Download TORQUE AND ANGULAR MOMENTUM 73. (11.3) Angular

TORQUE AND ANGULAR MOMENTUM
73. (11.3) Angular momentum of a particle.
v
l
v
r
The angular momentum of a particle,
v about the reference point O, is defined
p
v as the vector product of the position,
p
relative to the reference point, and
momentum of the particle
r r r
l ≡ r×p
O
74. (11.2, 10.6) Torque
v
F
v
r
r
τ
Torque τ , about the reference point O,
due to a force F exerted on a particle,
is defined as the vector product of the
position relative to the reference point
and force
r r r
τ ≡ r ×F
81
75. (11.3) Newton's second law for angular momentum.
a) The net torque τ , exerted on a particle, is equal to the
rate of change of its angular momentum l.
r
d l = τr
net
dt
proof
r
r
r
r
d l d r r dr r r dp r dp r r
r
= (r × p ) = × p + r ×
= r×
= r × Fnet = τ net
dt dt
dt
dt
dt
b) The net external torque τ , exerted on a system of
particles, is equal to the rate of change of its total angular
momentum L.
r
dL = τr
ext
dt
proof
r
r
dL d r
dl
r
r
r
r
r
= ∑ li = ∑ i = ∑ τ net,i = ∑ τint, i + ∑ τ ext ,i = ∑ τ ext,i = τ ext
dt dt i
i dt
i
i
i
i
r
τ ji
v
Fji
v
rj
v
p
v
ri
v
p
r
τ ij
82
v
Fij
76. (11.5) Conservation of angular momentum.
If the net external torque exerted on a system is a zero
vector, the total angular momentum of the system is
constant.
r
L( t ) = const .
proof.
r
dL = τr = 0r
ext
dt
Example. (14.5) Kepler's second law.
The gravitational torque (about the sun) exerted by the
sun on the planets is a zero vector. According to Newton's
v
second law, the angular
L
momentum (about the sun)
of the planets is conserved.
v dA
r
v
dr
r
r
dA = 1 r × dr = 1 rr × mvr = L = const .
dt 2 dt
2m
2m
The radius vector from the sun to any planet sweeps
out equal areas in equal times.
83
Example. What is the final angular velocity?
initial
v
ωa
Initial total angular momentum
a
La = 2 ⋅ a ⋅ mv a = 2 a 2mωa
Final total angular momentum
final
v
ωb
Lb = 2 ⋅ b ⋅ mv b = 2b2mω b
b
If the system is isolated, the total angular momentum
must be conserved. Therefore
2
a
ω b = 2 ωa
b
Puzzle. Who performed the work?
∆K tot
2

m b2 ω2b
ma 2 ω2a
a
2
= 2⋅
−2⋅
= m a  2 − 1ω2a > 0
2
2
b

84
KINETICS AND DYNAMICS OF A RIGID BODY
77. (10.intro) Rigid body
v
vi
v
rAi
A
v
ω
A system in which the relative
position of all particles is time
independent is called a rigid body.
All the particles have the same
angular velocity.
r
r
r r
vi = v A + ω × rAi
The motion can be considered as a superposition of the
translational motion of point A and the rotational motion
around point A. In practice, the center of mass or a point
on an instantaneous axis of rotations is chosen for the
description of translational motion.
Example
85
78. (11.4,) Relation between angular velocity and angular
momentum about a stationary point O.
In general, each component of the total angular
momentum depends on all the components of the angular
velocity.
Li = ∑ I ijω j
j
(The direction of the total angular momentum can be
different than the direction of the angular velocity.)
Coefficients Iij are called the components of rotational
inertia (moment of inertia) of the body about the reference
point.
v
ω
proof.
v
For simplicity's sake, let's choose the
ri '
v
z-axis along the angular velocity.
ri
ω = 0, 0, ω
We can express the total angular
momentum in terms of the z-component
of the angular velocity
r
r
r
r
r r
r rr r
L = ∑ ri × mi v i = ∑ ri × mi (ω × ri ) = ∑ mi ri2ω − (riω)ri
i
i
i
(
)
In scalar components
(
)
(
L z = ∑ m i ri2ω − z iω ⋅ zi =  ∑ m i ri2 − z2i
i
i
L x =  − ∑ m iz i x i ω
 i

)ω =  ∑ m r' ω
i
2
i i
L y =  − ∑ mi zi yi  ω
 i

;
86
78. All objects which are
symmetrical with respect to the
axis of rotation have the (total)
angular momentum parallel to the
angular velocity.
v
L
v
ω
v
L
v
ω
r
r
L = I ⋅ω
where the proportionality coefficient I = ∑ m i r '2i is called
i
the moment of inertia (rotational inertia) of the body about
the axis of rotation.
proof.
L x =  − ∑ m iz i x i ω = 0 ;
 i

