Ch 9. Rotational Dynamics

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Ch 9. Rotational Dynamics
The Action of Forces and Torques on Rigid Objects
(b) Combined translation and rotation
(a) Translation
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DEFINITION OF TORQUE
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)
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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 t (Greek letter tau), and its
magnitude is defined as the magnitude of the
force times the lever arm
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Forces of the same magnitude can produce
different torques, depending on the value of the
lever arm.
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Example 1.  Different Lever Arms, Different
Torques
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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.
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.
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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 3.6 102 m away
from point P.
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      (3.6    102 m) cos 55
The force F tends to produce a clockwise rotation
about the ankle joint, so the torque is negative
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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
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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.
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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
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Example 3.  A Diving Board 
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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
                               .
W 530 N
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Example 4.   Fighting a Fire
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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 ladders weight acts at the ladders center
and neglect the hoses weight. Find the forces
that the wall and the ground exert on the ladder.
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Force  Lever Arm  Torque
WL 355 N L (4.00 m) cos 50.0  WL L 
WF 875 N F (6.30 m) cos 50.0 WF F
P P (8.00 m) sin 50.0 P P
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                                          .
P 727 N
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Example 5.   Bodybuilding
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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 M (0.150) sin13.0 M M
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                                     .
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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
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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.
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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.
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Conceptual Example 7.  Overloading a Cargo Plane
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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.
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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.
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Newton's Second Law for Rotational Motion About a
Fixed Axis
FT maT t FTr t 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.
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ROTATIONAL ANALOG OF NEWTONS SECOND LAW FOR A
RIGID BODY ROTATING ABOUT A FIXED AXIS
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.
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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
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(a)
(r10, r2L)
(b)
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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 1.41 102 kg m2. What
net torque (assumed constant) must the motor
apply to the blade?
q  a w w0  t 
1508 rad (240.0 rev)  ? 503 rad/s (80.0 rev/s)  0 rad/s 
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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.
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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 archers aim.
Could there be any truth to this claim? Explain.
To the extent that I is larger, a given net
torque St 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 value of r in the equation I Smr2).
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Example 11. Hoisting a Crate
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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.
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m (4420 N)/(9.80 m/s2) 451 kg
ay ra (0.200 m)a.
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Analogies Between Rotational and Translational
Concepts
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Check Your Understanding 3
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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.
Lets say the length is 2a.
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Since is the same for all 3, the product
should be the same for all 3, hence large I
should have smaller .
IC gt IB gt IA
A, B, C
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Rotational Work and Energy
DEFINITION OF ROTATIONAL WORK
The rotational work WR done by a constant torque
t 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)
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kinetic energy is
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DEFINITION OF ROTATIONAL KINETIC ENERGY
The rotational kinetic energy KER of a rigid
object rotating with an angular speed w about a
fixed axis and having a moment of inertia I is
Requirement w must be expressed in rad/s. SI
Unit of Rotational Kinetic Energy joule (J)
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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.
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h h0, v0 0 m/s, w0 0 rad/s
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The solid cylinder, having the greater
translational speed, arrives at the bottom first.
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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?
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Solid ball I (2/5) mr2
For m1 and r1 , I1 (2/5) m1r12
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For m2 and r2 , I2 (2/5) m2r22
(a), (b) Both have the same translational speed.
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Angular Momentum
DEFINITION OF ANGULAR MOMENTUM The angular
momentum L of a body rotating about a fixed axis
is the product of the bodys moment of inertia I
and its angular velocity w with respect to that
axis
Requirement w must be expressed in rad/s. SI
Unit of Angular Momentum kgm2/s
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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.
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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.
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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.
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Keplers second law of planetary motion states
that a line joining a planet to the sun sweeps
out equal areas in equal time intervals.
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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.
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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 thin-walled spherical shell.
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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?
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Spherical shell I(2/3)MR2
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Conceptual Question 20
If the ice cap at the South Pole melted and the
water were uniformly distributed over the earth's
oceans, 1) earths angular velocity increases,
decreases, or same?
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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.
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Conceptual Question 21
River sediments, (Mississippi), towards equator.
What happens to angular velocity?
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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.
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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, , of the cloud
must remain constant. Since the cloud is
shrinking, its moment of inertia I decreases.
Since the product 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.
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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.
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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
This expression can be solved for x.
SOLUTION Solving the expression above for x, we
obtain
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Problem 19
cg
Fr
Ff
Axis
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.
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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)
where W is the weight of the plane, and ?w and ?f
are the lever arms for the forces W and Ff ,
respectively. Thus,
St -(1.00 106 N)(15.0 m - 12.6 m) Ff (15.0
m) 0
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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
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Problem 42
REASONING AND SOLUTION Newton's law applied to
the 11.0-kg object gives
a
a
T2
T1
T2 - (11.0 kg)(9.80 m/s2) (11.0 kg)(4.90 m/s2)
or T2 162 N
m2
m1
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For an axis about the center of the pulley
r
r
Clockwise rotation
T1
T2
T1r - T2r I(a)
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Problem 52
I total (stretched arms) 5.40 kg.m2
I red 3.8
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Problem 52
Angular momentum is conserved.