The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is ( Radius of the earth = 6400 km . At the poles g=10 m s- 2 )

Question

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is ( Radius of the earth = 6400 km . At the poles g=10 m s- 2 ) (A) 2.5× 10- 3 rad s- 1 (B) 5.0× 10- 1 rad s- 1 (C) 100 rad s- 1 (D) 7.8× 10- 2 rad s- 1

Tardigrade Admin · Accepted Answer

g prime=g-ω2 R cos 2 λ 0=g-ω2 R cos 2 60° 0=g-(ω2 R /4) ⇒ ω=2 √( barg/R)=(1/400) rad s -1=2.5 × 10-3 rad s-1

Q. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $6 0^{\circ}$ latitude becomes zero is ( Radius of the earth $= 6400 km$ . At the poles $g = 10 m s^{- 2}$ )

$2.5 \times 1 0^{- 3} r a d s^{- 1}$

$5.0 \times 1 0^{- 1} r a d s^{- 1}$

$100 r a d s^{- 1}$

$7.8 \times 1 0^{- 2} r a d s^{- 1}$

$g^{'} = g - ω^{2} R cos^{2} λ 0 = g - ω^{2} R cos^{2} 6 0^{\circ}$ $0 = g - \frac{ω ^{2} R}{4} \Rightarrow ω = 2 \frac{g ˉ}{R} = \frac{1}{400} r a d s^{- 1} = 2.5 \times 1 0^{- 3} r a d s^{- 1}$

Q. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60∘ latitude becomes zero is ( Radius of the earth =6400km . At the poles g=10ms−2 )

2.5×10−3rads−1

5.0×10−1rads−1

100rads−1

7.8×10−2rads−1