Chapter 11:  Angular Momentum
11.1: The Vector Product and Torque: vector product, cross product 11.2: Angular Momentum 11.3: Angular Momentum of a Rotating Rigid Object							11.4: Conservation of Angular Momentum 11.5: The Motion of Gyroscopes and Tops:  precessional motion & frequency 11.6: Angular Momentum as a Fundamental Quantity
11.1 Vector Product & Torque
Right Hand rule The vector product A x B is a third vector C having a magnitude AB sinθ equal to the area of the parallelogram described by AxB perpendicular the plane of A x B.										The direction of the Torque will be centered along the axis of rotation (the +z axis). Of course the rod will accelerate in the general direction of the applied force.
Example of Cross Product Lets say that a 3 meter rod is oriented in the x-y plane as seen to the right (pointing to the forward and right position so that, r = 2i + 3j + 0k).  A 5 Newton force is applied in the same plane, but with only a small component in the x-plane, F = 1i + 4j + 0k.  What is the resultant torque where t = r x F?
	i  (+)	j  (-)	k  (+)		t  =  +( 3*0 - 4*0 ) i   - ( 2*0 - 1*0 ) j   + (2*4 - 1*3 ) k       t  =  0 i  - 0j  + 5 k       t  =  5 k   (positive means counter clockwise)
r	2	3	0
F	1	4	0
Some Properties of Vector Products 1. A x B = -B x A 2. If A is || to B then A x B = 0 (AxA=0 also) 3. If A is ┴ to B then |A x B| = AB 4. A x (B + C) = A X B + A X C 5. d (A x B) / dt     = A x dB/dt   +   dA/dt x B
11.2 Angular Momentum of a Particle
Torque = r x F          or t = Fd = F sinφ r                      Angular momentum, L, is defined as r x p r x p  à  off centered momentum will cause rotation L = r x p           (both r & p are vectors) L = r x (mv) L = mv r sinφ F Δt = Δp         à     Impulse Δp = F Δt dp = F dt           Thus net force on a particle is the time rate of           change of the linear momentum                 ΣF = dp / dt Σt = r x   ΣF   Σt = r x dp/dt								Σt = r x dp/dt           Let's add ZERO to the above equation Σt = r x  dp / dt    +   0 Let 0 = dr/dt x p     dr/dt = v   ll   p Σt =  r x dp/dt   +   dr/dt x p d (A x B)/dt    = A x dB/dt   +   dA/dt x B apply Rule 5 which yields Σt = d(r x p) / dt Σt dt = d(r x p)                by def à  L = r x p Σt dt = dL Σt = dL / dt
Demo:  Fallling Bottle and keys:  ME-Q-FB
11.3 Angular Momentum of a Rotating Rigid Object
L        = r x p L        = r x (mv)       v = ωr Li        = miri^2ω          I = mr^2 L        = I     ω dL/dt = I dω/dt     where I is constant for a rigid object   Σt   = I     α								If you recall, we introduced this formula previously (Ch 10) and related it to Newton’s 2^nd law, we now can see the proof.  F  = m  a Σt = I  α
Example The angular momentum of a bowling ball spinning at 60 rpm is about ____.								mbowlingball ≈ 4 kg               rbowlingball ≈ 10 cm Lz = Iω                          Table 10.2 à I = 2/5 MR^2 Lz = 2/5 MR^2 ω         60 rpm = 1 rev / sec Lz = 2/5(4)(0.1)^2 (1 rev/sec)(2π / rev) Lz = 0.10 Nm^2
11.4 Conservation of Angular Momentum
Σtext = dLtot / dt Σtext = 0 Ltot = constant   so  Linit = Lfinal					Demo:  Euler’s Disk:  ME-M-ED
