Effect of centrifugal force

4. Effect of Centrifugal Force

Earth completes one revolution in 24 hours about its axis.

∴ Angular velocity of earth ω = ^2π/_T = ^{2 × 3.14}/_{24 × 60 × 60}
ω = 7.29 × 10⁻⁵ radian/s

Let a particle is situated at a point P on the earth whose latitude is λ. (The angle between the radial line joining the point P to the centre of the earth and the equatorial plane of earth is λ and is called latitude.) Assuming point P as origin draw a reference frame whose X' axis is towards east, Y' axis is towards north and Z' axis is vertically upwards as shown in Fig.(4.1). This reference frame rotates with an angular velocity ω due to the rotation of the earth. As shown in Fig.(4.1), the angular velocity ω can be divided into two components. One component ω cos λ is towards north i.e. along Y' axis and another component ω sin λ is vertically upwards i.e. along Z' axis.

If unit vectors along the axes of reference frame are î', ĵ', k̂', then

ω = (ĵ' ω cos λ + k̂' ω sin λ)             ......(1)

If the particle is at rest on the earth, then Coriolis force will be zero and only centrifugal force will appear to be acting on the particle.

Let actual acceleration due to gravity at point P is a = g and apparent acceleration due to gravity due to rotational motion of earth is a' = g_λ.

Hence from equation (13) of section 2,

g_λ = g − ω × (ω × r)             ......(2)

where vector r is a position vector of point P about centre of rotation O'. The component of r along Y'-axis is r sin λ and another component of r along Z'-axis is r cos λ.

∴ r = (−ĵ' r sin λ + k̂' r cos λ)

Fig. 4.1 : Rotational reference frame on earth

If radius of earth is R, then r = R cos λ

∴ r = (−ĵ' R sin λ cos λ + k̂' R cos² λ)             ......(3)

Substituting equations (1) and (3) in equation (2), we get

g_λ = −k̂'g − ω × [(ĵ' ω cos λ + k̂' ω sin λ) × (−ĵ' R sin λ cos λ + k̂' R cos² λ)]
g_λ = −k̂'g − ω × (î' ω R cos λ)
g_λ = −k̂'g − (ĵ' ω cos λ + k̂' ω sin λ) × (î' ω R cos λ)
or   g_λ = −k̂'g + k̂' ω² R cos² λ − ĵ' ω² R cos λ sin λ
or   g_λ = −k̂' (g − ω² R cos² λ) − ĵ' ω² R cos λ sin λ             ......(4)

∴ Magnitude of effective g

|g_λ| = √[(g − ω² R cos² λ)² + (ω² R cos λ sin λ)²]

Since ω is a small quantity, so the term containing ω⁴ can be neglected. Hence

|g_λ| = (g − ω² R cos² λ)             ......(5)

This equation (5) describes the effect of rotation of earth on acceleration due to gravity. If the direction of observed acceleration due to gravity makes an angle θ with the direction of real acceleration due to gravity, then

Fig.(4.2): Direction of acceleration due to gravity at latitude λ

tan θ = ^{ω2 R cos λ sin λ}/_{(g − ω² R cos² λ)}

∵ ω² R cos² λ is negligible in comparison to g, so it can be neglected. Hence

tan θ = ^{ω2 R cos λ sin λ}/_g = ^{ω2 R}/_2g sin 2λ             ......(6)

(i) At pole of the earth :

λ = 90°, cos λ = 0
∴ g_p = g

Thus the acceleration due to gravity at poles is equal to the real acceleration due to gravity and its direction is towards the centre of the earth.

(ii) At equator :

λ = 0, cos λ = 1
∴ g_e = g − ω² R
and θ₀ = 0

Thus at equator there is maximum effect of rotation of earth on acceleration due to gravity and the direction of acceleration due to gravity in this case also towards the centre of the earth.

(iii) At λ = 45°

g_λ = g − ^{ω2 R}/₂
and θ = tan⁻¹ (^{ω2 R}/_2g)

Thus at λ = 45° the direction of observed acceleration due to gravity is inclined maximum towards the direction of the real acceleration due to gravity.

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5. Effect of Coriolis Force on a Particles Moving Horizontally on Earth

In order to understand the effect of Coriolis force on the particle moving on earth, let us assume that a particle of mass m is moving horizontally on earth with a velocity v' at the place P whose angle of latitude is λ. Draw a reference frame S' at point P whose X'-axis is towards east, Y'-axis is towards north and Z'-axis is vertically upwards. This reference frame is rotating with an angular velocity ω as earth is rotating about its north-south axis. Therefore, from equation (1) of previous section,

ω = (ĵ' ω cos λ + k̂' ω sin λ)             ......(1)

and the velocity of particle is

v' = î' v_x' + ĵ' v_y'             ......(2)

Thus the Coriolis force

F_c = −2m (ω × v')

= −2m

î'	ĵ'	k̂'
0	ω cos λ	ω sin λ
vx'	vy'	0

F_c = î' 2m ω v_y' sin λ − ĵ' 2m ω v_x' sin λ + k̂' 2m ω v_x' cos λ             ......(3)

Thus horizontal component of Coriolis force is

F_ch = 2m ω (î' v_y' − ĵ' v_x') sin λ             ......(4)

and the magnitude of horizontal component of Coriolis force is

|F_ch| = 2m ω sin λ √(v_x'² + v_y'²)
or   |F_ch| = 2m ω v' sin λ             ......(5)

Thus the horizontal component of Coriolis force acts towards right in northern hemisphere and towards left in southern hemisphere.

