Ch 9 Rotational Dynamics

Report
Ch 9. Rotational Dynamics
The Action of Forces and Torques on Rigid Objects
(a) Translation
(b) Combined translation
and rotation
1
DEFINITION OF TORQUE
Torque  ( Magnitude
of the force )  ( Lever arm )
  F
Direction: The torque is positive when the force tends to
produce a counterclockwise rotation about the axis, and
negative when the force tends to produce a clockwise
rotation.
SI Unit of Torque: newton · meter
(N · m)
2
The line of action is an extended line drawn colinear with
the force. The lever arm is the distance between the line
of action and the axis of rotation, measured on a line that
is perpendicular to both. The torque is represented by
the symbol  (Greek letter tau), and its magnitude is
defined as the magnitude of the force times the lever arm
3
Forces of the same
magnitude can produce
different torques,
depending on the value
of the lever arm.
4
Example 1.
Different Lever Arms, Different
Torques
5
A force whose magnitude is 55 N is applied to a door. However,
the lever arms are different in the three parts of the drawing:
(a) = 0.80 m, (b) = 0.60 m, and (c) = 0 m. Find the
magnitude of the torque in each case.
(a)
(b)
(c)
In part c the line of action of F passes through the axis
of rotation (the hinge). Hence, the lever arm is zero, and
the torque is zero.
6
Example 2. The Achilles Tendon
The ankle joint and the Achilles tendon attached to the heel at
point P. The tendon exerts a force of magnitude F = 720 N.
Determine the torque (magnitude and direction) of this force
about the ankle joint, which is located 3.6 × 10–2 m away from
7
point P.

= (3.6 × 10–2 m) cos 55°
The force F tends to produce a clockwise rotation about
the ankle joint, so the torque is negative:
8
Check Your Understanding 1
The drawing shows an overhead view of a horizontal bar that is
free to rotate about an axis perpendicular to the page. Two
forces act on the bar, and they have the same magnitude.
However, one force is perpendicular to the bar, and the other
makes an angle f with respect to it. The angle f can be 90°,
45°, or 0°. Rank the values of f according to the magnitude of
the net torque (the sum of the torques) that the two forces
produce, largest net torque first.
0°, 45°, 90°
9
Rigid Objects in Equilibrium
CONCEPTS AT A GLANCE If a rigid body is in
equilibrium, neither its linear motion nor its rotational motion
changes. This lack of change leads to certain equations that
apply for rigid-body equilibrium.
10
EQUILIBRIUM OF A RIGID BODY: A rigid body is in
equilibrium if it has zero translational acceleration
and zero angular acceleration. In equilibrium, the sum
of the externally applied forces is zero, and the sum of
the externally applied torques is zero:
11
Example 3. A Diving Board
12
A woman whose weight is 530 N is poised at the right end of
a diving board with a length of 3.90 m. The board has
negligible weight and is bolted down at the left end, while
being supported 1.40 m away by a fulcrum. Find the forces
F1 and F2 that the bolt and the fulcrum, respectively, exert on
the board
F
.
F2 
y
  F1  F 2  W  0

  F2  2  W  W  0
WW
( 530 N )( 3 . 90 m )
2
W = 530 N

 1480 N
1 . 40 m
13
Example 4. Fighting a Fire
14
An 8.00-m ladder of weight WL = 355 N leans against a
smooth vertical wall. The term “smooth” means that the wall
can exert only a normal force directed perpendicular to the
wall and cannot exert a frictional force parallel to it. A
firefighter, whose weight is WF = 875 N, stands 6.30 m from
the bottom of the ladder. Assume that the ladder’s weight acts
at the ladder’s center and neglect the hose’s weight. Find the
forces that the wall and the ground exert on the ladder.
15
Force
Lever Arm
Torque
L
= (4.00 m) cos 50.0°
–WL L
WL = 355 N

WF = 875 N

F
= (6.30 m) cos 50.0°
–WF F

P
= (8.00 m) sin 50.0°
+P P
P
16

P 
 W L  L  W F  F  P  P  0
WL L  WF  F
P
.