L y =  − ∑ mi zi yi  ω = 0
 i

79. For symmetrical rigid bodies, the angular acceleration
is proportional to the net external torque. The
proportionality coefficient is equal to the rotational inertia
of the body about the axis of rotation.
r r
I ⋅ α = τext
proof.
r
r
r
dω dL r
I⋅α = I⋅
=
= τext
dt
dt
87
80. (10.7, 11.4) Rotation about a fixed or instantaneous
axis.
v
The angular acceleration, of an
ω
v object rotating about a fixed axis or
v
F
τω
instantenous, is proportional to the
v
component, along the axis of
τ
rotation, of the net external torque.
I ⋅ α = τ ext , ω
proof.
I⋅α = I⋅
dω dL ω
=
= τext , ω
dt
dt
81. (10.5) Moment of inertia (about an axis)
For a rigid body (system of particles),
the quantity
I A = ∑ mi r '2i
A
i
is called the moment of inertia about
axis A.
In the case of a continuous body,
calculating the rotational inertia requires
integration.
I A = ∫ r '2 dm
body
88
r’
mi
Example. Moment of inertia of a thin rod.
y
dx
x
L
z
L
M x3
1
2M
IA = ∫ x
dx = ⋅
= ML2
L
L 3 0 3
0
L
Example. Moment of inertia of a uniform circle.
dr
dm
dϕ
M
M R 3  2π 
M R4 1
I A = ∫ r dm = ∫ ∫ r ⋅ 2 rdϕdr =
r  ∫ dϕ dr = 2π ⋅ 2 ⋅
= MR 2
2 ∫
4 2
πR
πR 0  0 
πR
circle
0 0
2
R 2π
2
89
82. (10.5) Parallel - axis theorem
If the moment of inertia of a rigid body about an axis
through the center of mass is Ic , then the moment of inertia
about a parallel axis separated by distance D away from the
axis that passes through the center of mass is given by
I A = I C + MD 2
proof.
C
A
r r 2
I A = ∫ r dm = ∫ (D + r') dm =
2
body
v
D
v
r
Example.
body
r r
= ∫ D dm + ∫ r ' dm + 2 ∫ D ⋅ r ' dm =
2
v
rc '
body
mi
2
body
r r
= MD 2 + I C + 2 D ⋅ 0
A
body
C
2
1
1
L  1 3 
I A = ML2 + M  =  +  ML2 = ML2
12
3
 2   12 12 
90
83. (9.6,12.2) Center of a force.
If a certain body exerts a force on several particles of
a given system, the center of the force is defined by
r
position rcf such that for any point of reference
r
r
r r
rcf × ∑ f i = ∑ ri × f i
(
i
i
)
(In other words, the center of a force is such a point that
the effect of the total torque about any given point, exerted
on the system by another body, is as if the total force acted
on the center of the force.)
Example. Center of gravity.
v
ri
v
Wi
r
r
r
r
r
r r
r
r
τ = ∑ ri × mig =  ∑ miri  × g = Mrcm × g = rcm × Mg
i

i
The effect of gravitational torque is as if the entire weight
of the system were at the center of mass. The center of
gravity in a uniform gravitational field is at the center of
mass.
91
84. (10.4) Rotational kinetic energy of a rigid body
a) The total kinetic energy of
a system rotating about the
point of reference is called
the
rotational
(kinetic)
energy.
v
ω
Kω, o ≡ ∑ K o,i
i
(Rotational kinetic energy
depends on the choice of the reference point A for
consideration of the rotational motion.)
b) The rotational kinetic energy is related to the angular
speed and the moment of inertia of the body.
1
K ω,o = Iω2
2
proof.
1
1
1
1
2
K ω,o = ∑ K o,i = ∑ mi v 2i = ∑ m i (ωr 'i ) =  ∑ mi r 'i2 ω2 = Iω2
2 i
2

i
i 2
i 2
92
85. (11.1) Total kinetic energy of a rigid body. (theorem)
The total kinetic energy of a rigid body is equal to the
sum of its translational kinetic energy and its rotational
kinetic energy about the center of mass
K tot = K T + K ω,cm
proof.
1
1
r
r r
K tot = ∑ K i = ∑ m i v 2i = ∑ m i ( v cm + ω × ri )2 =
i
i 2
i 2
1
1
r r
1
r
r r
= ∑ m i v 2cm + ∑ m i (ω × ri )2 + ∑ m i ⋅ 2v cm ⋅ (ω × ri ) =
i 2
i 2
i 2
1
1
r
r
r
2
=  ∑ m i v cm
+  ∑ m i ri2 ω2 + v cm ⋅  ω × ∑ m i ri  =
2 i
2 i




