Fictitious forces

They are:
Centrifugal force,
Coriolis force, and
Euler force

Fictitious means created by the imagination. Here, in order to satisfy the Newton's second law.

1.Rotation of axes

The circular motion of a particle was described in the circular motion section.
We have found for an orbiting mass an expression of the radial and tangential acceleration and provided the definition of the centripetal force. The observer set on a moving mass m thought that it exists a centrifugal force just to justify the second Newton's law. The observer from the fixed inertial frame said that this force is just inertial or fictitious.

Now, we are going to set fixed in the the rotating frame B of the mass m to see what's really going on in the obverver at the frame A point of view. Notice that the observer in this frame B consider itself at rest. Wheras for the observer in the inertial frame A, the frame B is rotating and it is not an inertial frame

Now, if the mass m orbits at the angular speed ω, the frame B in which is attached spins at the same speed ω and in the same direction. We can then set the mass m on a rotating turntable spinning about the z-axis at the same angular velocity.

In the inertial frame A, the unit vector are i, j, and k; the vector position of the particle of mass m is r_A. Its velocity is v_A = dr_A/dt, and the acceleration is a_A = d²r_A/dt².

The particle of mass m rotates. Both r_A, and the three unit vector are u1, u2, and u3 rotate.

Let u1, u2 and u3 the three unit vectors of the non inertial frame B, and the mass m has the position vector r_B = x1 u1 + x2 u2 + x3 u3. The related velocity is v_B = dr_B/dt, and the acceleration is a_B = d²r_B/dt².

We can express any vector such as vector position, velocity, and acceleration of the particle in the frame B in the base (u1,u2,u3) and then transform then in the base (i,j,k) of the frame A.

After a rotation θ = ωt, about the z-axis,the unit vectors rotate counterclockwise:
r_A → x1 u1 + x2 u2 + x3 u3
u1 → u1 = cos ωt i + sin ωt j     (1)
u2 → u2 = - sin ωt i + cos ωt j    (2)
u3 → u3 = u3 (no change)     (3)

The derivative with respect to time gives:

du1/dt = - ω sin ωt i + ω cos ωt j
du2/dt = - ω cos ωt i - ωt sin ωt j

The vector Omega; = ω k can be written Ω = 0 i + 0 j + ω k .
Hence:
Ω x u1 = ω (0,0,1) x (cos ωt, sin ωt, 0)
ω(- sin ωt i + cos ωt j + 0 k) = ω du1/dt = u2
Ω x u1 = ω du1/dt = u2 (4)    

Similarly,
Ω x u2 = ω ω (0,0,1) x (- sin ωt, cos ωt, 0) =
ω ω(- cos ωt i - sin ωt j + 0 k) = - ω du2/dt = - u1
Ω x u2 = ω du2/dt = - u1     (5)

Similarly,
Ω x u3 = 0     (6)

2. Velocity of the particle

We will find the expression of the velocity of the particle in the frame B, and then in the frame A.

The corresponding velocity of the point mass m is
v_A = dr_A/dt = (dx1/dt )u1 + x1 du1/dt + (dx2/dt)u2 + x2 du2/dt + (dx3/dt)u3 + x3 du3/dt
v_A = v1 u1 + x1 du1/dt + v2 u2 + x2 du2/dt + v3 u3 + x3 du3/dt

The particle of mass m moves on the plane, so x3 = v3 = du3/dt = 0.

Using the relation (4) (5) and (6), we get:
dr_A/dt = v1 u1 + x1 (Ω x u1) + v2 u2 + x2 ( Ω x u2 )
v1 u1 + v2 u2 = v_B
x1 (Ω x u1) + x2 (Ω x u2) = ω(x1 u2 - x2 u1) = Ω x r

Therefore:
dr_A/dt = v_A = v_B + Ω x r     (7)

3. Acceleration of the particle

The corresponding acceleration is:

a_A = dv_A/dt + d(Ω x r)/dt
dv_A/dt = a1 u1 + v1 du1/dt + a2 u2 + v2 du2/dt + a3 u3 + v3 du3/dt =
= a1 u1 + a2 u2 + v1 du1/dt + v2 du2/dt + 0 =
= a_B + v1(Ω x u1 )+ v2(- Ω x u2 ) or =
a_B + v1(u2) + v2(- u1) = a_B+ Ω x v

d(Ω x r)/dt = Ω x (dr/dt) + (Ω/dt) x r
Using (7), we find:
Ω x ( v + Ω x r) + (Ω/dt) x r =
Ω x v + Ω x Ω x r + (Ω/dt) x r =


Therefore:
a_A = a_B + Ω x v + Ω x v + Ω x Ω x r + (Ω/dt) x r =
a_B + 2Ω x v + Ω x (Ω x r) + (Ω/dt) x r

4. Related forces

Let's multiply by the mass of the particle, the two hands of the found equality, we get the following forces:
F_A = F_B + F_co + F_ce + F_eu

Applying Newton's second law in the frame B, gives:
F_B = F_A + F_co + F_ce + F_eu
Where:

F_A = - m a_A is the force acting on the particle in the frame A
F_B = m a_B is the force acting on the particle in the frame B
F_co = - 2 mΩ x v is the force of Coriolis acting on the particle in the frame B
F_ce = - m Ω x (Ω x r) is the centrifugal force acting on the particle in the frame B
F_eu = - m (Ω/dt) x r is the Euler force acting on the particle in the frame B


The three forces F_co, F_ce, and F_eu are associated with rotation. They are called inertial forces, or pseudo-forces, or fictitious forces. They are real (in frame A), but they are fictitious forces in the rotating reference frame B. These forces are just apparent in a non-inertial reference frame and do not come from any physical interaction between objects.

4.1. Centrifugal force

The centrifugal force F_ce = - m Ω x (Ω x r) is directed along the r direction and opposite to the centripetal force.

For an observer on the mass m, the centrifugal force is an outward force that balance the centripetal force in order to satisfy Newton's second law.

4.2. Coroilis force

The Coriolis force F_co = - 2 mΩ x v curves the trajectory of the rotating particle and exerts a torque on it.

In our Earth rotating reference frame, we encountere all the time the effects of this force, but they are small, because the angular speed by which the earth spins is small, of value ω = 7.0 x 10^-5 rad/s.

For a n object of mass m = 10.0 kg moving horizontally at a speed v = 10.0 m/s, the outward Coriolis force curves its path toward a direction perpendiculary to the plane North x West-East by a force of 2 m Ωv = 2 . 10 . 7.0 x 10^-5 . 10 = 1.4 x 10^-2 kg.m/s² = 0.014 N, that is about the gravitational force exerted on an object of mass equal to 1 gram.

4.2. Euler force

The Euler force F_eu = - m (Ω/dt) x r appears when the rotation of the rotating reference frame is not uniforme, that is the angular velocity ω varies in time in magnitude or in direction such as a precession.