355 N 4 . 00 m  cos 50 . 0   875 N 6 . 30 m  cos 50 . 0 
8 . 00 m  sin 50 . 0 
 727 N
P = 727 N
17
Example 5. Bodybuilding

18
A bodybuilder holds a dumbbell of weight Wd. His arm is
horizontal and weighs Wa = 31.0 N. The deltoid muscle is assumed
to be the only muscle acting and is attached to the arm as shown.
The maximum force M that the deltoid muscle can supply has a
magnitude of 1840 N. The distances that locate where the various
forces act on the arm. The locations from where the various forces
act on the arm are shown in the figure. What is the weight of the
heaviest dumbbell that can be held, and what are the horizontal
and vertical force components, Sx and Sy, that the shoulder joint
applies to the left end of the arm?
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Force
Lever Arm
Torque
Wa = 31.0 N

a
= 0.280 m
–Wa a
Wd

d
= 0.620 m
–Wd d
M = 1840 N

= (0.150) sin13.0°
+M  M
M
20

Wd 
.

 W a  a  W d  d  M  M  0
 Wa a  M M
d
 31 . 0 N 0 . 280 m   1840 N 0 . 150 m  sin 13 . 0 
 86 . 1 N
0 . 620 m
21
Check Your Understanding 2
Three forces act on each of the thin, square sheets shown in the
drawing. In parts A and B of the drawing, the force labeled 2F
acts at the center of the sheet. The forces can have different
magnitudes (F or 2F) and can be applied at different points on
an object.
In which drawing is (a) the
translational acceleration equal to
zero, but the angular acceleration is
not equal to zero, (b) the
translational acceleration is not
equal to zero, but the angular
acceleration is equal to zero, and (c)
the object in equilibrium?
(a) C, (b) A, (c) B
22
Center of Gravity
DEFINITION OF CENTER OF GRAVITY: The center of
gravity of a rigid body is the point at which its weight can
be considered to act when the torque due to the weight is
being calculated.
23
24
Example 6. The Center of
Gravity of an Arm
The horizontal arm is
composed of three parts: the
upper arm (weight W1 = 17 N),
the lower arm (W2 = 11 N),
and the hand (W3 = 4.2 N).
The drawing shows the center
of gravity of each part,
measured with respect to the
shoulder joint. Find the center
of gravity of the entire arm,
relative to the shoulder joint.
25
26
Conceptual Example 7.
Overloading a Cargo Plane
27
A stationary cargo plane with its front landing gear 9
meters off the ground. This accident occurred because the
plane was overloaded toward the rear. How did a shift in
the center of gravity of the loaded plane cause the accident?
Because of the overloading, the center of gravity has shifted
behind the rear landing gear. The torque due to W is now
counterclockwise and is not balanced by any clockwise torque.
Due to the unbalanced counterclockwise torque, the plane
rotates until its tail hits the ground, which applies an upward
force to the tail. The clockwise torque due to this upward force
balances the counterclockwise torque due to W, and the plane
comes again into an equilibrium state, this time with the front
landing gear 9 meters off the ground.
28
The center of gravity of an object with an irregular shape
and a nonuniform weight distribution can be found by
suspending the object from two different points P1 and P2,
one at a time.
29
Newton's Second Law for Rotational
Motion About a Fixed Axis
FT = maT
 = FTr
 = maTr
aT = ra
The constant of proportionality is I =
mr2, which is called the moment of
inertia of the particle. The SI unit for
moment of inertia is kg · m2.
30
31
ROTATIONAL ANALOG OF NEWTON’S SECOND LAW
FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS

 Ia
Requirement: a must be expressed in rad/s2
Although a rigid object possesses a unique total mass, it does
not have a unique moment of inertia, for the moment of inertia
depends on the location and orientation of the axis relative to the
particles that make up the object.
32
Example 8.
The Moment of Inertia Depends
on Where the Axis Is
Two particles each have a mass m
and are fixed to the ends of a thin
rigid rod, whose mass can be
ignored. The length of the rod is L.
Find the moment of inertia when
this object rotates relative to an
axis that is perpendicular to the
rod at (a) one end and (b) the
center
33
(a)
(r1=0, r2=L)
(b)
34
35
36
37
38
Example 9.
The Torque of an Electric Saw Motor
The motor in an electric saw brings the circular blade from
rest up to the rated angular velocity of 80.0 rev/s in 240.0 rev.
One type of blade has a moment of inertia of 1.41 × 10–2
kg · m2. What net torque (assumed constant) must the motor
apply to the blade?
q
a

