PHY5210 W15

Chapter 9: Mechanics in Noninertial Frames

Reading:

In Taylor, read sections 9.4 to 9.5 for today, and 9.6 to 9.7 for Friday.

The Angular Velocity Vector

Recall

ω = ωû.

de/dt = ω×e.

Time Derivatives in a Rotating Frame

Consider again two coordinate systems, an inertial system Oxyz, and a non-inertial system O'x'y'z' whose origin coincides with the inertial frame, but is rotating with respect to it. Since the origins coincide, r = r', but since the axes don't

xi + yj + zk = x'i' + y'j' + z'k'.

Differentiate this to find the relation between velocities, noting that in the inertial frame, the axes are fixed, but in the non-inertial frame they aren't

(dx/dt)i + (dy/dt)j + (dz/dt)k = (dx'/dt)i' + (dy'/dt)j' + (dz'/dt)k' + x'(di'/dt) + y'(dj'/dt) + z'(dk'/dt)

The first three terms on the left are the velocity vector in the inertial frame, and the first three terms on the right are the velocity vector in the non-inertial frame, that is, the time derivative of the coordinates measured with respect to the axes in that frame. Therefore we can write

v = v' + x'(di'/dt) + y'(dj'/dt) + z'(dk'/dt).

We can use the "useful relation" to simplify the remaining three terms on the right. The primed coordinate unit vectors rotate with the rotating coordinate system, so the "useful relation" can be applied to yield di'/dt = Ω × i'. And likewise dj'/dt = Ω × j' and dk'/dt = Ω × k'. Therefore

x'(di'/dt) + y'(dj'/dt) + z'(dk'/dt) = x'(Ω × i') + y'(Ω × j') + z'(Ω × k')

or, rearranging terms

x'(di'/dt) + y'(dj'/dt) + z'(dk'/dt) = Ω × (x'i' + y'j' + z'k') = Ω × r'.

v = v' + Ω × r'

or more explicitly

(dr/dt)_inertial = (dr'/dt)_rot + Ω × r' = [(d/dt)_rot + Ω × ] r'.

Finally, since r = r', we have

(dr/dt)_inertial = [(d/dt)_rot + Ω × ] r

where the subscript rot reminds us that the derivative is taken with respect to the rotating coordinate axes. This applies equally to the time derivative of any vector Q:

(dQ/dt)_inertial = [(d/dt)_rot + Ω × ] Q'.

This result says that the time derivative operation in an inertial frame is equivalent to taking the time derivative in a rotating frame plus Ω × Q'.

Acceleration in a Rotating Frame

We can apply the above result to the velocity vector to find the acceleration:

(dv/dt)_inertial = (dv/dt)_rot + Ω × v

But we know that v = v' + Ω × r', so

(dv/dt)_inertial = (d/dt)_rot(v' + Ω × r') + Ω × (v' + Ω × r')

= (dv'/dt)_rot + [d(Ω × r')/dt]_rot + Ω × v' + Ω × (Ω × r')

= (dv'/dt)_rot + (dΩ/dt)_rot × r' + Ω × (dr'/dt)_rot + Ω × v' + Ω × (Ω × r')

Now we need to simplify these terms. Let's use our rule for taking time derivatives to evaluate the derivative of Ω :

(dΩ/dt)_inertial = (dΩ/dt)_rot + Ω × Ω = (dΩ/dt)_rot = Ω[dot]

since the cross product of a vector with itself is zero. That is, all observers measure the same angular velocity of the rotating reference frame, a convenient result. We can also use the relations v' = (dr'/dt)_rot and a' = (dv'/dt)_rot. The result becomes

a = a' + Ω[dot] × r' + 2Ω × v' + Ω × (Ω × r')

This is the result we want. The terms on the right have names:

Ω[dot] × r' is the transverse acceleration,
2Ω × v' is the Coriolis acceleration, and
Ω × (Ω × r') is the centripetal acceleration.

Newton's Second Law in a Rotating Frame

Multiplying both sides of the acceleration relation by m and rearranging terms, we arrive at a form of Newton's second law suitable to use in a rotating frame of reference:

ma' = F_real - mΩ[dot] × r' - 2mΩ × v' - mΩ × (Ω × r')

where

F_real is the sum of all the real forces on the object (forces that exist in an inertial reference frame,
F'_trans = -mΩ[dot] × r' is the transverse force,
F'_cor = 2mv' × Ω is the Coriolis force, and
F'_cent = -mΩ × (Ω × r') is the centrifugal force.

Let's take a look at each of these forces in turn.