Ta k e i t F u rt h e r
Demonstration
Tangential Speed Versus
Angular Speed
Purpose Show that tangential speed
depends on radius.
TAKE IT FURTHER
Tangential Speed
and Acceleration
This feature explores the concepts of tangential speed and acceleration in greater
detail. Be sure you have also read the feature titled “Angular Kinematics.”
Materials two tennis balls attached to
different lengths of string (approx. 1.0 m
and 1.5 m)
Procedure Outside on an athletic field,
hold the ends of both strings and whirl
the tennis balls at constant angular
speed over your head. Point out the
equal angular speeds of the tennis
balls. Ask students to predict the flights
of the tennis balls when the strings
are released.
Aiming away from students and any
breakable items, release the strings and
have students observe the flights.
Discuss the longer horizontal displacement of the outer ball as a function of
its greater tangential speed.
Tangential Speed
Imagine an amusement-park carousel rotating about its center. Because a
carousel is a rigid object, any two horses attached to the carousel have the
same angular speed and angular acceleration. However, if the two horses
are different distances from the axis of rotation, they have different
tangential speed. The tangential speed of a horse on the carousel is its
speed along a line drawn tangent to its circular path.
FIGURE 1
Tangential Speed Horses on a
carousel move at the same angular speed
but different tangential speeds.
vt,inside
A
vt,outside
The tangential speeds of two horses at different distances from the
center of a carousel are represented in Figure 1. Note that the two horses
travel the same angular displacement during the same time interval. To
achieve this, the horse on the outside must travel a greater distance (∆s)
than the horse on the inside. Thus, the outside horse at point B has a
greater tangential speed than the inside horse at point A. In general, an
object that is farther from the axis of a rigid rotating body must travel at a
higher tangential speed to cover the same angular displacement as would
an object closer to the axis.
If the carousel rotates through an angle θ, a horse rotates through an
arc length ∆s in the interval ∆t. To find the tangential speed, start with
the equation for angular displacement:
B
∆s
∆θ = _
r
Next, divide both sides of the equation by the time it takes to travel ∆s :
TEACH FROM VISUALS
HRW • Holt Physics
PH99PE-C07-002-002-A
FIGURE 1 Point out to students that
each horse is a different distance away
from the center of the carousel.
Ask Which horse would travel farther
before hitting the ground if the horses
were released from the carousel?
The left side of the equation equals ωavg. Also, ∆s is a linear distance,
so ∆s divided by ∆t is a linear speed along an arc length. If ∆t is very
short, then ∆s is so small that it is nearly tangent to the circle; therefore,
∆s/∆t is the tangential speed, vt.
Tangential Speed
vt = r ω
tangential speed = distance from axis × angular speed
Answer: The outer horse would travel
farther because it has a greater
tangential speed.
252
Untitled-32 252
252 Chapter 7
∆s
∆θ = _
_
r∆t
∆t
Chapter 7
5/9/2011 8:18:00 AM
Untitled-32 253
Classroom Practice
Here, ω is the instantaneous angular speed, rather than the average
angular speed, because the time interval is so short. This equation is valid
only when ω is measured in radians per unit of time. Other measures of
angular speed must not be used in this equation.
Tangential Speed
A golfer has a maximum angular speed
of 6.3 rad/s for her swing. She can
choose between two drivers, one
placing the club head 1.9 m from her axis
of rotation and the other placing it 1.7 m
from the axis.
a. Find the tangential speed of the
club head for each driver.
Tangential Acceleration
If a carousel speeds up, the horses experience an angular acceleration.
The linear acceleration related to this angular acceleration is tangent to
the circular path and is called the tangential acceleration. If an object
rotating about a fixed axis changes its angular speed by ∆ω in the interval
∆t, the tangential speed of a point on the object has changed by the
amount ∆vt. Dividing the equation for tangential speed by ∆t results in
b. All other factors being equal, which
driver is likely to hit the ball farther?
∆vt = r∆ω
∆vt
∆ω
_
= r_
∆t
∆t
If the time interval ∆t is very small, then the left side of this relationship gives the tangential acceleration of the point. The angular speed
divided by the time interval on the right side is the angular acceleration.
Thus, the tangential acceleration (at) of a point on a rotating object is
given by the following relationship:
Answers
a. 12 m/s, 11 m/s
b. The longer driver will hit the ball
farther because its club head has
a higher tangential speed.
Tangential Acceleration
at = rα
tangential acceleration = distance from axis × angular acceleration
The angular acceleration in this equation refers to the instantaneous
angular acceleration. This equation must use the unit radians to be valid.
