/12$ radian per second, At t=0 the object starts at an angle $\theta=0$ What is the angular displacement of the particle after 4 s?

Answer 1

$\frac{\pi}{3}$ radians

Answer 2

Given:

Angular speed, $\omega = \frac{\pi}{12}$ rad/s Time, $t = 4$ s Initial angular position, $\theta_0 = 0$

For uniform circular motion, the angular displacement ($\Delta\theta$) is given by the product of angular speed ($\omega$) and time ($t$): $\Delta\theta = \omega \times t$ Substitute the given values: $\Delta\theta = \left(\frac{\pi}{12} \text{ rad/s}\right) \times (4 \text{ s})$ $\Delta\theta = \frac{4\pi}{12} \text{ rad}$ $\Delta\theta = \frac{\pi}{3} \text{ rad}$

Angular displacement is the change in angular position. For uniform circular motion, where angular speed ($\omega$) is constant, the angular displacement ($\Delta\theta$) over a time interval ($t$) is simply the product of angular speed and time, i.e., $\Delta\theta = \omega t$. Substituting the given values of $\omega = \pi/12$ rad/s and $t = 4$ s yields an angular displacement of $\pi/3$ radians.