Transcript Lecture 11

Chapter 11: Angular Momentum
Sect. 11.1: Vector Product & Torque
Recall Ch. 7:
Scalar Product of Two Vectors
• If A & B are vectors, their Scalar Product is defined as:
AB ≡ AB cosθ
• In terms of vector components & unit vectors i,j,k are
along the x,y,z axes:
A = Axi + Ayj + Azk B = Bxi + Byj + Bzk
• Using ii = jj = kk = 1, ij= ik = jk = 0 gives
AB = AxBx + AyBy + AzBz
• Dot Product clearly a SCALAR.
• Another kind of product of 2 vectors, useful in physics is called
the Vector Product or Cross Product.
• Look closely at the relationship between a torque τ & the force F
which produces it.
Ch. 10: We saw: a force F acting on
a body at position r produces a
torque with magnitude:   rF sin 
p
Figure: F causes a torque τ that rotates
the object about an axis perpendicular
to BOTH r AND F. Mathematicians
have taught us that the torque τ is a
VECTOR in the direction of the axis
of rotation & can be written
  Fr
Vector Product Definition
• If A & B are vectors, their Vector (Cross)
Product is defined as:
A third vector
C  A B
• C is read as “A cross B”
• The magnitude of vector C is AB sinθ
where θ is the angle between A & B
Vector Product
• The magnitude of C, which is AB sinθ is equal to
the area of the parallelogram formed by A and B.
• The direction of C is perpendicular to the plane
formed by A and B
• The best way to
determine this
direction is to use
the right-hand rule
Vector Product Properties
• The vector product is not commutative! Unlike scalars, the
order in which the vectors are multiplied is important
By the way it’s defined,
• If A is parallel to B (θ = 0o or 180o), then
A  B  B  A
A B  0 A  A  0
• If A is perpendicular to B, then
A  B  AB
• The vector product is distributive:
A x (B + C) = A x B + A x C
Vector Product Derivative Properties
• The derivative of the cross product with respect to
some variable, such as t, obeys the “chain rule” of
calculus:


d
dA
dB
A B 
B  A 
dt
dt
dt
Note! It is important to preserve the multiplicative
order of A and B
Vector Products of Unit Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
Contrast with scalar products of unit vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  1
ˆi  ˆj  ˆi  kˆ  ˆj  kˆ  0
Signs are interchangeable in cross products
 
A  -B   A  B
 
ˆi   ˆj  ˆi  ˆj
Vector Products Using Determinants
• The cross product can be expressed as
ˆi
A  B  Ax
Bx
ˆj
Ay
By
kˆ
Ay
Az 
By
Bz
Az
ˆi  Ax
Bz
Bx
Az ˆ Ax
j
Bx
Bz
Ay
kˆ
By
• Expanding the determinants gives
A  B   Ay Bz  Az By  ˆi   Ax Bz  Az Bx  ˆj   Ax By  Ay Bx  kˆ
Example 11.1
A  2ˆi  3ˆj; B  ˆi  2ˆj
• Given
• Find
• Result is
A B
A  B  (2ˆi  3ˆj)  ( ˆi  2ˆj)
 2ˆi  ( ˆi )  2ˆi  2ˆj  3ˆj  ( ˆi )  3ˆj  2ˆj
 0  4kˆ  3kˆ  0  7kˆ
Example 11.2: Torque Vector
• Given the force F and position r:
F  (2.00 ˆi  3.00 ˆj) N
r  (4.00 ˆi  5.00 ˆj) m
• Find the torque τ produced
• Result is
  r  F  [(4.00ˆi  5.00ˆj)N]  [(2.00ˆi  3.00ˆj)m]
 [(4.00)(2.00)ˆi  ˆi  (4.00)(3.00)ˆi  ˆj
(5.00)(2.00)ˆj  ˆi  (5.00)(3.00)ˆi  ˆj
 2.0 kˆ N  m
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: v = rω,
θ
ω
α
τ
I
∑τ = Iα
(½)Iω2
τθ
?
at= rα
ac = (v2/r) = ω2r , τ = dF , I = ∑(mr2)
Sect. 11.2: Angular Momentum
Consider a rigid, vertical pole
through a frozen ice pond, as in the
figure. As a skater moves in a
straight line past it, she grabs &
holds the pole & suddenly she is
going in a circle around the pole. We
can analyze this motion by using the
concept of
Angular Momentum
Angular Momentum
• Consider a particle of mass m located at the vector position
r and moving with linear momentum p. See figure 
• Find the net torque
r   F    r 
dp
dt
dr
Add the term
 p  sinceit  0 
dt
d (r  p )



dt
• The instantaneous angular momentum L of a particle
relative to the origin O is defined as the cross product of
the particle’s instantaneous position vector r and its
instantaneous linear momentum p:
L  r p
Torque and Angular Momentum
• Recall Newton’s 2nd Law in Momentum Form: ∑F = (dp/dt)
• The net torque is related to the angular momentum in a
manner similar to the way that the net force is related to
the linear momentum. That is:
dL
  dt
• The net torque acting on a particle is equal to the time rate
of change of the particle’s angular momentum
• This is the most general form of rotational analog of
Newton’s 2nd Law or Newton’s 2nd Law for Rotations.
∑τ & L must be measured about the same origin.
Can show that this reduces to ∑τ = Iα if I is time independent.
Angular Momentum
• SI units of angular momentum are (kg.m2)/s
• Both the magnitude and direction of the angular
momentum depend on the choice of origin
• The magnitude is L = mvr sin
 is the angle between p and r
• The direction of L is
perpendicular to the plane
formed by r and p.
Ex. 11.3: Angular Momentum of a Particle in Circular Motion
• Particle of mass m moving in a
circular path of radius r.
• The vector L = r  p is pointed
out of the diagram
• The magnitude is
L = mvr sin 90o = mvr
sin 90o is used since v is
perpendicular to r
• A particle in uniform circular
motion has a constant angular
momentum about an axis
through the center of its path
Angular Momentum of a Particle System
• Angular Momentum of a System of Particles = vector sum of
the angular momenta of the each particle:
Ltot  L1  L 2 
 Ln   Li
i
• Differentiating with respect to time gives Newton’s 2nd Law for
Rotations in a many particle system: dLtot
dLi
dt

i
dt
  i
i
• System has internal forces & external forces. Can show: All torques
coming from internal forces add to zero. So, ∑τi becomes ∑τext
where now the sum includes external torques only.
• Therefore, Newton’s 2nd Law for Rotations in a many particle
system becomes:
dL tot


 ext dt
– The net external torque acting on a system about some axis passing through an
origin in an inertial frame equals the time rate of change of the total angular
momentum of the system about that origin
Example 11.4: A System of Objects
• A sphere, mass m1 & a block, mass m2 are
connected by a light cord passing over a pulley
which is a thin ring of radius R & mass M. Block
slides on a flat, frictionless surface. Find the
acceleration a of the sphere & the block using
angular momentum & torque methods.
• Angular momentum about pulley rotation axis: Pulley rotates, while other 2
objects translate. At time where m1 & m2, moving together, have speed v,
angular momentum of m1 is m1vR & that of m2 is m2vR. At that same time,
angular momentum of pulley is MvR.
Total angular momentum: Ltot = m1vR + m2vR + MvR
• External torque comes solely from weight of m1, m1g. So,
dL tot
∑τext = m1gR.
Newton’s 2nd Law:

ext

dt
 m1gR = d(m1vR + m2vR + MvR)/dt = (m1+ m2 + M)R(dv/dt)
or m1g = (m1+ m2 + M)a. So,
a = (m1g)/(m1+ m2 + M)