## Discuss the effect of Coriolis force in case of a freely falling body on the surface of the earth and calculate the deviation from the vertical suffered by a body dropped from rest from a height h.( or Deflection of a body, falling freely, due to Coriolis force) :

Let P be a point at height h from the surface of earth at latitude λ. Consider a rectangular Cartesian co-ordinate system fixed to the earth at p.

The z-axis is taken vertically upward

x axis is taken towards east

y asis is taken towards north

The earth rotates from west to east. The angular velocity **w ** lies in y-z plane in a direction parallel to the polar axis about which the earth rotates. Hence **w** has no component along east-west direction.

Now

where,

————–1

Let a body of mass m is dropped from p and attains a velocity **u**‘ after a time t, The velocity will be in the vertical direction (z- -axis)

————-2

as

Coriolis force acting on the body at that instant is given by

————3

Let

Hence magnitude of Coriolis forces on the body is 2mwvCosλ and acts towards **i** direction ice towards east direction in northern hemisphere.

Deviation calculation:-

Let x be the amount of deviation due to Coriolis force .

According to. Newton’s Second laws of motion,

or

Integrating,

Again integrating

Above equation describes the deviation along the x-axis

As the body is dropped from a height h, , is if t be the time in which it touches the ground, then

Substituting the value of t in equation 4, we have.

———-4

or

or,

———–5

East ward deviation of a freely falling body N/m wl IS ( due to Coriolis force).

Note: At the equation λ= 0 9-0 hence deflection is maximum

## Centre of mass

The concept of centre of gravity or weight was first introduced by the ancient Greek mathematician, physicist, and engineer Archimedes of Syracuse. He worked with simplified assumptions about gravity that amount to a uniform field, thus arriving at the mathematical properties of what we now call the centre of mass. Archimedes showed that the torque exerted on a lever by weights resting at various points along the lever is the same as what it would be if all of the weights were moved to a single point—their centre of mass.

**Linear moment**

We consider a system of n particles of masses

having position vectors

respectively Then linear moment of the system =

Where **m**i is the mass of its particle and **r**i is the position vector of the i’th particle relative to origin 0 – Linear moment also known as first moment of the mass.

Position of centre of mass of a system of particle:–

Let us consider a systers of particles having N particles. Let

be the centre of mass of the system of particles.

If **R** be the position vector of centre of mass then

to of to be the position vector of i’th particle of mass **m**i then

Now

Since linear moment (or moment of mass) with respect to centre of mass is zero Thus

using equation 1

or

where

or

NOTE : For continuous distribution of matter-

Centre of mass of a triangle containing masses m1, m2, m3 at the corner :–,

Let us consider a triangle containing masses m1, m2, m3 at the corners A, B, C respectively.

Let the coordinates of the corners are

If coordinate of Centre of mass is

then

Special case—- If

then

**Calculation of centre of mass :**

The centre of mass is a point in a system that responds to external forces as if the total mass of the system were concentrated at this point. The centre of mass can be calculated by taking the masses you are trying to find the centre of mass between and multiplying them by their positions. Then, you have to add these together and divide that by the sum of all the individual masses.

**Centre of mass of a uniform semicircular wire**

Let us us consider a semicircular wire of radius a radius a, mass m and length l

Hence mass per unit length =

Mass of elementary length (dm) =

If (xo, yo) be the co-ordinate of centre of mass then ,

**Centre of mass semicircular disc**

Let us consider a semicircular disc of radius a, mass m

Hence mass per unit area =

Elementary area =

Mass of elementary area =

If (xo,yo) is the coordinate of centre of mass then,

**Centre of mass** of a solid circular cone will be

Let us consider a solid circular having base radius a and height h.

From symmetry centre of mass must lie along OA (along z axis)

Let us take a section at z and z+dz with radius r

Volume of elementary section =

Mass per unit volume =

Mass of elementary section dm =

From figure

*** WE HAVE NOT GIVEN ALL THE CENTRE OF MASS CALCULATION.ACCORDING TO THE DEMAND WE WILL GIVE THE SAME.

Read more –

CLASSICAL MECHANICS | Virtual displacement| D’Alembert’s principle| Frame of Reference

CLASSICAL MECHANICS| Acceleration of a particle| Coriolis force and Centripetal force

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