Question

A particle is moving along a circular path whose angular position changes from ${\theta }_i=\frac{\mathrm{\pi }}{6}$ rad to ${\theta }_f=\frac{\mathrm{\pi }}{2}$ rad in 5s. Determine the angular displacement and the average angular velocity of the particle for this 5s interval

Answer 1

Step 1: Given data:
At t=0s, initial position is ${\theta }_i=\frac{\mathrm{\pi }}{6}$
At t=5s, the final position is ${\theta }_f=\frac{\mathrm{\pi }}{2}$

Step 2 : To find the angular displacement:

The angular displacement of the particle in 5s is,

$\Delta \theta ={\theta }_f-{\theta }_i$, where ${\theta }_f,{\theta }_i$ are the final and initial position.

$\Delta \theta =\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{6}=\frac{3\mathrm{\pi }-\mathrm{\pi }}{6}$

$\Delta \theta =\frac{2\mathrm{\pi }}{6}\mathrm{rad}$
$\Delta \theta =\frac{\mathrm{\pi }}{3}\mathrm{rad}$

The angular displacement of the particle in 5s is $\frac{\mathrm{\pi }}{3}$ rad

To find angular velocity:

The angular velocity of the particle in 5s is,

$\omega =\frac{\Delta \theta }{\Delta t}$
$\omega =\frac{\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{5}$

$\omega =\frac{\mathrm{\pi }}{15}$rad/s

The angular velocity of the particle in 5s is $\frac{\mathrm{\pi }}{15}$rad/s

The angular displacement and the average angular velocity of the particle for this 5s interval is $\frac{\mathrm{\pi }}{3}rad,\frac{\mathrm{\pi }}{15}$rad/s