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Question: The angular velocity of earth with which it has to rotate so that the acceleration due to gravity on...

The angular velocity of earth with which it has to rotate so that the acceleration due to gravity on 60o{{60}^{o}}latitude zero is:
A.) ω=2.48\omega =2.48×10−3rad/s{{10}^{-3}}rad/s
B.) ω=3.58\omega =3.58×10−3rad/s{{10}^{-3}}rad/s
C.) ω=6.48\omega =6.48×10−3rad/s{{10}^{-3}}rad/s
D.) ω=4.25\omega =4.25×10−3rad/s{{10}^{-3}}rad/s

Explanation

Solution

Hint: Apply formula of acceleration due to gravity at latitude. Focus on word latitude. Get formula. Values of terms given in formula are given in questions. Effective acceleration due to gravity is zero.

Complete step by step solution:
Acceleration due to gravity Effect of rotation of earth or latitude is given by,

gθ=g−Rω2cos⁡2θ{{g}_{\theta }}=g-R{{\omega }^{2}}{{\cos }^{2}}\theta ---------(1)
Where,
gθ{{g}_{\theta }}= effective acceleration due to gravity
ω\omega = angular velocity of rotation of earth about its axis
θ\theta = latitude angle
g = acceleration due to gravity

It is given in question that effective acceleration due to gravity will be zero at 60 degree.
So to calculate angular velocity of earth =?
Therefore,

gθ=0,θ=60o{{g}_{\theta }}=0,\theta ={{60}^{o}}

Put above value in equation one, we get

0=g−Rω2cos⁡260 g=Rω2cos⁡260 \begin{aligned} & 0=g-R{{\omega }^{2}}{{\cos }^{2}}60 \\\ & g=R{{\omega }^{2}}{{\cos }^{2}}60 \\\ \end{aligned}

We know that, R= 64×105m64\times {{10}^{5}}m

ω=gR14 ω=9.8×464×105 ω=2.48×10−3rad/s \begin{aligned} & \omega =\dfrac{g}{R\dfrac{1}{4}} \\\ & \omega =\dfrac{9.8\times 4}{64\times {{10}^{5}}} \\\ & \omega =2.48\times {{10}^{-3}}rad/s \\\ \end{aligned}
So the correct option is A here.

Additional information:
At equator

θ=0o so, gθ=g−Rω2cos⁡0 gθ=g−Rω2 \begin{aligned} & \theta ={{0}^{o}} \\\ & so, \\\ & {{g}_{\theta }}=g-R{{\omega }^{2}}\cos 0 \\\ & {{g}_{\theta }}=g-R{{\omega }^{2}} \\\ \end{aligned}

At poles,

θ=90o so, gθ=g−Rω2cos⁡90 gθ=g \begin{aligned} & \theta ={{90}^{o}} \\\ & so, \\\ & {{g}_{\theta }}=g-R{{\omega }^{2}}\cos 90 \\\ & {{g}_{\theta }}=g \\\ \end{aligned}

The value of the acceleration due to gravity increases from equator to pole due to rotation of earth. It means that the value of g increases with latitude. The value of g remains the same at pole. It is not affected by rotation of earth.

Note: Value of angular velocity can be deflected by some point. Remember the value of acceleration due to gravity and radius of earth along with the unit. Do not get confused between angular velocity and acceleration due to gravity. If you do not get the idea what to calculate in question then look for options units.