
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). $\dfrac{\pi }{1800}rad\,{{s}^{-1}}$
Answer
465.9k+ 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).
Additional Information:
The axis on which the Earth rotates is slightly tilted. The geographical North Pole is the point where the Earth’s axis meets its surface and geographical South Pole is the other point where the axis meets the surface. The rotation of the Earth is in an anti-clockwise direction.
Note:
The time period is the time taken to complete one rotation. The time taken to complete one rotation by Earth is the length of one day on Earth whereas the time taken to complete one revolution by the Earth around the sun is the length of one year. The Earth’s orbit about which it rotates is nearly circular.
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).
Additional Information:
The axis on which the Earth rotates is slightly tilted. The geographical North Pole is the point where the Earth’s axis meets its surface and geographical South Pole is the other point where the axis meets the surface. The rotation of the Earth is in an anti-clockwise direction.
Note:
The time period is the time taken to complete one rotation. The time taken to complete one rotation by Earth is the length of one day on Earth whereas the time taken to complete one revolution by the Earth around the sun is the length of one year. The Earth’s orbit about which it rotates is nearly circular.
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