Question:
Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth $=12800 \mathrm{~km}$ and it takes 24 hours for the earth to complete one revolution about its axis.
Solution:
Speed of particle at equator=distance/time $=\frac{2 \pi R}{T}$
$=\frac{2 \times 3.14 \times 6400 \times 10^{3}}{24 \times 60 \times 60}$
$=465.1 \mathrm{~m} / \mathrm{s}$
Acceleration of particle $=$
$\mathrm{A}_{\Gamma}=\frac{v^{2}}{R}$
$A_{r}=\frac{(465.1)^{2}}{\left(6400 \times 10^{3}\right)}$
$A_{r}=0.038 \mathrm{~m} / \mathrm{s}^{2}$
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