Question

A dust particle lying at the rim of a wheel has an angular speed of 22 Radian per second find its time period

Answer 1

we know, $\omega=\frac{2\pi}{T}$
where $\omega$ is angular speed , T is time period.

Given, $\omega=22rad/s$
T = 2π/22 = π/11 sec

hence, A dust particle lying at rim of wheel has angular speed 22 rad/sec then, time period = π/11 sec

Answer 2

Answer: $T = \dfrac{\pi}{11} sec$

Explanation:

Given that,

$\omega = 22\rps$

We know that,

Angular speed is define as,

$\omega = \dfrac{2\pi}{T}$

Where,

$\omega$ = angular speed

T = time period

$T = \dfrac{2\pi}{\omega}$

$T = \dfrac{2\pi}{22}$

$T = \dfrac{\pi}{11} sec$

Hence, the time period is $T = \dfrac{\pi}{11} sec$.