In Taylor, read sections 9.6 to 9.7 for today, and 9.7 to 9.8 for Friday.

The angular velocity vector

A useful relation:

d**e**/dt = **ω**×**e**.

The time derivative of vector in a rotating coordinate system:

(d**Q**/dt)_{inertial} = [(d/dt)_{rot} + **Ω** × ] **Q**

In deriving this we had the expression:

and I said that the last three terms are equal to **Ω** × **r'**.
Here is the demonstration.
From the "useful relation" we get that d**i'**/dt = **Ω** × **i'**.
And likewise d**j'**/dt = **Ω** × **j'** and d**k'**/dt = **Ω** × **k'**.
Therefore

x'(d**i'**/dt) + y'(d**j'**/dt) + z'(d**k'**/dt) = x'(**Ω** × **i'**) + y'(**Ω** × **j'**) + z'(**Ω** × **k'**)

or, rearranging terms

x'(d**i'**/dt) + y'(d**j'**/dt) + z'(d**k'**/dt) = **Ω** × (x'**i'** + y'**j'** + z'**k'**) = **Ω** × **r'**.

We ended the last lecture with the acceleration in a rotating coordinate system:

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

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

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:

m**a'** = **F**_{real} - m**Ω**[dot] × **r'** - 2m**Ω** × **v'** - m**Ω** × (**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}= -m**Ω**[dot] ×**r'**is the transverse force,**F'**_{cor}= 2m**v'**×**Ω**is the Coriolis force, and**F'**_{cent}= -m**Ω**× (**Ω**×**r'**) is the centrifugal force.

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

To get a feel for the centrifugal force, imagine a mass m a distance r_{⊥} from the axis of rotation.
This mass feels a force -mΩ²r_{⊥} directed away from the axis of rotation (centrifugal means away from the center).
This is the force "felt" by an object rotating around an axis that must be counteracted by a centripetal force (centripetal means towards the center).

As an example, consider the following amusment park ride. People enter a cylindrical "room" and stand against the outer wall. The cylindircal room begins to rotate. The riders feel a centrifugal force directed away from the center of the cylinder, that pushes them against the outer wall. When the cylinder is rotating sufficiently fast, the floor drops away, leaving the riders held to the wall by friction (the normal force of the wall balances the centrifugal force and the friction is proportional to the normal force).

From the point of view of an outside, inertial viewer, the wall pushes against the riders to give them the centripetal acceleration necessary for their circular motion. From the point of view of a rider, a centrifugal force pushes them against the wall.

To get a feel for the Coriolis force, consider a puck sliding across a frictionless, rotating surface (this was a homework problem from chapter 1).
An observer in an inertial frame would see the puck slide along a straight line across the surface, as the surface rotates beneath it.
An observer in a non-inertial frame, rotating with the surface, would see the puck follow a curved path.
The curvature is produced by the Coriolis force 2m**v'** × **Ω**.
The Coriolis force is directed perpendicular to **Ω**, and since **Ω** is perpendicular to the rotating surface, the Coriolis force must lie parallel to the surface.
It is also perpendicular to the velocity (in the rotating frame), so it will cause the puck to move in a curved path.
Since both Ω and v will be constant in magnitude, the force will also be constant, and the resulting path is circular.

On earth, the Coriolis force has dramatic effects, being largely responsible for the rotation of storm systems, including tonadoes, hurricanes, and typhoons.

The transverse force appears when the rate of rotation changes **Ω**[dot]&neq; 0.
It is ignored by Taylor.
I will only discuss this if I include a homework problem that requires it.

A smooth rod of length l rotates in a plane with a constant angular velocity **Ω** about an axis fixed at the end of the rod and perpendicular to the plane of rotation.
A bead of mass m can slide along the rod without friction is initially positioned at the stationary end of the rod and given a slight push such that its initial speed directed down the rod is ε = Ωl (draw figure).
Calculate how long it takes for the bead to reach the other end of the rod.

Choose a coordinate system that rotates with the rod. Let xhat be along the rod and yhat be perpendicular to the rod. Since the bead is constrained to the rod, the problem becomes one dimensional. The only regular force is the one constraining the bead to the rod, and since this force is perpendicular to the rod, it points in the y direction. There is no regular foce in the x direction.

We need to find the fictitious forces on the bead.
To do this, assume the bead is a distance x from the origin, that is, it is located at position (x, 0, 0) in the rotating frame, and assume it has velocity (xdot, 0, 0).
(The y and z components of the velocity must be zero in this frame, because the bead can only move along the rod, in x.)
Using **Ω** = Ωkhat = (0, 0, Ω), the fictitious forces are

The Coriolis force points in the y direction and will be canceled by the force that constains the bead to the rod. The centrifugal force will contribute to the motion of the bead along the wire. The equation of motion in x is

m(xddot) = mΩ²x

or
xddot = Ω²x.

This is a second order linear equation in x, and the solution is

x(t) = Ae^{Ωt} + Be^{-Ωt}.

xdot(t) = ΩAe^{Ωt} - ΩBe^{-Ωt}.

The initial conditions are that x(0) = 0, and xdot(0) = ε = Ωl. Applying these yields A + B = 0, and ε = Ω(A - B), or A = -B = ε/2Ω. Therefore, x(t) is

x(t) = (ε/2Ω)(e^{Ωt} - e^{-Ωt}) = (ε/Ω)sinh(Ωt).

The time, T, for the bead to reach the end of the rod is found by solving
x(T) = (ε/Ω)sinh(ΩT) = l.
Recall that ε = Ωl, therefore sinh(ΩT) = 1, or T = (1/Ω)sinh^{-1}(1) = 0.88/Ω.