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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!