Question

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on ${60}^o$ latitude becomes zero is

• A. $2.5\times {10}^{-3}rad{s}^{-1}$
• B. $1.5\times {10}^{-3}rad{s}^{-1}$
• C. $4.5\times {10}^{-3}rad{s}^{-1}$
• D. $0.5\times {10}^{-3}rad{s}^{-1}$

Answer 1

The correct option is A $2.5\times {10}^{-3}rad{s}^{-1}$

the acceleration due to gravity to rotate in terms of latitude$\left(\theta \right)$

$g^{\prime }=g-w^{2}R{cos}^2\theta$

$0=g-w^{2}R{cos}^{2}60$

$g=\frac{w^{2}R}{4}$

$w=\sqrt{\frac{4g}{R}}=2\sqrt{\frac{g}{R}}=\frac{2}{800}$

$w=2.5\times {10}^{-3}\frac{rad}{sec}$