really mean Iaa

Week 6: Vector Torque and Angular Momentum

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for some axis a such that at least one of the two conditions above are satisfied, but math/physics/engineering students, at least, should try to remember that it really ain’t so116.

All of the (scalar) I’s we computed in the last chapter satisfied these symmetry conditions:

•A ring rotating about an axis through the center perpendicular to the plane of the ring has both symmetries. So does a disk.

•A rod rotating about one end in the plane perpendicular to ω~ has mirror symmetry in the plane but not mirror symmetry across the axis of rotation.

•A hollow or filled sphere have both symmetries.

•A disk around an axis o to the side (evaluated using the parallel axis theorem) has the plane symmetry.

•A disk around an axis that lies in the plane of the disk with a pivot in the perpendicular plane through the center of the disk has at least the planar symmetry relative to the perpendicular plane of rotation.

and so on. Nearly all of the problems we consider in this course will be su ciently symmetric that we can use:

L = Iω

(578)

with the pivot and direction of the rotation and the symmetry of the object with respect to the axis and/or plane of rotation “understood”.

Let us take a quick tour, then, of the angular momentum we expect in these cases. A handful of examples should su ce, where I will try to indicate the correct direction as well as show the “understood” scalar result.

Example 6.3.1: Angular Momentum of a Point Mass Moving in a Circle

For a point mass moving in a circle of radius r in the x-y plane, we have the planar symmetry. ω~ = ωzˆ is in the z-direction, and I = Izz = mr2. The angular momentum in this direction is:

v2
L = Lz = (~r × p~)z = mvr = mr2 r2	= Iω	(579)

The direction of this angular momentum is most easily found by using a variant of the right hand rule. Let the fingers of your right hand curl around the axis of rotation in the direction of the motion of the mass. Then your thumb points out the direction. You should verify that this gives

~

the same result as using L = ~r ×p~, always, but this “grasp the axis” rule is much easier and faster to use, just grab the axis with your fingers curled in the direction of rotation and your thumb has got it.

Example 6.3.2: Angular Momentum of a Rod Swinging in a Circle

To compute the angular momentum of a rod rotating in a plane around a pivot through one end, we choose coordinates such that the rod is in the x-y plane, rotating around z, and has mass M

116Just FYI, in case you care: The correct rule for computing ~ from is

L ω~

X

LA = IAB ωB

B

for a, b = x, y, z.

284

+y

dL z

Week 6: Vector Torque and Angular Momentum

v	M
	L
	dm
	dr
r	+x

Figure 81: The geometry of a rod of mass M and length L, rotating around a pivot through the end in the x-y plane.

and length l (note that it is now tricky to call its length L as that’s also the symbol for angular momentum, sigh). From the previous example, each little “point-like” bit of mass in the rod dm has an angular momentum of:

dLz = |~r × dp~| = r(dm v) =	¡r2dm¢	³ r ´		= dI ω	(580)
			v

so that if we integrate this as usual from 0 to l, we get:

~	1	2
|L| = Lz =	3	M l	ω = Iω	(581)

Example 6.3.3: Angular Momentum of a Rotating Disk

Suppose a disk is rotating around its center of mass in the x-y plane of the disk. Then using exactly the same argument as before:

L = Lz = Z	r2dmω = Iω = 2 M R2		ω	(582)
	1

The disk is symmetric, so if we should be rotating it like a spinning coin or poker chip around (say) the x axis, we can also find (using the perpendicular axis theorem to find Ix):

	L = Lx = Ixω =	1	M R2	ω	(583)
		4

and you begin to see why the direction labels are necessary. A disk has a di erent scalar moment of inertia about di erent axes through the same pivot point. Even when the symmetry is obvious, we may still need to label the result or risk confusing the previous two results!

We’re not done! If we attach the disk to a massless string and swing it around the z axis at a distance ℓ from the center of mass, we can use the parallel axis theorem and find that:

	L = Lznew = Iznew(M ℓ2 +	1	M R2)ω	(584)
		2

That’s three results for a single object, and of course we can apply the parallel axis theorem to the x-rotation or y-rotation as well! The L = Iω result works for all of these cases, but the direction

~

of L and ω~ as well as the value of the scalar moment of inertia I used will vary from case to case, so you may want to carefully label things just to avoid making mistakes!