0
1508 rad
(240.0 rev)
?
503 rad/s
(80.0 rev/s)
0
rad/s
t
39
40
The lever arm is just the radius of the circular rail,
which is designed to be as large as possible. Thus, a
relatively large torque can be generated for a given force,
allowing the rider to accelerate quickly.
41
Conceptual Example 10.
Archery and Bow Stabilizers
Archers can shoot with amazing accuracy, especially using
modern bows. Notice the bow stabilizer, a long, thin rod that
extends from the front of the bow and has a relatively massive
cylinder at the tip. Advertisements claim that the stabilizer
helps to steady the archer’s aim. Could there be any truth to
this claim? Explain.
To the extent that I is larger, a given net torque S will create a
smaller angular acceleration and less disturbance of the aim. It is
to increase the moment of inertia of the bow that the stabilizer has
been added. The relatively massive cylinder is particularly
effective in increasing the moment of inertia, because it is placed
at the tip of the stabilizer, far from the axis of rotation (a large
42
value of r in the equation I = Smr2).
Example 11. Hoisting a Crate
43
A crate that weighs 4420 N is being lifted by the mechanism.
The two cables are wrapped around their respective pulleys,
which have radii of 0.600 and 0.200 m. The pulleys are
fastened together to form a dual pulley and turn as a single
unit about the center axle, relative to which the combined
moment of inertia is I = 50.0 kg · m2. A tension of magnitude T1
= 2150 N is maintained in the cable attached to the motor. Find
the angular acceleration of the dual pulley and the tension in
the cable connected to the crate.

2150
 T1  1  T 2  2  I a
N 0 . 600 m   T 2 0 . 200 m   50 . 0 kg  m a
2
44
m = (4420 N)/(9.80 m/s2) = 451 kg:
ay = ra = (0.200 m)a.
45
Analogies Between Rotational and
Translational Concepts
Physical Concept
Rotational
Translational
Displacement
q
s
Velocity
v
Acceleration

a
The cause of acceleration
Torque 
Force F
Inertia
Moment of inertia I
Mass m
Newton’s second law
Work
S = Ia
q
SF = ma
Fs
Kinetic energy
½I2
½mv2
Momentum
L = I
p = mv
a
46
Check Your Understanding 3
47
Three massless rods are free to rotate about an axis at
their left end (see the drawing). The same force F is
applied to the right end of each rod. Objects with
different masses are attached to the rods, but the total
mass (3m) of the objects is the same for each rod. Rank
the angular acceleration of the rods, largest to smallest.
Let’s say the length is 2a.
All three has the same Torque around axis
to page.
  F  2a
  Ia
48
Since is the same for all 3, the product I a should be
the same for all 3, hence large I should have smaller a .
IA 

IB 

IC 

mr
2
 m a   m  2 a   5 ma
mr
2
 2 m a   m 2 a   6 ma
mr
2
2
2
2
2
 3 m 2 a   12 ma
2
2
2
2
IC > I B > I A
aC  aB  a A
 in descending
order
A, B, C
a A  aB  aC
49
Rotational Work and Energy
DEFINITION OF ROTATIONAL WORK
The rotational work WR done by a constant torque  in
turning an object through an angle q is
Requirement: q must be expressed in radians. W = Fs = Frq
SI Unit of Rotational Work: joule (J)
50
kinetic energy is
51
DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy KER of a rigid object rotating
with an angular speed  about a fixed axis and having a
moment of inertia I is
Requirement:  must be expressed in rad/s.
SI Unit of Rotational Kinetic Energy: joule (J)
52
53
Example 12. Rolling Cylinders
A thin-walled hollow cylinder (mass =
mh, radius = rh) and a solid cylinder
(mass = ms, radius = rs) start from rest
at the top of an incline . Both cylinders
start at the same vertical height h0. All
heights are measured relative to an
arbitrarily chosen zero level that passes
through the center of mass of a
cylinder when it is at the bottom of the
incline. Ignoring energy losses due to
retarding forces, determine which
cylinder has the greatest translational
speed upon reaching the bottom.
54
h = h0, v0 = 0 m/s, 0 = 0 rad/s
55
I  MR
I 
1
MR
2
2
2
The solid cylinder, having the greater translational
speed, arrives at the bottom first.
56
Check Your Understanding 4
Two solid balls are placed side by side at the top of an incline
plane and, starting from rest, are allowed to roll down the
incline. Which ball, if either, has the greater translational
speed at the bottom if (a) they have the same radii but one is
more massive than the other, and (b) they have the same
mass but one has a larger radius?
57
vf 
2 mgh
m
Solid ball I= (2/5) mr2
I
r
2
For m1 and r1 , I1= (2/5) m1r12
v1 
2 m 1 gh
m1 

