Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The angular speed of the Earth around its own axis is-(A). $\dfrac{\pi }{43200}rad\,{{s}^{-1}}$(B). $\dfrac{\pi }{3600}rad\,{{s}^{-1}}$(C). $\dfrac{\pi }{86400}rad\,{{s}^{-1}}$(D). $\dfrac{\pi }{1800}rad\,{{s}^{-1}}$

Last updated date: 06th Sep 2024
Total views: 409.8k
Views today: 8.09k
Verified
409.8k+ views
Hint: The angular velocity in circular motion is analogous to velocity in translational motion. A body covers one full angle while completing one rotation. The Earth follows two kinds of circular motion; rotation about its axis and revolution about the sun. The time period of rotation is 24 hours or one day.

Formulae used:
$\omega =\dfrac{\theta }{t}$
$T=\dfrac{2\pi }{\omega }$

Complete step-by-step solution:
The angular speed is the angular displacement covered in unit time. It is given by-
$\omega =\dfrac{\theta }{t}$
$\theta$ is the angular displacement
$t$ is the time taken

The Earth rotates about its own axis.One rotation is completed in about 24 hours. Therefore, time period is,
\begin{align} & T=24\,hrs \\ & \Rightarrow T=24\times 3600s \\ & \therefore T=86400s \\ \end{align}

The time period is given by-
$T=\dfrac{2\pi }{\omega }$
Here, $2\pi$ is the total angular displacement covered in one rotation, we substitute values for Earth in the above equation to get,
\begin{align} & \Rightarrow 86400=\dfrac{2\pi }{\omega } \\ & \therefore \omega =\dfrac{2\pi }{86400}rad\,{{s}^{-1}} \\ \end{align}

Therefore, the magnitude of the angular momentum of the Earth about its own axis is $\dfrac{2\pi }{86400}\,rad\,{{s}^{-1}}$.

Hence, the correct option is (C).