Calculate angular velocity of earth so that acceleration due to gravity at $60^{o}$ latitude becomes zero. (Radius of earth $= 6400$ km, gravitational acceleration at poles $= 10 m / s^{2}, cos 60^{o} = 0.5$ )

7.8 \times 10^{- 2} r a d / s

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0.5 \times 10^{- 3} r a d / s

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1 \times 10^{- 3} r a d / s

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2.5 \times 10^{- 3} r a d / s

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Solution

The correct option is D $2.5 \times 10^{- 3} r a d / s$
We have, $g_{1} = g - ω^{2} R {cos}^{2} ϕ$
According to the question, $ω = \frac{π}{3}$ and $g_{1} = 0$ .
So, we must have $ω = \sqrt{\frac{4 g}{R}} = 2.5 \times 10^{- 3} r a d / s$

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Calculate angular velocity of earth so that acceleration due to gravity at 60o latitude becomes zero. (Radius of earth =6400km, gravitational acceleration at poles=10m/s2,cos60o=0.5)

The correct option is D 2.5×10−3rad/sWe have, g1=g−ω2Rcos2ϕAccording to the question, ω=π3 and g1=0.So, we must have ω=√4gR=2.5×10−3rad/s

Calculate angular velocity of earth so that acceleration due to gravity at $60^{o}$ latitude becomes zero. (Radius of earth $= 6400$ km, gravitational acceleration at poles $= 10 m / s^{2}, cos 60^{o} = 0.5$ )

The correct option is D $2.5 \times 10^{- 3} r a d / s$
We have, $g_{1} = g - ω^{2} R {cos}^{2} ϕ$
According to the question, $ω = \frac{π}{3}$ and $g_{1} = 0$ .
So, we must have $ω = \sqrt{\frac{4 g}{R}} = 2.5 \times 10^{- 3} r a d / s$