PHY5210 W08

Chapter 9: Mechanics in Noninertial Frames


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

The Angular Velocity Vector

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,

ω = ωû.
The magnitude of the vector is the angular velocity, and the direction is parallel to the axis of rotation (use the right hand rule).

A Useful Relation

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

v = ω×r.

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

de/dt = ω×e.

Addition of Angular Velocities

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

v31 = v32 + v21.

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

ω31×r = ω32×r + ω21×r = (ω32 + ω21r.


ω31 = ω32 + ω21

So relative angular velocities add just like relative velocities.

Notation for Angular Velocities

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

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).

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

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 ×.

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. 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:

© 2008 Robert Harr