5
7
I1
r1
2

2 m 1 gh
m1 
2
5
m1

2 gh
1
2
5
( 2 gh )
58
For m2 and r2 , I2= (2/5) m2r22
v2 
2 m 2 gh
m2 

5
I2
2
2
r

2 m 2 gh
m2 
2
5
m2

2 gh
1
2
5
( 2 gh )
7
(a), (b) Both have the same translational speed.
59
Angular Momentum
DEFINITION OF ANGULAR MOMENTUM The
angular momentum L of a body rotating about a fixed
axis is the product of the body’s moment of inertia I and
its angular velocity  with respect to that axis:
L  I
Requirement:  must be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
60
PRINCIPLE OF CONSERVATION OF ANGULAR
MOMENTUM The total angular momentum of a system remains
constant (is conserved) if the net average external torque acting on
the system is zero.
61
Conceptual Example 13.
A Spinning Skater
An ice skater is spinning with
both arms and a leg
outstretched. She pulls her arms
and leg inward. As a result of
this maneuver, her spinning
motion changes dramatically.
Using the principle of
conservation of angular
momentum, explain how and
why it changes.
62
Example 14.
A Satellite in an Elliptical Orbit
An artificial satellite is placed into
an elliptical orbit about the earth.
Telemetry data indicate that its
point of closest approach (called
the perigee) is rP = 8.37 × 106 m
from the center of the earth, and
its point of greatest distance
(called the apogee) is rA = 25.1 ×
106 m from the center of the earth.
The speed of the satellite at the
perigee is vP = 8450 m/s. Find its
speed vA at the apogee.
63
Kepler’s second law of
planetary motion states
that a line joining a planet
to the sun sweeps out equal
areas in equal time
intervals.
64
Concepts & Calculations
Example 15. Torque and Force
A crate resting on a horizontal
surface. It has a square cross section
and a weight of W = 580 N, which is
uniformly distributed. At the bottom
right edge is a small obstruction that
prevents the crate from sliding when
a horizontal pushing force P is
applied to the left side. However, if
this force is great enough, the crate
will begin to tip or rotate over the
obstruction. Determine the minimum
pushing force that leads to tipping.
65

  P P  W  W  0
66
Concepts & Calculations
Example 16. Which Sphere
Takes Longer to Stop?
Two spheres are each rotating
at an angular speed of 24
rad/s about axes that pass
through their centers. Each
has a radius of 0.20 m and a
mass of 1.5 kg. However, one
is solid and the other is a thinwalled spherical shell.
67
Suddenly, a net external torque due to friction (magnitude =
0.12 N · m) begins to act on each sphere and slows the motion
down. How long does it take each sphere to come to a halt?
  0  at
a 

I
  0 

t
I
68
t 
  0
(  ) / I

I (   0 )

Solid sphere I=(2/5)MR2
2
t
2
t 5
I (   0 )

 5
MR (   0 )
2

(1 . 5 kg )( 0 . 20 m ) ( 0 rad / s )  ( 24 rad / s ) 
2
 0 . 12 N  m
 4 .8 s
69
Spherical shell I=(2/3)MR2
2
t
2
t 3
I (   0 )

 3
MR (   0 )
2

(1 . 5 kg )( 0 . 20 m ) ( 0 rad / s )  ( 24 rad / s ) 
2
 0 . 12 N  m
 8 .0 s
70
Conceptual Question 20
If the ice cap at the South Pole melted and the water were
uniformly distributed over the earth's oceans,
1) earth’s angular velocity increases, decreases, or same?
I increases
but
L  I  constant (no external torques)
  decreases .
71
REASONING AND SOLUTION Consider the earth to be
an isolated system. Note that the earth rotates about an
axis that passes through the North and South poles and is
perpendicular to the plane of the equator. If the ice cap at
the South Pole melted and the water were uniformly
distributed over the earth's oceans, the mass at the South
Pole would be uniformly distributed and, on average, be
farther from the earth's rotational axis. The moment of
inertia of the earth would, therefore, increase. Since the
earth is an isolated system, any torques involved in the
redistribution of the water would be internal torques;
therefore, the angular momentum of the earth must
remain the same. If the moment of inertia increases, and
the angular momentum is to remain constant, the angular
velocity of the earth must decrease.
72
Conceptual Question 21
River sediments, (Mississippi), towards equator.
What happens to angular velocity?
Main distribution from top towards equator (r---larger)
I increases.
Since
L  I

