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Week 6: Vector Torque and Angular Momentum

285

Example 6.3.4: Angular Momentum of Rod Sweeping out Cone

Ha! Caught you! This is a rotation that does not satisfy either of our two conditions. As we shall

~

see below, in this case we cannot write L = or L = Iω~ – they simply are not correct!

6.4: Angular Momentum Conservation

We have derived (trivially) the Law of Conservation of Angular Momentum: When the total external torque acting on a systems is zero, the total angular momentum of the system is constant, that is, conserved. As you can imagine, this is a powerful concept we can use to understand many everyday phenomena and to solve many problems, both very simple conceptual ones and very complex and di cult ones.

The simplest application of this concept comes, now that we understand well the relationship between the scalar moment of inertia and the angular momentum, in systems where the moment of inertia of the system can change over time due to strictly internal forces. We will look at two particular example problems in this genre, deriving a few very useful results along the way.

Example 6.4.1: The Spinning Professor

 

ω 0

 

ω f

D

D

D/2

D/2

Figure 82: A professor stands on a freely pivoted platform at rest (total moment of inertia of professor and platform I0) with two large masses m held horizontally out at the side a distance D from the axis of rotation, initially rotating with some angular velocity ω0.

A professor stands on a freely pivoted platform at rest with large masses held horizontally out at the side. A student gives the professor a push to start the platform and professor and masses rotating around a vertical axis. The professor then pulls the masses in towards the axis of rotation, reducing their contribution to the total moment of inertia as illustrated in figure 82

If the moment of inertia of the professor and platform is I0 and the masses m (including the arms’ contribution) are held at a distance D from the axis of rotation and the initial angular velocity is ω0, what is the final angular velocity of the system ωf when the professor has pulled the masses in to a distance D/2?

The platform is freely pivoted so it exerts no external torque on the system. Pulling in the masses exerts no external torque on the system (although it may well exert a torque on the masses themselves as they transfer angular momentum to the professor). The angular momentum of the system is thus conserved.

Initially it is (in this highly idealized description)

Li = Iiω0 = (2mD2 + I0)ω0

(585)

286

 

 

 

 

Week 6: Vector Torque and Angular Momentum

Finally it is:

2

+ I0)ωf = (2mD2 + I0)ω0 = Li

(586)

Lf = If ωf = (2m µ

 

D

 

 

2

 

 

 

 

 

Solving for ωf :

 

 

(2mD2

 

 

 

 

ωf =

 

+ I0)

ω0

(587)

 

(2m ¡

D

2

+ I0)

 

 

 

 

 

 

 

 

2 ¢

 

 

 

From this all sorts of other things can be asked and answered. For example, what is the initial kinetic energy of the system in terms of the givens? What is the final? How much work did the professor do with his arms?

Note that this is exactly how ice skaters speed up their spin when performing their various nifty moves – start spinning with arms and legs spread out, then draw them in to spin up, extend them to slow down again. It is how high-divers control their rotation. It is how neutron stars spin up as their parent stars explode. It is part of the way cats manage to always land on their feet – for a value of the world “always” that really means “usually” or “mostly”117.

Although there are more general ways of a system of particles altering its own moment of inertia, a fairly common way is indeed through the application of what we might call radial forces. Radial forces are a bit special and worth treating in the context of angular momentum conservation in their own right.

6.4.1: Radial Forces and Angular Momentum Conservation

One of the most important aspects of torque and angular momentum arises because of a curious feature of two of the most important force laws of nature: gravitation and the electrostatic force. Both of these force laws are radial, that is, they act along a line connecting two masses or charges.

Just for grins (and to give you a quick look at them, first in a long line of glances and repetitions that will culminate in your knowing them, here is the simple form of the gravitational force on a “point-like” object (say, the Moon) being acted on by a second “point-like” object (say, the Earth) where for convenience we will locate the Earth at the origin of coordinates:

~

MmMe

 

 

F m = −G

r2

rˆ

(588)

In this expression, ~r = rrˆ is the position of the moon in a spherical polar coordinate system (the direction is actually specified by two angles, neither of which a ects the magnitude of the force). G is called the gravitational constant and this entire formula is a special case of Newton’s Law of Gravitation, currently believed to be a fundamental force law of nature on the basis of considerable evidence.

A similar expression for the force on a charged particle with charge q located at position ~r = rrˆ exerted a charged particle with charge Q located at the origin is known as a (special case of) Coulomb’s Law and is also held to be a fundamental force law of nature. It is the force that binds electrons to nuclei (while making the electrons themselves repel one another) and hence is the dominant force in all of chemistry – it, more than any other force of nature, is “us”118. Coulomb’s Law is just:

~

qQ

 

 

F q = −ke

r2

 

rˆ

(589)

117I’ve seen some stupid cats land flat on their back in my lifetime, and a single counterexample serves to disprove the absolute rule...

