University Physics I - Classical Mechanics, 2019

9.4. TORQUE 201
momentum involves the particle’s distance to a point. For particles at different “heights” along the
axis of rotation, these quantities are different. It can be shown that, in the general case, all we can
say is that L z = Iω z ,ifwecallz the axis of rotation and calculate ⃗ L relative to a point on that
axis.
On the other hand, if the axis of rotation is an axis of symmetry of the object, then ⃗ L has only a z
component, and the result ⃗ L = I⃗ω holds as a vector equation. Most of the systems we will consider
this semester will be covered under this clause, or under the “essentially flat” clause mentioned
above.
In what follows we will generally assume that I has only a z component, and we will drop the
subscript z in the equation L z = Iω z ,sothatL and ω will not necessarily be the magnitudes of
their respective vectors, but numbers that could be positive or negative, depending on the direction
of rotation (clockwise or counterclockwise). This is essentially the same convention we used for
vectors in one dimension, such as ⃗a or ⃗p, in the early chapters; it is fine for all the cases in which
the (direction of the) axis of rotation does not change with time, which are the only situations we
will consider this semester.
9.4 Torque
We are finally in a position to answer the question, when is angular momentum conserved? To do
this, we will simply take the derivative of ⃗ L with respect to time, and use Newton’s laws to find
out under what circumstances it is equal to zero.
Let us start with a particle and calculate
dL
⃗
dt = d d⃗v
(m⃗r × ⃗v) =md⃗r × ⃗v + m⃗r ×
dt dt dt
(9.19)
The first term on the right-hand side goes as ⃗v ×⃗v, which is zero. The second term can be rewritten
as m⃗r × ⃗a. But, according to Newton’s second law, m⃗a = ⃗ F net . So, we conclude that
d ⃗ L
dt = ⃗r × ⃗ F net (9.20)
So the angular momentum, like the ordinary momentum, will be conserved if the net force on the
particle is zero, but also, and this is an important difference, when the net force is parallel (or
antiparallel) to the position vector. For motion on a circle with constant speed, this is precisely
what happens: the force acting on the particle is the centripetal force, which can be written as
⃗F c = m⃗a c = −mω 2 ⃗r (using Eq. (9.18)), so ⃗r × ⃗ F c = 0, and the angular momentum is constant.