In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10 m s^{- 2} a n d r a d i u s o f e a r t h i s 6400 k m)$

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Solution

The correct option is B
$g^{^{'}} = g - ω^{2} R c o s^{2} λ a t e q u a t o r λ = 0 F o r w e i g h t l e s s n e s s g^{^{'}} = 0 0 = g - ω^{2} R \Rightarrow ω = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{r a d}{s}$

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Q. In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be

(g = 10 m s^{- 2} a n d r a d i u s o f e a r t h i s 6400 k m)

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

Q. The angular velocity of rotation of the earth in order to make the effective acceleration due to gravity equal to zero at equator should be

g = 10 m / s^{2}, R = 6400 k m

Q. The speed of earth 's rotation about its axis is

ω

.Its speed is increased to

x

times to make the effective acceleration due to gravity equal to zero at the equator, then

x

is around (

g = 10 m s^{- 2}

R = 6400

km)

Q. A uniform circular disc of mass

1 kg

and radius

0.5 m

is rotating about its axis with a constant angular velocity

5 rad/sec

. Two birds, each of mass

0.25 kg

sit gently at the opposite ends of the diameter of the disc. The final angular velocity of the disc will be

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In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ms−2 and radius of earth is 6400 km)

The correct option is B g′=g−ω2Rcos2λat equator λ=0For weightlessness g′=00=g−ω2R⇒ω=√gR=1800rads

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10 m s^{- 2} a n d r a d i u s o f e a r t h i s 6400 k m)$

The correct option is B
$g^{^{'}} = g - ω^{2} R c o s^{2} λ a t e q u a t o r λ = 0 F o r w e i g h t l e s s n e s s g^{^{'}} = 0 0 = g - ω^{2} R \Rightarrow ω = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{r a d}{s}$