118Modulated by quantum principles, especially the notion of quantization and the Pauli Exclusion Principle, both

beyond the scope of this course. Pauli is arguably co-equal with Coulomb in determining atomic and molecular structure.

Week 6: Vector Torque and Angular Momentum

287

where ke is once again a constant of nature.

Both of these are radial force laws. If we compute the torque exerted by the Earth on the moon:

τm = ~r × µ−G

MmMe

rˆ = 0

(590)

 

r2

If we compute the torque exerted by Q on q:

 

 

 

rˆ = 0

 

τq = ~r × µke r2

(591)

 

 

qQ

 

 

 

~

 

 

 

 

 

 

Indeed, for any force law of the form F (~r) = F (~r)rˆ the torque exerted by the force is:

 

τ = ~r × F (~r)rˆ = 0

(592)

and we can conclude that radial forces exert no torque!

 

In all problems where those radial forces are the only (significant) forces that act:

 

A radial force exerts no torque and the angular momentum of the object upon which the force acts is conserved.

Note that this means that the angular momentum of the Moon in its orbit around the Earth is constant – this will have important consequences as we shall see in two or three weeks. It means that the electron orbiting the nucleus in a hydrogen atom has a constant angular momentum, at least as far as classical physics is concerned (so far). It means that if you tie a ball to a rubber band fastened to a pivot and then throw it so that the band remains stretches and shrinks as it moves around the pivot, the angular momentum of the ball is conserved. It means that when an exploding star collapses under the force of gravity to where it becomes a neutron star, a tiny fraction of its original radius, the angular momentum of the original star is (at least approximately, allowing for the mass it cast o in the radial explosion) conserved. It means that a mass revolving around a center on the end of a string of radius r has an angular momentum that is conserved, and that this angular momentum will remain conserved as the string is slowly pulled in or let out while the particle “orbits”.

Let’s understand this further using one or two examples.

Example 6.4.2: Mass Orbits On a String

mr

v

F

Figure 83:

A particle of mass m is tied to a string that passes through a hole in a frictionless table and held. The mass is given a push so that it moves in a circle of radius r at speed v. Here are several questions that might be asked – and their answers:

a)What is the torque exerted on the particle by the string? Will angular momentum be conserved if the string pulls the particle into “orbits” with di erent radii?

288

Week 6: Vector Torque and Angular Momentum

This is clearly a radial force – the string pulls along the vector ~r from the hole (pivot) to the mass. Consequently the tension in the string exerts no torque on mass m and its angular momentum is conserved. It will still be conserved as the string pulls the particle in to a new “orbit”.

This question is typically just asked to help remind you of the correct physics, and might well be omitted if this question were on, say, the final exam (by which point you are expected to have figured all of this out).

b)What is the magnitude of the angular momentum L of the particle in the direction of the axis of rotation (as a function of m, r and v)?

Trivial:

L = |~r × p~| = mvr = mr2(v/r) = mr2ω =

(593)

By the time you’ve done your homework and properly studied the examples, this should be instantaneous. Note that this is the initial angular momentum, and that – from the previous question – angular momentum is conserved! Bear this in mind!

c)Show that the magnitude of the force (the tension in the string) that must be exerted to keep the particle moving in a circle is:

L2 F = T = mr3

This is a general result for a particle moving in a circle and in no way depends on the fact that the force is being exerted by a string in particular.

As a general result, we should be able to derive it fairly easily from what we know. We know two things – the particle is moving in a circle with a constant v, so that:

F =

mv2

 

(594)

 

 

r

We also know that L = mvr from the previous question! All that remains is to do some algebra magic to convert one to the other. If we had one more factor of m on to, and a factor of r2 on top, the top would magically turn into L2. However, we are only allowed to multiply by one, so:

F =

mv2

×

mr2

=

m2v2r2

=

L2

r

mr2

mr3

mr3

as desired, Q.E.D., all done, fabulous.

d) Show that the kinetic energy of the particle in terms of its angular momentum is:

L2

K = 2mr2

More straight up algebra magic of exactly the same sort:

 

mv2

 

mv2

mr2

 

 

L2

K =

 

=

 

×

 

=

 

 

2

2

mr2

2mr2

(595)

(596)

Now, suppose that the radius of the orbit and initial speed are ri and vi, respectively. From under the table, the string is slowly pulled down (so that the puck is always moving in an approximately circular trajectory and the tension in the string remains radial) to where the particle is moving in a circle of radius r2.

e)Find its velocity v2 using angular momentum conservation. This should be very easy, and thanks to the results above, it is:

L1 = mv1r1 = mv2r2 = L2

(597)