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Transcript
EFFECT OF CENTRIFUGAL AND CORIOLIS FORCES DUE TO EARTH’S ROTATION ON ‘g’ 1. Effect of centrifugal force The acceleration of a particle in frame of reference S’ rotating with a constant angular velocity ω is given by: ….(1) Here r is position vector of the particle and the other terms have their usual meaning. When a particle is at rest on the surface of earth which rotates with constant angular velocity ω about its polar axis, then: Then, Coriolis acceleration=0 i.e. and Therefore, from equation (1), the acceleration is given by …(2) Acceleration due to gravity at P acts along PO. The components of g are: This acceleration is given by eqn. (2) as where λ is latitude of particle P. This equation shows that centrifugal acceleration decreases the effective acceleration due to gravity. Hence §there is no effect of rotation of earth on the value of acceleration due to gravity at its poles §there is maximum effect of rotation of earth on the value of acceleration due to gravity at its equator. 2. Effect of Coriolis Force Consider a point P at latitude λ on the surface of earth. Let the earth rotate with constant angular velocity ω about its polar axis. ω makes an angle λ with y- axis at point P Imagine a point P’ vertically above P and drop a body of mass ‘m’, then velocity v acquired by it at time ‘t’ is given by: Vx and vy are taken as zero because body has velocity only along negative z-axis. The coriolis force acting on the particle is given by: This shows that coriolis force acts on the particle along positive x-axis. From Newton’s second law of motion Using the relation, v= u+at along negative z-direction. Integration w. r. t. time ‘t’ gives Again integrating w.r.t. t, we get: Hence, due to coriolis force, the particle dropped vertically downwards suffers a deviation along positive x-axis i.e. towards east. The displacement of the particle will be maximum for λ=0 i.e. at the equator. THANX!
