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Question: With what angular velocity the earth should spin in order that a body lying at \({37^{th}}\) latitud...

With what angular velocity the earth should spin in order that a body lying at 37th{37^{th}} latitude may become weightless?
A. 54gR\dfrac{5}{4}\sqrt {\dfrac{g}{R}}
B. 2516gR\dfrac{{25}}{{16}}\sqrt {\dfrac{g}{R}}
C. 53gR\dfrac{5}{3}\sqrt {\dfrac{g}{R}}
D. 259gR\dfrac{{25}}{9}\sqrt {\dfrac{g}{R}}

Explanation

Solution

In order to solve this question we need to understand angular speed and centripetal force. So angular velocity is defined as velocity with which the body rotates, it is mathematically defined as the division of linear velocity with the radius in which it rotates. Centripetal force is defined as the pseudo force which acts in the opposite direction of radius, this force is the reason why a body moving on a circle is in equilibrium.

Complete step by step answer:
Latitude is defined as the angle that a body or object subtends with the equator axis; it is used to denote the position of an object on earth along with longitude.Consider earth rotates with angular velocity ω\omega and let the mass of body be mm.Also let the angle that it makes with equator axis be θ\theta and the radius of earth be RR.So the centripetal force that acts in outward direction is given by,
f=mR2ωf = m{R^2}\omega

If the earth is not rotating then weight of body acts in downward direction and it is equal to, f1=mg{f_1} = mg
Since the earth is rotating so let the weight of the body denoted by, f2=mg{f_2} = mg'.
Since we know variation of acceleration due to gravity gg with θ\theta as,
g=gRω2cos2θg' = g - R{\omega ^2}{\cos ^2}\theta
So the force f2{f_2} becomes after putting value,
f2=m(gRω2cos2θ){f_2} = m(g - R{\omega ^2}{\cos ^2}\theta )
f2=mgmRω2cos2θ\Rightarrow {f_2} = mg - mR{\omega ^2}{\cos ^2}\theta

So for the body to be weightless we know, f2=0{f_2} = 0.
Putting values we get,
m(gRω2cos2θ)=0m(g - R{\omega ^2}{\cos ^2}\theta ) = 0
ω2=gRcos2θ\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}\theta }}
ω2=gRcos237\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}37}}
Since, cos37=45\cos 37 = \dfrac{4}{5}
So, ω2=25g9R{\omega ^2} = \dfrac{{25g}}{{9R}}
ω=54gR\therefore \omega = \dfrac{5}{4}\sqrt {\dfrac{g}{R}}

So the correct option is A.

Note: It should be remembered that earth is also tilted at its own axis and due to this climate changes on earth. Also it should be noted here that with some angular velocity we can feel weightlessness or in other words we can eliminate gravity, so gravity is not a force it’s merely due to space time curvature.