Download Lecture 11a

Sect. 11.3: Angular Momentum Rotating Rigid Object
Translation-Rotation Analogues & Connections
Translation
Rotation
Displacement
x
Velocity
v
Acceleration
a
Force (Torque)
F
Mass (moment of inertia) m
Newton’s 2nd Law
∑F = ma
Kinetic Energy (KE) (½)mv2
Work (constant F,τ)
Fd
Momentum
mv
CONNECTIONS:
θ
ω
α
τ
I
∑τ = Iα
(½)Iω2
τθ
?
s = rθ, v = rω, at= rα
ac = (v2/r) = ω2r , τ = Fdsin, I = ∑(mr2)
• Recall: Angular Momentum L of point mass m at
position r moving with momentum p = mv (see figure) the
cross product:
L =r p
• Units: (kg.m2)/s
• Magnitude: L = mvr sin
 = angle between p & r
• Direction: Perpendicular to
the plane formed by r & p.
 If r & p are in the xy-plane, L is in the z-direction.
• Special Case: Particle of mass m
moving with velocity v in a circular path
of radius r (see figure). Momentum p = mv.
Note also that for this case, v = rω, where
ω = angular velocity of m around the
center of the circle.
• Angular Momentum Magnitude
L = mvr sin(90o) = mvr = mr2ω
v is perpendicular to r  angle between
p & r is  = 90º  sin = sin(90o) = 1
• Angular Momentum Direction
Vector L = r  p is pointed out of the
diagram (towards the reader).
• So, a particle in uniform circular motion has a
constant angular momentum L = mvr = mr2ω
about an axis through the center of its path.
L =r p
• Generalize to a Rigid Object
rotating about an axis passing
through it’s center of mass (see
figure). Each particle in the
object rotates in the xy plane
about the z axis with angular
speed w. Angular momentum
of an individual particle is
Li = mi ri2w.
The vectors L & w are
directed along the z axis.
• To find the total angular momentum
L of the object, sum up the angular
momenta Li of the individual
particles:
(1) Lz =  Li =  mi ri 2 w = Iw

i

i
From Ch. 10, I  ∑miri2 is the
object’s moment of inertia.
Combining (1) with Newton’s 2nd
Law for Rotations:
dL tot

=
 ext dt
Gives:
Newton’s 2nd Law for Rotations:
(for rigid objects)
α = angular acceleration of the object

ext
dLz
dw
=
=I
= I
dt
dt
Example 11.5: Bowling Ball
• A bowling ball (assume a perfect sphere,
mass M = 7 kg, radius R = 0.12 m), spins at
frequency f = 10 rev/s
 ω = 2πf = 20π rad/s
From table (Ch. 10) moment of inertia is
I = (2/5)MR2
• Calculate the magnitude of the angular
momentum, Lz = Iω
• Direction of L is clearly the +z direction.
Lz = Iω = (2/5)MR2ω = (2/5)(7)(.12)2(20π ) = 2.53 kg m2/s
Example 11.6: Seesaw
• A father, mass mf & his daughter, mass md sit on
opposite sides of a seesaw at equal distances from
the pivot point at the center. Assume seesaw is a
perfect rod, mass M & length l. At some time,
their angular speed is ω.
• Table (Ch. 10) gives moment of inertia of seesaw:
Is = (1/12)Ml2. Moments of inertia of the father
& daughter about pivot point. Treat them like
point masses: If = mf[(½)l]2, Id = md[(½)l]2
(A) Find an expression for the magnitude of the angular momentum
(about pivot point) at time when angular speed is ω. Note: Total moment of
inertia is: I = Is + If + Id = (1/12)Ml2 + (¼)mfl2 + (¼)mdl2

The total angular momentum is:
L = Iω = (¼)l2[(⅓)M + mf + md]ω
Example 11.6: Seesaw continued
(B) Find an expression for the magnitude of the angular
acceleration when the seesaw makes an angle θ with the
horizontal. Newton’s 2nd Law for Rotations: ∑τext = Iα
∑τext = [(1/12)Ml2 + (¼)mfl2 + (¼)mdl2]α
Left side:
(1)
mdgcosθ


 mfgcosθ
∑τext = τs + τf + τd
Torque definition: τ = rFsin = Fd;  = angle between force F & vector r
from point of application to pivot point. Moment arm d = rsin
Seesaw: τs = Mgds; Mg passes through pivot point. Moment arm ds = 0  τs = 0
Father: τf = wfdf; wf =  component of father’s weight: wf = mfgcosθ
df = (½)lsin;  = 90º, sin = 1,
 df = (½)l

τf = (½)lmfgcosθ (counterclockwise, positive torque!)
Daughter: τd = wddd; wd =  component of daughter’s weight:
wg = mfgcosθ; dd = (½)lsin;  = 90º, sin = 1,  dd = (½)l

τd = - (½)lmfgcosθ (clockwise, negative torque!)
So, from (1), α = [∑τext]/I = [2(mf – md)cosθ](l)[(⅓)M +mf + md]