University Physics I - Classical Mechanics, 2019

9.3. THE CROSS PRODUCT AND ROTATIONAL QUANTITIES 199
product ⃗r × ⃗v will always point upwards, along the positive z axis. Furthermore, since ⃗r and ⃗v
always stay perpendicular, the magnitude of L,byEq.(9.9), ⃗ will always be | L| ⃗ = mR|⃗v|. Taking
note of I = mR 2 and of Eq. (8.36), we see we have then
| L| ⃗ = ( mR 2) |⃗v|
= I|ω| (9.12)
R
z
L
ω
y
+
r
θ
R cos θ
v
P (x,y)
R sin θ
x
Figure 9.5: A particle moving on a circle in the x-y plane. For the direction of rotation shown, the vectors
⃗L = m⃗r × ⃗v and ⃗ω lie along the z axis, in the positive direction.
This suggests that we should define the angular velocity vector, ⃗ω, as a vector of magnitude |ω|,
pointing along the positive z axis if the motion in the x-y plane is counterclockwise as seen from
above (and in the opposite direction otherwise). Then this will hold as a vector equation:
⃗L = I⃗ω (9.13)
It may seem a very strange choice to have the angular velocity point along the z axis, when the
particle is moving in the x-y plane, but in a certain way it makes sense. Suppose the particle is
moving with constant angular velocity: the directions of ⃗r and ⃗v are constantly changing, but ⃗ω is
pointing along the positive z direction, which does remain fixed throughout.
There are some other neat things we can do with ⃗ω as defined above. Consider the cross product
⃗ω ×⃗r. Inspection of Figure 9.5 and of Eq. (8.36) shows that this is nothing other than the ordinary
velocity vector, ⃗v:
⃗v = ⃗ω × ⃗r (9.14)
Wecanalsotakethederivativeof⃗ω to obtain the angular acceleration vector ⃗α, sothatEq.(8.33)
will hold as a vector equation:
⃗ω(t +Δt) − ⃗ω(t)
⃗α = lim
= d⃗ω
(9.15)
Δt→0 dt dt