Transcript Chapters 10&11 - Texas Christian University
Chapters 10, 11 Rotation and angular momentum
Rotation of a rigid body
•
We consider rotational motion of a rigid body a fixed axis about
•
Rigid body rotates with all its parts locked together and without any change in its shape
•
Fixed axis : it does not move during the rotation
•
This axis is called axis of rotation
•
Reference line is introduced
Angular position
•
Reference line is fixed in the body, is perpendicular to the rotation axis, intersects the rotation axis, and rotates with the body
•
Angular position – the angle (in radians or degrees) of the reference line relative to a fixed direction ( zero angular position )
Angular displacement
•
Angular displacement position. – the change in angular
•
Angular displacement is considered CCW positive in the direction and holds for the rigid body as a whole and every part within that body
f
i
Angular velocity
•
Average angular velocity
avg
t f f
i
t i
•
Instantaneous angular velocity in angular position – the rate of change
t
lim
t
0
t
d
dt
Angular acceleration
•
Average angular acceleration
avg
t f f
t
i i
t
•
Instantaneous angular acceleration – the rate of change in angular velocity
lim
t
0
t
d
dt
Rotation with constant angular acceleration
•
Similarly to the case of 1D motion with a constant acceleration we can derive a set of formulas:
Relating the linear and angular variables: position
•
For a point on a reference line at a distance
r
the rotation axis: from
s
r
•
θ
is measured in radians
s
r
Relating the linear and angular variables: speed
v
ds
dt
•
ω
is measured in rad/s
d
(
r
)
dt
r d
dt
r
•
Period
T
2
r v
2
a t
Relating the linear and angular
dv dt
variables: acceleration
d
(
r
)
dt
r d
dt
r
•
α
is measured in rad/s 2
•
Centripetal acceleration
a c
v
2
r
(
r
) 2
r
2
r
Rotational kinetic energy
•
We consider a system of particles rotational motion participating in
•
Kinetic energy of this system is
•
Then
K
i m i v i
2 2
K
i m i v i
2 2
i m i
(
i r i
) 2 2 2 2
i m i
(
r i
) 2
Moment of inertia
•
From the previous slide
K
2 2
i m i
(
r i
) 2 •
Defining moment of inertia ( rotational inertia ) as
I
i m i
(
r i
) 2 •
We obtain for rotational kinetic energy
K
I
2 2
Moment of inertia: rigid body
•
For a rigid body with volume
V
and density generalize the definition of a rotational inertia:
ρ(V)
we
I
volume r
2
dV
r
2
dm
•
This integral can be calculated for different shapes and density distributions
•
For a constant density and the rotation axis going through the center of mass the rotational inertia for 8 common body shapes is given in Table 10-2 (next slide)
Moment of inertia: rigid body
Moment of inertia: rigid body
•
The rotational inertia of a rigid body depends on the position and orientation of the axis of rotation relative to the body
Chapter 10 Problem 25 Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.
Parallel-axis theorem
•
Rotational inertia of a rigid body with the rotation axis, which is perpendicular to the
xy
plane and going through point
P
:
I
r
2
dV
volume r
2
volume dm
•
Let us choose a reference frame, in which the center of mass coincides with the origin
I
r
2
dm
Parallel-axis theorem
[(
x
(
x
2
a
) 2 (
y
b
) 2 ]
dm y
2 2
a
)
dm xdm
(
a
2 2
b
b
2
ydm
)
dm
r CM
(
i
ˆ
x
i
ˆ
xdm
ˆ
j
j
ˆ
y
)
dm
/
ydm
/
M M
0
I
r
2
dm
R
Parallel-axis theorem
[(
x
(
x
2
a
) 2 (
y
b
) 2 ]
dm y
2 )
dm
(
a
2
b
2 )
dm
(
R
2 )
dm
(
h
2 )
dm
I CM
Mh
2
I
I CM
Mh
2
Parallel-axis theorem
I
I CM
Mh
2
Chapter 10 Problem 51 A uniform rectangular flat plate has mass M and dimensions a by b. Use the parallel-axis theorem in conjunction with Table 10.2 to show that its rotational inertia about the side of length b is Ma 2 /3.
Torque
•
We apply a force at point
P
to a rigid body that is free to rotate about an axis passing through
O
•
Only the tangential component
F t = F sin φ
force will be able to cause rotation of the
Torque
•
The ability to rotate will also depend on how far from the rotation axis the force is applied
•
Torque (turning action of a force):
(
F t
)(
r
) (
F
sin )(
r
) •
SI unit : N*m (don’t confuse with J)
Torque
•
Torque:
(
F t
)(
r
) (
F
sin )(
r
) (
F
)(
r
sin ) •
Moment arm :
r
┴
= r sinφ
•
Torque can be redefined as: force times moment arm
τ = F r
┴
Newton’s Second Law for rotation
•
Consider a particle rotating under the influence of a force
•
For tangential components
F t r
ma t r
m
(
r
)
r
(
mr
2 )
I
I
•
Similar derivation for rigid body
Newton’s Second Law for rotation
I
i
i
Chapter 10 Problem 57 A 2.4-kg block rests on a slope and is attached by a string of negligible mass to a solid drum of mass 0.85 kg and radius 5.0 cm, as shown in the figure. When released, the block accelerates down the slope at 1.6 m/s 2 . Find the coefficient of friction between block and slope.
