Question

What should be the angular velocity of earth about own is so that a person weight at equator will be $ \dfrac{1}{3} $ of his weight at poles:-
(1) $ \sqrt {\dfrac{g}{R}} $
(2) $ \sqrt {\dfrac{{2g}}{{3R}}} $
(3) $ \sqrt {\dfrac{{2g}}{{5R}}} $
(4) $ \sqrt {\dfrac{{3g}}{{2R}}} $

Answer 1

Hint: According to the question, we need to find the angular velocity of earth about own is so that a person weight at equator will be $ \dfrac{1}{3} $ of his weight at poles. The formula that will be used here is $ {g'} = g - R{\omega ^2}{\cos ^2}\theta $ .

Complete answer:
Angular velocity or rotational velocity, also known as angular frequency vector, is a vector measure of rotation rate, that refers to how fast an object rotates or revolves relative to another point, i.e. how fast the angular position or orientation of an object changes with time. The amount of change of angular displacement of the particle at a given period of time is called angular velocity. The track of the angular velocity vector is vertical to the plane of rotation, in a direction which is usually indicated by the right-hand rule.
The formula for angular velocity is
 $ \omega = \dfrac{{d\theta }}{{dt}} $
An object weighs differently on the equator and at the poles. An object weighs more on poles than equator because the gravity is maximum at poles and minimum at equator.
Now by applying the formula, we get,
 $ {g'} = g - R{\omega ^2}{\cos ^2}\theta $
At equator $ \theta = {0^ \circ } $
So, $ {g'} = g - R{\omega ^2} $
Let the weight at pole be mg and let the weight at equator be mg’.
Now according to the question, weight at equator is $ \dfrac{1}{3} $ of his weight at poles.
 $ mg' = \dfrac{{mg}}{3} $
 $ g' = \dfrac{g}{3} $
Now putting the value of $ {g'} = g - R{\omega ^2} $ ,
 $ g - R{\omega ^2} $ = $ \dfrac{g}{3} $
Now by further solving, we get,
 $ R{\omega ^2} = g - \dfrac{g}{3} $
 $ R{\omega ^2} = \dfrac{{2g}}{3} $
Taking R on the other side, we get,
 $ {\omega ^2} = \dfrac{{2g}}{{3R}} $
 $ \omega = \sqrt {\dfrac{{2g}}{{3R}}} $
So, the final answer is (2) $ \sqrt {\dfrac{{2g}}{{3R}}} $ .

Note:
Be careful while applying the formula. Remember that an object weighs more on poles than equator because the gravity is maximum at poles and minimum at equator. Calculation mistakes are possible, so try to avoid them and be sure of the final answer. Also remember the values of trigonometric functions.