What should be the angular velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes 35 of the present weight at the equator. Equatorial radius of the earth is 6400 km, ω=13N×10−5rad/s , then the value of N is

Weight at equator (W)=mgObserved weight at equator (W')=mg'=35mg , R=6400kmNow, Latitude λ=0We know, mg'=mg−mRω2cos2λ35mg=mg−mRω2cos200=mg−mRω2 mRω2=mg−35mg∴ω=25gR1/2 Putting the values,ω=2×9.85×6.4×1061/2=7.8×10−4rad/s

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What should be the angular velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is 6400 km, $ω = 13 N \times 10^{- 5} r a d / s$ , then the value of $N$ is

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Detailed Solution

Weight at equator $(W) = m g$

Observed weight at equator $(W') = m g'$ $= \frac{3}{5} m g$ , $R = 6400 k m$

Now, Latitude $(λ) = 0$

We know, $m g' = m g - m R ω^{2} \cos^{2} λ$

$\frac{3}{5} m g = m g - m R ω^{2} \cos^{2} 0^{0} = m g - m R ω^{2}$

$m R ω^{2} = m g - \frac{3}{5} m g$

$∴ ω = {(\frac{2}{5} \frac{g}{R})}^{1 / 2}$

Putting the values,

$ω = {(\frac{2 \times 9.8}{5 \times 6.4 \times 10^{6}})}^{1 / 2}$

$= 7.8 \times 10^{- 4} r a d / s$

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