Rotational work
•
Work
dW
F t ds
F t rd
d
• •
Power
P
dW dt
d dt
Work – kinetic energy theorem
K
I
2
f
2
I
i
2 2
W W
i
f
d
Corresponding relations for translational and rotational motion
Smooth rolling
•
Smooth rolling – object is rolling without slipping or bouncing on the surface
•
Center of mass is moving at speed
v CM
•
Point of momentary contact between the two surfaces is moving at speed
v CM s = θR ds/dt = d
(
θR
)
/dt = R dθ/dt v CM = ds/dt = ωR
Rolling: translation and rotation combined
•
Rotation – all points on the wheel move with the same angular speed
ω
•
Translation – all point on the wheel move with the same linear speed
v CM
Rolling: translation and rotation combined
K
I CM
2 2 2
Mv CM
2
Rolling: pure rotation
•
Rolling can be viewed as a pure rotation around the axis
P
moving with the linear speed
v com
•
The speed of the top of the rolling wheel will be
v top =
(
ω
)(2
R
)
=
2(
ωR
)
=
2
v CM
Friction and rolling
•
Smooth rolling is an idealized mathematical description of a complicated process
•
In a uniform smooth rolling,
P
is at rest, so there’s no tendency to slide and hence no friction force
•
In case of an accelerated smooth rolling
a CM = α R f s
opposes tendency to slide
Rolling down a ramp
Mgh i
Mv
2
f
2
I
2
f
2
gh i
v
2
f
2
Iv f
2 2
MR
2
v f
2
gh i MR
2
MR
2
I
Chapter 10 Problem 39 What fraction of a solid disk’s kinetic energy is rotational if it’s rolling without slipping?
Vector product of two vectors
•
The result of the vector ( cross ) multiplication of two vectors is a vector
a
b
c
•
The magnitude of this vector is
•
c
and
a
ab
sin
φ
is the smaller of the two angles between
b
Vector product of two vectors
•
Vector vectors
a c
is perpendicular and
b
to the plane that contains and its direction is determined by the right-hand rule
•
Because of the right-hand rule, the order of multiplication is important (commutative law does not apply)
b
a
(
a
b
)
i
ˆ •
i
ˆ
For unit vectors
0
j
ˆ ˆ
j i
ˆ
j
ˆ
k
ˆ
j
ˆ
k
ˆ
i
ˆ
k
ˆ
k
ˆ
k
ˆ
i
ˆ
j
ˆ
a
b
Vector product in unit vector notation
(
a x i
ˆ
a y
ˆ
j
a z k
ˆ ) (
b x i
ˆ
b y j
ˆ
b z k
ˆ )
a x i
ˆ
a x i
ˆ
b y j
ˆ
b x i
ˆ
a x b x
(
i
ˆ
i
ˆ )
a x b y
(
i
ˆ ˆ
j
) 0
a x b y k
ˆ
a
b
(
a y b z
b y a z
)
i
ˆ (
a z b x
b z a x
)
j
ˆ (
a x b y
b x a y
)
k
ˆ
Torque revisited
•
Using vector product, we can redefine torque ( vector ) as:
r
F
r
F
r
F
rF
sin
r
sin
F
Angular momentum
L
•
Angular momentum velocity
v
of a particle of mass
m
with respect to the origin
O
and is defined as
r
p
m
(
r
v
) •
SI unit: kg*m 2 /s
Newton’s Second Law in angular form
L
r
p
m
(
r
v
)
d L
m dt m
r
i
i r
a
d v
dt r
net d r
dt
m a
v
r
m
F net
r
a
i
d L dt
net r
v
v
F i
Angular momentum of a system of
d L dt
n
L d
L n dt
particles
n
n
L n
net
,
n
net
d L
net dt
Angular momentum of a rigid body
•
A rigid body (a collection of elementary masses
Δ
m i
) rotates about a fixed axis with constant angular speed
ω
•
For sufficiently symmetric objects:
L
I
Conservation of angular momentum
•
From the Newton’s Second Law
d L dt
net
•
If the net torque acting on a system is
d L dt
0
L
const
zero , then
•
If no net external torque acts on a system of particles, the total angular momentum of the system is conserved (constant)
•
This rule applies independently to all components
net
,
x
0
L x
const
Conservation of angular momentum
L
I
const I
i
i
I f
f
Conservation of angular momentum
L
const
More corresponding relations for translational and rotational motion
Chapter 11 Problem 28 A skater has rotational inertia 4.2 kg·m 2 with his fists held to his chest and 5.7 kg·m 2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his rotation axis, how fast will he be spinning?
Questions?
Answers to the even-numbered problems Chapter 10 Problem 24 0.072 N
⋅
m
Answers to the even-numbered problems Chapter 10 Problem 30 2.58 × 10 19 N
⋅
m
Answers to the even-numbered problems Chapter 10 Problem 40 hollow
Answers to the even-numbered problems Chapter 11 Problem 16 69 rad/s; 19 ° west of north
Answers to the even-numbered problems Chapter 11 Problem 18 (a) 8.1 N
⋅
m kˆ (b) 15 N
⋅
m kˆ
Answers to the even-numbered problems Chapter 11 Problem 24 1.7 × 10 -2 J
⋅
s
Answers to the even-numbered problems Chapter 11 Problem 26 (a) 1.09 rad/s (b) 386 J
Answers to the even-numbered problems Chapter 11 Problem 30 along the x-axis or 120 ° clockwise from the x-axis
Answers to the even-numbered problems Chapter 11 Problem 42 26.6
°