Example of exploding galaxy What was the angular velocity of our galaxy when it was (a) 10^14 meters in diameter (b) 10^10 m in diameter? Need to look up the following information online 10^11 stars in a galaxy mstar = 2 x 10^30 kg rgalaxy = 50,000 light years = 5x10^20 m Tgalaxy = 2x10^8 years = 6x10^15 sec v = wr = 2pr/T          à wf = 2p/T						I              ω      =      If             ωf         ½mri^2        ω      = ½mrf^2            2p/T (a)              assuming a disk         ½m(5x10^13)^2 ω = ½m(5x10^20)^2   2p/6x10^15         ω = (5x10^20)^2/(5x10^13)^2   * 2p/6x10^15         ω = 10^14  * 2p/6x10^15         ω = 0.1 rad / sec (b)         ω = (5x10^20)^2/(5x10^9)^2   * 2p/6x10^15         ω = 10^7 rad / sec
Example 11.9 The spinning bicycle wheel Use the right hand rule... L = r x p The wheel is spinning in the horizontal plane, so L = Linitial Rotate the wheel through its center by 180°.  Now inverted wheel has Linvert = - Linitial
No external torques have been applied (only one torque internal to the system) so Angular momentum should be conserved.						The wheel started at +Linit to the inverted -Linit. The student on the stool (the only other thing internal to the system) went from 0 to 2 Li (to conserve angular momentum).
Lecture Example During lecture I spin three times around on the platform in 3 seconds.  In each arm is a 2 kg mass.  These are brought in to the center of my body. Then I extend my arms in to 1 meters…and now 3 rotations take about 6 seconds.  What is my inertia?				1^st point:  calc the inertia at the center of rotation and when extended to 1 meter.  Remember I’ve two arms!!!							At the center: I = mr^2 = (2(0.02m)^2 ) Icenter = 0.0008 kg m^2 Extended: I = mr^2 = 2(1)^2 Iextend = 2 kg m^2
				Given: wf = ½w (same angular distance in half the time)         I              w      =     If wf (I+ Icenter)         w      = (I+ Iextend) ½w (I+ 2(0.0008))  w      = (I+ 2(2))   ½w 2I                            ≈ (I+ 4) I ≈ 4 kg m^2
Demo:  Angular Momentum Platform:  ME-Q-AM
11.5 Gyroscopes
As a Gyroscope is spinning it wobbles (precesses) at a certain frequency, ω. The weight vector (mg) of the gyroscope is off centered (this perpendicular distance is the lever arm, r) from the pivot point so that, t = r x mg The Torque produces a change in dL in the direction of the Torque, (t = dL / dt).   Just like t, dL must also be perpendicular to L. Since dL is perpendicular to L, we know that dL is independent of L Note: dL & t only change in the x-y plane. The z component is constant during the rotation										The gyroscope is precessing about the pivot point with an angular velocity, ω, so that L = Iω. From the above diagram sinφ          = ΔL         / L            (φ << 10°) dφ            = dL          / L dφ            = dt   dt   / L dφ            = mgr^ dt         / L dφ/dt       = mgr^      / I ω d ω           = mgr^      / I ω ωprec = mgr^ / I ω
Demo:  Bicycle Wheel, Gyroscope:  ME-Q-GB Demo: Gyroscope:  ME-Q-GY Demo: Gyroscopic Stabilizer:  ME-Q-GS
11.6 Angular Momentum as a Fundamental Quantity
L          = ħ              ħ = 1.0546 x 10^-34 J s ICMω     = ħ           ħ  = h / 2π ω = ħ /ICM
Example What is the angular velocity of a N2 molecule? mN-atom = 14 amu                mamu = 1.66 x 10^-27 kg dN-molecule = 3.4 Angstroms							ω = ħ      /       ICM ω = ħ      / (mr^2 + mr^2)    (two nitrogen atoms) ω = ħ      / 2         m                   r^2 ω = 10^-34 / 2 (14)( 1.66x10^-27)  (1.7 x 10^-10)^2 ω = 7.7 x 10^10 rad / sec