i
= K T + K ω ,cm + 0
93
Example.
A uniform ball is
released from height h along a
ramp. (a) What is the speed of the
rolling ball at the bottom of the
ramp? (b) What is the magnitude
of the friction force? (c) What is
the magnitude of acceleration of
the ball?
v
N
v
fs
M = 100g
v
W
h = 1m
α = 30°
(a) Consider the change in the total kinetic energy of the ball
∆K tot = K tot ,f − K tot ,i = (K T, f + K ω,f ) − (0 + 0) =
2
1
12
7
 v 
2
= Mv 2cm +  MR 2  cm  = Mv cm
2
2 5
 R  10
There are three forces exerted on the ball: gravitational,
frictional and normal. Only gravitational force performs work on the
particles of the system. (Not on the center of mass!). The work done
by the gravitational force is related to the displacement of the center
of mass.
r r r
r
r r
Wg = ∑ m ig ⋅ ∆ri = g ⋅ ∑ m i ∆ri = Mg ⋅ ∆ rcm = Mgh
i
i
From the work-energy theorem for the total kinetic energy we
can find the final speed.
vcm
10Wg
10∆K tot
10Wtot
10
m
=
=
=
=
gh = 3.7
7M
7M
7M
7
s
94
(b) Consider the work-energy theorem for the translational kinetic
energy. The change in the translational kinetic energy is
1
5
2
∆K T = Mv cm
− 0 = Mgh
2
7
There are two forces performing work on the center of mass,
gravitational force and frictional force. (Compare this with the work
done on the particles.) Since both forces are constant we can perform
the integration in our head.
Wcm = Mgh −
fs ⋅ h
sin α
According to the work-energy theorem
∆K T = Wcm
Therefore the frictional force has a magnitude of
fs =
sin α
(Mgh − Wcm ) = sin α  Mgh − 5 Mgh  = 2 Mg sin α = 0.14 N
h
h 
7
 7
95
(c) We can find the magnitude of the acceleration from Newton's
second law for translational motion.
2
r
Mg sin α − Mg sin α
r
F
5
m
7
a cm = acm = net =
= g sin α = 3.5 2
M
M
7
s
(c’ - Alternative solution) We can find the acceleration considering
Newton's second law for rotational motion of a rigid body.
Consider torque about the instantaneous axis.
τ z = 0 + 0 − MgR sin α
The moment of inertia about this axis can be found from the parallel
axis theorem
2
I = MR 2 + MR 2
5
The angular acceleration is related to the z-component of angular
momentum which, according to Newton's second law for rotational
motion, depends on the external torque
1 dL
1
− MgR sin α
5 g
αz = ⋅ z = ⋅ τz =
=−
sin α
7
I dt
I
2
7
R
MR
5
The magnitude of the acceleration of the center of the sphere is
therefore
5
a cm = αR = g sin α
7
96
86. (10.8) Work and power in rotational motion.
v
dr
v
F
r
dθ
v
ω
v
τ
a) The differential work in a rotational motion depends on
the torque about the point of rotation
r r
dW = τ ⋅ dθ
proof.
r r r r r
r r r r r
dW = F ⋅ dr = F ⋅ (dθ × r ) = dθ ⋅ (r × F) = τ ⋅ dθ
b) In rotational motion, the power of the force depends on
the torque of the force and the angular velocity of the point
where the force is applied
r r
P = τ⋅ω
proof.
r
r
r r
P = dW = τ ⋅ dθ = τ ⋅ ω
dt
dt
97
v
Fi
87. Transformation of torque
v
τ A, i
v
rA , i
v
rB,i
v
τB,i
A
B
The torque exerted by one system on another about
two different reference points, depends on the net force and
the relative position of the reference points.
→
r
r
r
τA = AB× Fnet + τB
proof.
→
→
r
r
r
r
r
r
r
r
 → r  r
τ A = ∑ rA ,i × Fi = ∑  AB+ rB, i  × Fi = AB× ∑ Fi + ∑ rB,i × Fi = AB× Fnet + τ B
i
i 
i
i

Comment. If the net force applied to a body is zero, the
torque of this force about any point has the same value.
88. (12.intro, 12.1) Equilibrium of a rigid body.
A rigid object is in equilibrium, if and only if the following
conditions are satisfied: (a) the net external force is a zero
vector and (b) the net external torque is a zero vector.
r
r
F
=
0
∑ ext
and
98
r
r
τ
=
0
∑ ext
Example. (Civil engineering problem) Determine the internal
torque distribution in the beam.
y
L
v
F2
d
A
O
τz
x
x
v
F1
L
d
x
Consider the part on the beam to the right of point A. It is in
equilibrium, therefore the internal torque balances the all external
torques applied to the beam. Hence the problem requires to find the
(z-component of) torque about point A of the (external) forces to the
right of point A.
Between point O and the point where force F 1 is applied, there
are two forces to the right. The z-component of the torque of these
two forces is
τz(x) = F1⋅(d − x) – F2⋅(L − x) = (F2 − F1)⋅x + (F1d − F2L) = (F2 − F1)⋅x
(The second term has a zero value because the external torque about
the origin must be zero.)
The torque about points to the right of force F 1 depends only
on the force F 2.
τz(x) = − F2 ⋅ (L − x)
99
ELASTIC PROPERTIES OF MATTER
89. (15.intro) States of matter
Matter can exist in one of the following four states
a) Solid state - The atoms forming the
solid maintain their relative average
position. In crystalline solids, the atoms
show long range order - crystals have a
periodic structure. In amorphous solids
there is no long range order.
b) Liquid state - The atoms or
molecules change their arrangement
constantly but there is strong
interaction between them.
c) Gaseous state - The atoms or
molecules change their arrangement
constantly and the interaction between
the particles is very weak.
d) Plasma state - The substance is a collection of
electrons and positively charged ions.
Comment. The last three forms of matter are called fluids.
100