is constant, (why?)
decreases.
73
Conceptual Question 21
REASONING AND SOLUTION Note that the earth rotates
about an axis that passes through the North and South poles
and is perpendicular to the plane of the equator. When rivers
like the Mississippi carry sediment toward the equator, they
redistribute the mass from a more uniform distribution to a
distribution with more mass concentrated around the equator.
This increases the moment of inertia of the earth. If the earth
is considered to be an isolated system, then any torques
involved in the redistribution of mass are internal torques.
Therefore, the angular momentum of the earth must remain
constant. The moment of inertia increases, and the angular
momentum must remain the same; therefore, the angular
velocity must decrease.
74
Conceptual Question 22
REASONING AND SOLUTION Let the system be the
rotating cloud of interstellar gas. The gravitational force
that pulls the particles together is internal to the system;
hence, the torque resulting from the gravitational force is an
internal torque. Since there is no net external torque acting
on the collapsing cloud, the angular momentum, I  , of
the cloud must remain constant. Since the cloud is
shrinking, its moment of inertia I decreases. Since the
product I  remains constant, the angular velocity
must increase as I decreases. Therefore, the angular velocity
of the formed star would be greater than the angular
velocity of the original gas cloud.

75
Problem 12
REASONING When the board just begins to tip, three forces
act on the board. They are the weight W of the board, the
weight WP of the person, and the force F exerted by the right
support.
x
F
1.4 m
x
W = 225 N
W
P =
450 N
76
Since the board will rotate around the right support, the lever
arm for this force is zero, and the torque exerted by the right
support is zero. The lever arm for the weight of the board is
equal to one-half the length of the board minus the overhang
length: 2.5 m  1.1 m = 1.4 m
The lever arm for the weight of the person is x. Therefore,
taking counterclockwise torques as positive, we have
W P x  W (1.4 m )  0
This expression can be solved for x.
SOLUTION Solving the expression above for x, we obtain
x
W (1.4 m )
WP

(225 N )(1.4 m )
450 N
 0.70 m
77
Problem 19
cg
Fr
Ff

Axis


f
REASONING The jet is in equilibrium, so the sum of the
external forces is zero, and the sum of the external torques is zero.
We can use these two conditions to evaluate the forces exerted on
the wheels.
78
SOLUTION
a. Let Ff be the magnitude of the normal force that the ground
exerts on the front wheel. Since the net torque acting on the
plane is zero, we have (using an axis through the points of
contact between the rear wheels and the ground)
S = W w + Ff
f = 0
where W is the weight of the plane, and w and f are the
lever arms for the forces W and Ff , respectively. Thus,
S = 1.00 *106 N)(15.0 m  12.6 m) + Ff (15.0 m) = 0
Solving for Ff gives Ff =
1 . 60  10 5 N
79
b. Setting the sum of the vertical forces equal to zero yields
SFy = Ff + 2Fr  W = 0
where the factor of 2 arises because there are two rear
wheels. Substituting in the data,
SFy = 1.60 *105 N + 2Fr  1.00 * 106 N = 0
Fr =
4.20  10 5 N
80
Problem 42
a
a
T2
T1
REASONING AND SOLUTION Newton's
law applied to the 11.0-kg object gives
T2  m 2 g  m 2 a
T2  (11.0 kg)(9.80 m/s2) = (11.0 kg)(4.90 m/s2)
or T2 = 162 N
m2=
m1 =
A similar treatment for the 44.0-kg object yields
T1  m 1 g  m 1 a
T1  (44.0 kg)(9.80 m/s2) = (44.0 kg)(4.90 m/s2)
or T1 = 216 N
81
For an axis about the center of the pulley
r
r
Clockwise rotation
T2
T1
T1r  T2r = I(a
a
(T1  T 2 ) r   I  
r
T1  T 2   
1
2
Mr
2
a
r
2
82
 T1  T 2  
M 
2
a
Ma
2
(T1  T 2 )
Solving for the mass M we obtain
M = (2/a)(T2  T1) = [2/(4.90 m/s2)](162 N  216 N)
=22.0 kg
83
Problem 52
  5 . 00 rad / s
I total (stretched arms) 5.40 kg.m2
I red =3.8
84
Problem 52
Angular momentum is conserved.
L  I 0 0  I 1 1
5 . 4  0  3 . 8
 
I0
I
0 
5 . 40 kg  m
2
3 . 80 kg  m
2
5 . 00 rad / s  7 . 11 rad / s
85

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