In Taylor, read sections 9.3 to 9.5 for today, and 9.5 to 9.6 for Friday.

We denote the magnitude of the angular velocity of an object by ω.
For the purposes of this chapter, we can assume that an object is rotating about a fixed axis, whose direction is given by a unit vector we can call **û**.
To express both the magnitude of the angular velocity and the direction of the axis of rotation, we can combine these quantities into an angular velocity vector,

Assume we have a system with fixed point O, and an angular velocity **ω**.
What is the velocity of a point located a distance **r** from O?
The rotation of the system means that the point moves in a circle around the axis, and the radius of the circle is r_{⊥} = |**r**| sinθ where θ is the angle between **r** and **ω**.
The velocity of an object moving around a circle of radius r_{⊥} with angular velocity ω is v = ωr_{⊥}, and the direction is tangential to the circle (perpendicular to **r** and **ω**).
Using the expression for r_{⊥}, and the relation **ω**×**r** = |**r**| ω sinθ, we can write

v = |**ω**×**r**|

Finally, since the velocity vector is perpendicular to **r** and **ω**, we can write

This relation is more general, and works for any vector and its derivative:

d**e**/dt = **ω**×**e**.

Relative angular velocity vectors add just like relative velocity vectors. That is, if we know that the velocity of object 2 with respect to object 1 is v21, and that the velocity of object 3 with respect to object 2 is v32, then the velocity of object 3 with respect to object 1 is

Now use the relation derived above to show how angular velocities add.

therefore

So relative angular velocities add just like relative velocities.

I'll denote the angular velocity of the coordinate system (its rotation about the origin) by **Ω**.

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

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

The remaining three terms on the right represent the rotation of the coordinate axes.

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

Finally, since r = r', we have
(d**r**/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 (d**Q**/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 We can apply the above result to the velocity vector to find the acceleration:

(d**v**/dt)_{inertial} = (d**v**/dt)_{rot} + **Ω** × **v**

But we know that (d**v**/dt)_{inertial} = (d/dt)_{rot}(**v'** + **Ω** × **r'**) + **Ω** × (**v'** + **Ω** × **r'**)

= (d**v'**/dt)_{rot} + [d(**Ω** × **r'**)/dt]_{rot} + **Ω** × **v'** + **Ω** × (**Ω** × **r'**)

= (d**v'**/dt)_{rot} + (d**Ω**/dt)_{rot} × **r'** + **Ω** × (d**r'**/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:

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

since the cross product of a vector with itself is zero.
We can also use the relations 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.