April 23

The assignment for next week is posted.

First, discuss any rotation problems that you'd like to go over.

Second, I have in mind to try to explain in a bit more detail than I did on Tuesday two ideas from this 3D rotation stuff.

(This is all culture - cool stuff that's mostly outside the scope of this class.)

1. matrices as rotations : linear algebra 101

It turns out that the math behind 3D rotations is pretty interesting. There are all sorts of tricky bits, such as the fact that a rotation around the x axis followed by a rotation around the y axis gives a different result than the other way around.

In other words, rotations don't commute : a b ≠ b a .

Another sort of thing that doesn't commute are matrices. So ... maybe it isn't that surprising that the two are deeply related.

See for example

• wikipedia: matrix multiplication
• wikipedia: rotation matrix

Why does that matter for the physics we're doing?

Well, for linear momentum, the equation is

\[ \vec{p} = m \vec{v} \]

and those two vectors always point in the same direction since the mass is a scalar.

But for angular momentum, the equation is

\[ \vec{L} = \left[ I \right] \vec{\omega} \]

where the moment of inertia is a matrix ... and so the spin vector and angular momentum vector can point in different directions. The moment of inertia matrix (well, tensor actually) may rotate the spin vector.

It turns out that there is also a special set of of three perpendicular directions call the "principle axes" for any solid, even irregular ones like a potato. These axes can be found from the moment of inertia tensor. And further, if you use those axes as (x,y,z), then the moment of inertia is diagonal :

Ixx   0     0
0     Iyy   0
0     0     Izz

It's almost magic.

2. gyroscope equation starting from \( \vec{\tau} = d \vec{L} / dt \)

• ... which is analogous to \( \vec{F} = d \vec{p} / dt \)
• First review position, velocity, acceleration for circular motion.
• Then consider a gyro rotating at right angles to gravity,
• and precessing in a circle with \( \Omega \) precession rate.
• The gyro spins at \( \omega \) , has inertia \( M r^2 \) if r is the lever arm from pivot, |L| = I \( \omega \)
• and experiences a torque which is gravity times lever arm, |\(\tau\) = M g r
• ... so we can put all that all together to get : \( \tau = M g r = \Omega M r^2 \omega \)
• ... which lets you calculate the rate of precession.
• Using the right hand rule, it can also explain the direction of precession given direction of spin.
• And for a tilted gyro, we only use the component of L in the plane of precession.

Maybe that'll make more sense on the blackboard. ;)

See for example this explanation in University Physics

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My blackboard scribblings are posted.

And I found my 1D billiard ball solution from 1992 for the 'interesting' problem that I posed previously.

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