In SI, angular acceleration is expressed as radians per second per second.
Finding Total Acceleration
Any object moving in a circle has a centripetal acceleration. When both
components of acceleration exist simultaneously, the tangential acceleration is tangent to the circular path and the centripetal acceleration points
toward the center of the circular path. Because these components of
acceleration are perpendicular to each other, the magnitude of the total
acceleration can be found using the Pythagorean theorem, as follows:
atotal =
a 2t + a 2c
√����
FIGURE 2
Total Acceleration The
direction of the total acceleration of a
rotating object can be found using the
inverse tangent function.
at
Tangential Acceleration
A yo-yo has a tangential acceleration of
0.98 m/s2 when it is released. The string
is wound around a central shaft of radius
0.35 cm. What is the angular acceleration
of the yo-yo?
Answer: 2.8 × 102 rad/s2
a total
Teaching Tip
ac
To simulate the large accelerations
involved in spaceflight, Mercury
astronauts rode in the U. S. Navy’s
centrifuge in Johnsville, Pennsylvania.
The astronauts sat in a gondola at the
end of a 15.2 m arm that spun around a
central axis. During the spin, the astronauts experienced a combination of
centripetal and tangential accelerations
of the gondola that ranged from 8 to 10
times the acceleration due to gravity.
The direction of the total acceleration, as shown in Figure 2, depends
on the magnitude of each component of acceleration and can be found
using the inverse of the tangent function. Note that when there is a
tangential acceleration, the tangential speed is changing, and thus this
situation is not an example of uniform circular motion.
253
Circular Motion and Gravitation
HRW • Holt Physics
PH99PE-C07-002-007-A
5/9/2011 8:18:02 AM
Circular Motion and Gravitation 253
Misconception Alert!
Point out that objects rotate around
their center of mass in the absence
of other forces. For example, a ruler
thrown through the air will rotate
around its own center of mass because
air resistance is evenly distributed and
produces zero net torque. However,
a ruler with an index card taped to one
end will not rotate around its own
center of mass when it is thrown
through the air because the card
provides air resistance, which in turn
produces a torque.
Teaching Tip
Let students examine objects that have
a center of mass outside the object
itself, such as a doughnut, a coat hanger,
or a boomerang.
TAKE IT FURTHER
Rotation and Inertia
In this feature, you will explore the concept of rotational inertia.
Center of Mass
You have learned that torque measures the ability of a force to rotate an
object around some axis, such as a cat-flap door rotating on a hinge.
Locating the axis of rotation for a cat-flap door is simple: It rotates on its
hinges because the house applies a force that keeps the hinges in place.
Now imagine you are playing fetch with your dog, and you throw a
stick up into the air for the dog to retrieve. Unlike the cat-flap door, the
stick is not attached to anything. There is a special point around which
the stick rotates if gravity is the only force acting on the stick. This point is
called the stick’s center of mass.
The center of mass is also the point at which all the mass of the body
can be considered to be concentrated (for translational motion). This
means that the complete motion of the stick is a combination of both
translational and rotational motion. The stick rotates in the air around its
center of mass. The center of mass, in
FIGURE 1
turn, moves as if the stick were a point
Center of Mass The point around which this hammer
mass, with all of its mass concentrated
rotates is the hammer’s center of mass. The center of mass
at that point for purposes of analyzing
traces out the parabola that is characteristic of projectile motion.
its translational motion. For example,
the hammer in Figure 1 rotates about its
center of mass as it moves through the
air. As the rest of the hammer spins, the
center of mass moves along the parabolic path of a projectile.
For regularly shaped objects, such
as a sphere or a cube, the center of
mass is at the geometric center of the
object. For more complicated objects,
finding the center of mass is more
difficult. Although the center of mass is
the position at which an extended
object’s mass can be treated as a point
mass, the center of gravity is the position at which the gravitational force
acts on the extended object as if it were
a point mass. For many situations, the
center of mass and the center of gravity
are equivalent.
254
Untitled-32 254
254 Chapter 7
©2008 Richard Megna/Fundamental Photographs
Ta k e i t F u rt h e r
Chapter 7
5/9/2011 8:18:05 AM
Untitled-32 255
Moment of Inertia
Demonstration
Did YOU Know?
You may have noticed that it is easier to rotate a baseball bat around some
axes than others. The resistance of an object to changes in rotational
motion is measured by a quantity called the moment of interia.
