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Inertial Force

Suppose there are two frames, an inertial frame S and a non-inertial frame S’. We first consider that co-ordinate system in S’ is translating with a constant acceleration A with respect to S; the co-ordinate system S’ does not rotate in space with respect to S. Let us now observe the motion of a particle P from both the frames S and S’ and inter-relate the dynamics.

At any instant t, let the position vectors P as observed from origins of S and S’ are respectively r (t) and r’ (t). Hence, we have

r (t) = r’ (t) + R (t)                                     (i)

where R (t) is the instantaneous position vector of origin O’ of the co-ordinate system in S’, with respect to origin O of the co-ordinate system in S. In particular, if O and O’ coincide at t = 0, then we have

R (t) = V_0t + ½ A t₂

Where V₀ is the velocity of O’ at t = 0; A is constant acceleration of O’.

Differentiating equation (i) with respect to t, we find

v (t) = v’ (t) + V (t)                                      (ii)

and, a (t) = a’ (t) + A                                 (iii)

where (v, a) and (v’, a’) are the velocity and acceleration of particle P as observed from S and S’ at time t; V (t) is the instantaneous velocity of origin O’ relative to O, given by

V (t) = V₀ (t) + A t

Equation (i) give the relationship between kinematical variables as observed from frames S and S’. Note that acceleration a and a’ of the particle P observed from the inertial frame S and non-inertial frame S’ are different.

Let us now multiply eq. (iii) by m, mass of the particle:

m a = m a’ + m A                                        (iv)

The left-hand side is equal to the ‘ordinary’ force F acting on the particle as seen by an observer on inertial frame S. Hence, we find

F = m a = m a’ + m A                                  (v)

How does an observer on the accelerating frame S’ interprets eq. (v)? In other words, what force F’ gives rise to acceleration a’ of the particle according to observer in S’? If the observer in S’ believes that, according to Newton’s law, an acceleration is always produced by a force then he would write the following:

F’ = m a’ = F – m A

= F + F_i                                                       (vi)

The net force acting on the particle, according to S’, consists of two parts: an ‘ordinary’ force F and another force F_i = - m A. The force F arises because of interaction of particle with some other objects; the force F_i arises because frame S’ itself has acceleration A. F_i is called the inertial force.

According to observer in S’, the inertial force F_i is as real as F in the sense that its effect on the motion of a particle can be observed. Typical examples are the backward push we feel inside a car accelerating with respect to earth regarded as an inertial frame; that’s a ‘real’ push. Or, the bob of a pendulum suspended inside from the ceiling of car is backward and is brought to rest in the car only when (horizontal component of) tension T in the string balances the inertial force F_i. That is, as seen in S’ (accelerating car), bob is in rest because net horizontal and vertical forces are zero:

T sin θ – F_i = 0

and, T cos θ – mg = 0

where θ is the angle string makes with vertical at the equilibrium position of bob. Hence, by observing θ, we can estimate the acceleration of car.

or, A = g tan θ                                             (vii)

Thus we see that the acceleration of a non-inertial frame can be determined by dynamics of objects observed from this frame.