The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400\, km$. At the poles $g = 10\,m{s^{ - 2}})$
The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400\, km$. At the poles $g = 10\,m{s^{ - 2}})$
$2.5 \times {10^{ - 3}}\,rad/s$
$5.0 \times {10^{ - 1}}\,rad/s$
$10 \times {10^1}\,rad/s$
$7.8 \times {10^{ - 2}}\,rad/s$
The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400\, km$. At the poles $g = 10\,m{s^{ - 2}})$
$g' = g - {\omega ^2}R{\cos ^2}\lambda $ $⇒$ $0 = g - {\omega ^2}R{\cos ^2}{60^o}$
$0 = g - \frac{{{\omega ^2}R}}{4}\, \Rightarrow \,\omega = 2\sqrt {\frac{g}{R}} = \frac{1}{{400}}\frac{{\,rad}}{{\sec }}$
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