A baseball bat can be modeled as a
rotating thin rod. When a bat is held
at its end, its length is greatest with
respect to the rotation axis, so its
moment of inertia is greatest. Thus,
the bat is easier to swing if you hold
the bat closer to the center. Baseball
players sometimes do this either
because a bat is too heavy (large M) or
too long (large �).
The moment of inertia, which is abbreviated as I, is similar to mass
because they are both forms of inertia. However, there is an important
difference between them. Mass is an intrinsic property of an object, and the
moment of inertia is not. The moment of inertia depends on the object’s
mass and the distribution of that mass around the axis of rotation. The farther
the mass of an object is, on average, from the axis of rotation, the greater is
the object’s moment of inertia and the more difficult it is to rotate the object.
According to Newton’s second law, when a net force acts on an object, the
resulting acceleration of the object depends on the object’s mass. Similarly,
when a net torque acts on an object, the resulting change in the rotational
motion of the object depends on the object’s moment of inertia. (This law is
covered in more detail in the feature “Rotational Dynamics.”)
FIGURE 2
THE MOMENT OF INERTIA FOR VARIOUS RIGID OBJECTS OF MASS M
Moment
of inertia
R
thin hoop about
symmetry axis
HRW • Holt R
Physicsthin hoop about
diameter
PH99PE-C08-002-012a-A
R
point mass
HRW • Holt Physicsabout axis
PH99PE-C08-002-012h-A
HRW • Holt Physics
PH99PE-C08-002-012b-A
disk or cylinder about
R
symmetry axis
ℓ
MR 2
1
__
MR 2
2
MR 2
1
__
MR 2
2
Moment
of inertia
Shape
thin rod about
perpendicular axis
through center
HRW • Holt Physics thin rod about perℓ
PH99PE-C08-002-012d-A
pendicular axis
through end
HRW • Holt Physics
solid sphere about
R
PH99PE-C08-002-012e-A
diameter
hp06se_apx00j030a.eps
thin spherical shell
about diameter
R LLCooper
Nanda Patel/
9/10/04
Materials broomstick or dowel
Procedure If possible, let student
volunteers assist by trying the various
demonstrations.
Thin rod about perpendicular axis
through center: Hold the rod in the
center with one hand. Rotate the rod
back and forth through half rotations at
regular time intervals. Note the force
required to change the motion.
Thin rod about perpendicular axis
through end: Hold rod at one end.
Rotate the rod through half circles in the
same time interval as previously used.
Note the increased force required
(corresponding to the larger moment
of inertia).
Cylinder: Hold the rod vertically
between your palms with your fingers
extended. Rotate the cylinder by
moving your palms back and forth in
the same regular time interval as used
previously. Note the much smaller force
required in this case.
Some simple formulas for calculating the moment of inertia of common
shapes are shown in Figure 2. The units for moment of inertia are kg•m2. To
get an idea of the size of this unit, note that bowling balls typically have
moments of inertia about an axis through their centers ranging from about
0.7 kg•m2 to 1.8 kg•m2, depending on the mass and size of the ball.
Shape
Moment of Inertia of a Rod
Purpose Give visual examples of the
two cases of a thin rod and the case
of a cylinder described in Figure 2.
1
__
Mℓ2
12
1
__
Mℓ2
3
2
__
MR 2
5
2
__
MR 2
3
2nd pass
HRW • Holt Physics
PH99PE-C08-002-012c-A
hp06se_apx00j031a.eps
Nanda Patel/ LLCooper
9/10/04
2nd pass
Circular Motion and Gravitation
255
5/9/2011 8:18:07 AM
Circular Motion and Gravitation 255
Ta k e i t F u rt h e r
Misconception Alert
Students will likely confuse zero net
torque with zero rotation. State explicitly that an object rotating at constant
speed is experiencing zero net torque.
This situation is analogous to linear
motion: an object moving at a constant
velocity has zero net force acting on it.
The Language of
Physics
TAKE IT FURTHER
Rotational Dynamics
The feature “Angular Kinematics” developed the kinematic equations for rotational
motion. Similarly, the feature “Rotation and Inertia” applied the concept of inertia
to rotational motion. In this feature, you will see how torque relates to rotational
equilibrium and angular acceleration. You will also learn how momentum and
kinetic energy are described in rotational motion.
FIGURE 1
Equal and Opposite Forces
The two forces exerted on this table
are equal and opposite, yet the table
moves. How is this possible?
If you and a friend push on opposite sides of a table, as shown in Figure 1,
the two forces acting on the table are equal in magnitude and opposite in
direction. You might think that the table won’t move because the two
forces balance each other. But it does; it rotates in place.
