Lecture 22: Energy Conservation in 3D, Cen-
▯ relevant text sections: Morin 5.3, 7.1-7.2
Conservative 3D Forces
▯ What 3D forces are conservative? MUST only be functions of position:
F(~r) but must also meet the requirement that:
r ▯ F = 0 (1)
i.e. that the CURL of the function is 0. In order to prove this, you need
to use Stoke’s Theorem (which you will learn in multivariable calculus
near the end of the course). I will assume you haven’t seen this yet and
therefore you are not responsible for proving the above. But you are
responsible for being able to use Equation (1) to test whether a force
is conservative. You can read about the proof in the text if you like.
▯ Since some of you haven’t seen the \Curl" before, its just a cross prod-
uct of the gradient operator and a function. I’ve provided the formula
in the supplementary notes and Deepak will cover some examples in
next Monday’s lecture. Also check out Appendix B.6.
▯ It turns out that one class of 3D forces: \Central Forces" are ALWAYS
conservative. A central force is de▯ned to be a force that points radially
and whose magnitude depends only on the distance from the origin (i.e.
on the magnitude jrj = r). This means you can write a central force in
polar coordinates as:
F = F(r)r^ (2)
Notice that there is no ▯ component to the force and the force only
depends on the scalar r. You can prove that all central forces are
conservative by showing that r ▯ F = 0 for all central forces. This is
done on pg 151 of Morin, but I’ve done a more detailed version of this
in the supplementary notes so please look at that.
1 ▯ The fact that central forces are all conservative is a very important
result because it says we can de▯ne potential energy functions for any
central force. Some important ones you might remember: GRAVITY
FORCE, ELECTRIC FORCE. Basically, many important forces have
this form and we can de▯ne potentials for all of them and therefore
study their energy equation fairly simply.
▯ We will not be covering sections 5.4-5.8 on gravity and momentum
conservation in this course so you are not responsible for the types of
questions in these sections for the exam (e.g. ▯nding the gravitational
force for di▯erent mass distributions, using momentum conservation to
solve problems, collision problems).
Angular Momentum Conservation with Central Forces
▯ Since central forces are conservative, we can write them in terms of a
potential V (r) with:
F = F(r)r ^ = ▯rV (r) = ▯ dV ^
so taking Rhr r 0omp0nent gives: F(r) = ▯ dr and integrating:
V (r) = ▯ r0dr F(r )+V . 0ere the gradient has a component only in
the radial direction.
▯ Central forces have special properties. The ▯rst is that the angular
momentum L of a particle in a central force ▯eld is conserved. If we
de▯ne the linear momentum p ~ = m~ v, then the angular momentum is
L = ~r ▯ ~:
L out of page
▯ The vector L is perpendicular to the instantaneous plane of motion of
the ▯eld, which contains r and ~. In the ▯gure,r and ~ are in the y ▯z
plane and L points in the x direction, using the right hand rule.
▯ In a central ▯eld, the angular momentum of a particle is conserved. We
can show this by taking the time derivative of angular momentum and
showing that it is 0:
_ _ _ ~
L = ~r ▯ ~ + r ▯ ~ = v ▯ (m~v) + r ▯ F = 0:
The ▯rst term ~v ▯ (m~v) is zero by de▯nition of the cross product, and
the second term ~r ▯ F is zero because F points in the ~ r direction.
This tells us that the magnitude and the orientation of L are constant
following the particle in a central ▯eld.
▯ Conservation of angular momentum provides a big constraint on the
motion. One thing it means is that a particle in a central ▯eld remains
in the plane of motion, because the DIRECTION of the angular mo-
mentum vector cannot change. Since the angular momentum vector
points perpendicular to the plane of motion, this means the plane of
motion can’t change.
Energy Equation and E▯ective Potential
▯ Since we have a central force, it turns out to be easiest to work in polar
coordinates to derive an energy equation. It also turns out we can
derive an energy equation that only involves r and r _ (so no ▯ stu▯).
This means it e▯ectively becomes a 1D equation. Here is how:
3 1. Start with our de▯nition for mechanical energy:
E = K + V = mv + V (r)
2. Express v in terms of the derivative of the position coordinate.
Recall that position in polar