The vertical component of Coriolis force is

F_c = k̂' 2m ω v_x' cos λ

The vertical component of Coriolis force always acts upwards in both hemispheres.

Natural erosion of river banks, cyclones, etc. can be explained by the effect of Coriolis force on particles moving on the earth.

(i) Erosion of river banks

Suppose that water is flowing with a velocity v = î' v_x' from west to east in the river in northern hemisphere.

The horizontal component of Coriolis force acting on the water particles—

F_ch = − ĵ' 2m ω v_x' sin λ

It is evident from this equation that erosion takes place on the right bank of the river flowing from west to east in northern hemisphere due to Coriolis force and left bank of the river in the southern hemisphere.

(ii) Cyclones

Some times cyclone is generated in the atmosphere due to the inward spiraling motion of the air. The Cyclones can be explained on the basis of Coriolis force. When air of the atmosphere becomes hot at certain place, it goes up and creates low pressure at that place. As a result air rushes with a very high speed from high pressure area toward low pressure area. Hence Coriolis force starts acting on the air particles which rotate the air anti-clockwise spiraling upward in the northern hemisphere and clockwise in the southern hemisphere of the earth. This generates the cyclone as shown in fig.(5.1).

Fig.(5.1): Formation of cyclone in northern hemisphere

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6. Effect of Coriolis Force on Bodies falling Vertically Downward on Earth

Let a body of mass m is falling downward from a height h due to gravity. After t seconds its velocity will be

v' = − k̂' g t             ......(1)

where k̂' is the unit vector vertically upward.

Fig. 6.1 : Displacement a body falling on earth

If angle of latitude at this place is λ and the angular velocity of rotational reference frame is ω, then

ω = (ĵ' ω cos λ + k̂' ω sin λ)             ......(2)

Thus, Coriolis force acting on the body is

F_c = −2m (ω × v')

= −2m

î'	ĵ'	k̂'
0	ω cos λ	ω sin λ
0	0	−g t

= î' 2 m g t ω cos λ             ......(3)

Since the direction of î' is towards east, therefore the body falling downward in the northern hemisphere moves toward east.

Thus from Newton second law the equation of motion of a body towards east will be—

m ^{d2 x'}/_dt² = 2 m g t ω cos λ
or   ^{d2 x'}/_dt² = 2 g t ω cos λ

Integrating with respect to time t, we have

^dx'/_dt = (g ω cos &lambda) t² + C

where C is an integral constant.

Since ^dx'/_dt = 0 at t = 0, therefore C = 0

Hence   ^dx'/_dt = (g ω cos &lambda) t²

Again integrating,

x' = ¹/₃ (g ω cos &lambda) t³ + C'

where C' is another integral constant.

Let the coordinates of point P at t = 0 is (0, 0, h)

Hence x' = 0 at t = 0. ∴ C' = 0.

∴ x' = ¹/₃ (g ω cos &lambda) t³             ......(4)

If the body is falling from a height h, then time taken by the body in falling is—

t = (^2h/_g)^1/2      [By h = ¹/₂ g t²] ......(5)

Substituting equation (5) in equation (4), we have

x' = ¹/₃ (g ω cos &lambda) (^2h/_g)^3/2
x' = ²/₃ (h ω cos &lambda) √(^2h/_g)             ......(6)

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7. Effect of Coriolis Force on Bodies Thrown Vertically Upward from Earth

If a body of mass m is thrown vertically upward with a velocity u from a certain place of the earth.

After t seconds the velocity of a body will be—

v' = k̂' (u − g t)             ......(1)

If angle of latitude at this place is λ and the angular velocity of rotational reference frame is ω, then

ω = (ĵ' ω cos λ + k̂' ω sin λ)             ......(2)

Coriolis force acting on the body is

F_c = −2m (ω × v')

= −2m

î'	ĵ'	k̂'
0	ω cos λ	ω sin λ
0	0	u − g t

= − î' 2 m (u − g t) ω cos λ             ......(3)

Fig.(7.1): Displacement of a body on earth when thrown vertically upward

Thus the displacement of the body due to Coriolis force acting toward west in the northern hemisphere is toward west.

From Newton second law,

m ^{d2 x'}/_dt² = −2 m (u − g t) ω cos λ
or   ^{d2 x'}/_dt² = −2 (^u/_g − t) g ω cos λ             ......(4)

Integrating with respect to time t, we have

^dx'/_dt = −2 (g ω cos &lambda) (^{u t}/_g − ^t2/₂) + C