The piece of furniture can move even though the net force acting on it
is zero because the net torque acting on it is not zero. If the net force on
an object is zero, the object is in translational equilibrium. If the net
torque on an object is zero, the object is in rotational equilibrium. For an
object to be completely in equilibrium, both rotational and translational,
there must be both zero net force and zero net torque. The dependence of
equilibrium on the absence of net torque is called the second condition
for equilibrium.
Another way to state the second
equilibrium condition is to say that
the sum of the clockwise torques
must equal the sum of the counterclockwise torques.
Newton’s Second Law for Rotation
Teaching Tip
Just as net force is related to translational acceleration according to
Newton’s second law, there is a relationship between the net torque on an
object and the angular acceleration given to the object. Specifically, the
net torque on an object is equal to the moment of inertia times the
angular acceleration. This relationship is parallel to Newton’s second law
of motion and is known as Newton’s second law for rotating objects. This
law is expressed mathematically as follows:
Students may be confused about which
axis of rotation to choose when
applying the second condition for
equilibrium to an object. Tell students
that any axis can be used; the resultant
torque acting on an object in rotational
equilibrium is independent of where the
axis is placed. This fact is useful in
solving rotational equilibrium problems
because an unknown force that acts
along a line passing through this axis of
rotation will not produce any torque.
Beginning a diagram by arbitrarily setting
an axis where a force acts can eliminate
an unknown in the problem.
Newton’s Second Law for Rotating Objects
τnet = Iα
net torque = moment of inertia × angular acceleration
This equation shows that a net positive torque corresponds to a
positive angular acceleration, and a net negative torque corresponds to a
negative angular acceleration. Thus, it is important to keep track of the
signs of the torques acting on the object when using this equation to
calculate an object’s angular acceleration.
256
Untitled-32 256
256 Chapter 7
Rotational Equilibrium
Chapter 7
5/9/2011 8:18:09 AM
Angular Momentum
Demonstration
FIGURE 2
Conserving Angular Momentum
Because a rotating object has inertia, it also possesses momentum
associated with its rotation. This momentum is called angular momentum.
Angular momentum is defined by the following equation:
When this skater brings his hands and feet
closer to his body, his moment of inertia
decreases, and his angular speed increases
to keep total angular momentum constant
Angular Momentum
L = Iω
angular momentum = moment of inertia × angular speed
Colliding Magnetic Marbles
Purpose Illustrate conservation of
angular momentum and show that
objects moving linearly can also have
angular momentum.
Materials 2 magnetic marbles,
overhead projector
The unit of angular momentum is kg•m2/s. When the net external
torque acting on an object or objects is zero, the angular momentum of
the object(s) does not change. This is the law of conservation of angular
momentum. For example, assuming the friction between the skates and
the ice is negligible, there is no torque acting on the skater in Figure 2, so
his angular momentum is conserved. When he brings his hands and feet
closer to his body, more of his mass, on average, is nearer to his axis of
rotation. As a result, the moment of inertia of his body decreases. Because
his angular momentum is constant, his angular speed increases to
compensate for his smaller moment of inertia.
Procedure Place one magnetic marble
in the center of the overhead projector.
Send the second magnetic marble
toward the first to cause a glancing
collision. The marbles will stick together
and spin about their center of mass.
Point out to students that the linear
momentum is not converted to angular
momentum. Repeat the demonstration
and have students observe the slight
translation of the two marbles after the
collision. This is the linear momentum,
which is lost due to friction. The spinning
is due to conservation of the angular
momentum of the moving marble
relative to the stationary marble.
Angular Kinetic Energy
Rotating objects possess kinetic energy associated with their angular
speed. This is called rotational kinetic energy and is expressed by the
following equation:
Rotational Kinetic Energy
KErot = __12 Iω2
1 × moment of inertia × (angular speed)2
rotational kinetic energy = _
2
As shown in Figure 3, rotational kinetic energy is analogous to the
1 mv2.
translational kinetic energy of a particle, given by the expression _
2
The unit of rotational kinetic energy is the joule, the SI unit for energy.
FIGURE 3
(tr), (c) ©David Madison
COMPARING TRANSLATIONAL AND ROTATIONAL MOTION
Untitled-32 257
Translational motion
Rotational motion
Equilibrium
∑F = 0
∑τ = 0
Newton’s second law
∑F = ma
∑τ= Iα
Momentum
p = mv
L = Iω
Kinetic energy
KE = __12 mv2
KE = __12 Iω2
Circular Motion and Gravitation
257
5/9/2011 8:18:11 AM
Circular Motion and Gravitation 257