Topic 6 – Rotation
Chapter 9
I.
Introduction.
a. Rotational motion is all around us. Earth rotates about its
axis. Wheels, gears, propellers, motors, the drive shaft in a
car, a CD in its player, a pirouetting ice skater, all rotate.
b. In this chapter, we consider rotation about an axis that is
fixed in space, as in a merry-go-round, or about an axis that
is moving without changing its direction in space, as in a
rolling wheel on a car that is traveling in a straight line. The
study of rotational motion is continued in Chapter 10 where
more general examples of rotational motion are considered.
II.
Rotational Kinematics: Angular Velocity and Angular
Acceleration.
a. Every point of a rigid object rotating
about a fixed axis moves in a circle
whose center is on the axis and whose
radius is the radial distance from the axis
of rotation to that point. A radius drawn
from the rotation axis to any point on the
body sweeps out the same angle in the
same time.
b. Let ri be the distance from the center of the disk
to the ith particle (Figure 9-2), and let θi be the
angle measured counterclockwise from a fixed
reference line in space to a radial line from the
axis to the particle. As the disk rotates through
an angle dθ, the particle moves through a circular
arc of directed length dsi.
i. If counterclockwise is designated as the positive
direction, then dθ, θi, and dsi, shown in Figure 9-2,
are all positive.
ii. If clockwise is designated the positive direction, they
are all negative.
iii. The angle θi, the directed length dsi, and the distance ri
vary from particle to particle, but the ratio dsi/ri, called
the angular displacement dθ, is the same for all
particles of the disk.
iv. For one complete revolution, the arc length si is 2 ri.
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Topic 6 – Rotation
Chapter 9
c. The time rate of change of the angle is the same for all
particles of the disk, and is called the angular velocity ω of
the disk.
i. The instantaneous angular velocity ω is an angular
displacement of short duration divided by the time.
ii. All points on the disk undergo the same angular
displacement during the same time, so they all have
the same angular velocity.
iii. The SI units of ω are rad/s. Because radians are
dimensionless, the dimension of angular velocity is that
−1
of reciprocal time, T .
iv. The magnitude of the angular velocity is called the
angular speed.
v. Practice Problem 9 – 1: A compact disk is rotating at
3000 rev/min. What is its angular speed in radians per
second?
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Topic 6 – Rotation
Chapter 9
d. Angular acceleration is the rate of change of angular velocity.
i. If the rotation rate of a rotating object increases, the
angular speed |ω| increases.
ii. If |ω| is increasing, and if the angular velocity ω is
clockwise, then the change in the angular velocity Δω
is also clockwise.
iii. If the rotation rate decreases, then both Δω and αav are
in the opposite direction to ω.
e. The instantaneous rate of change of angular velocity is called
the angular acceleration α.
2
i. The SI units of α are rad/s . α is positive if ω is
increasing, and α is negative if ω is decreasing.
f. The angular displacement θ, the angular velocity ω, and
angular acceleration α are analogous to the linear
displacement x, linear velocity
in one-dimensional motion.
, and linear acceleration ax
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Topic 6 – Rotation
Chapter 9
g. Example 9 – 1: A compact disk rotates from rest to 500
rev/min in 5.5 s. (a) What is its angular acceleration,
assuming that it is constant? (b) How many revolutions does
the disk make in 5.5 s? (c) How far does a point on the rim
6.0 cm from the center of the disk travel during the 5.5 s it
takes to get to 500 rev/min?
h. Practice Problem 9 – 2: (a) Convert 500 rev/min to rad/s.
(b) Check the result of Part (b) in the example using
.
i. The linear velocity
of a particle on the disk is tangent to
the circular path of the particle and has magnitude dsi/dt. We
can relate this tangential velocity to the angular velocity of
the disk.
j. The tangential acceleration of a particle on the disk is
.
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Topic 6 – Rotation
Chapter 9
k. Each particle of the disk also has a centripetal acceleration,
which points inward along the radial line.
i. Practice Problem 9 – 3: A point on the rim of a
compact disk is 6.00 cm from the axis of rotation. Find
the tangential speed
, tangential acceleration at, and
centripetal acceleration ac of the point when the disk is
rotating at a constant angular speed of 300 rev/min.
ii. Practice Problem 9 – 4: Find the linear speed of a
point on the CD in Example 9 – 1 at (a) r = 2.40 cm,
when the disk rotates at 500 rev/min, and (b) r = 6.00
cm, when the disk rotates at 200 rev/min.
5
Topic 6 – Rotation
Chapter 9
Where:
Variable
Symbol Units
Rot Variable
Symbol
Units
Position
x
m
Angle
rad
Displacement
s
m
Angular Displacement
rad
Velocity (initial)
vo
m/s
Angular Velocity (Initial)
o
rad/s
Velocity (final)
v1
m/s
Angular Velocity (Final)
1
rad/s
Acceleration
a
m/s2
Angular Acceleration
Time
t
s
Time
t
s
Force (Total)
F
N
Torque
T
Nm
Mass
M
kg
Mass Moment of Inertia
I
kgm2
rad/s2
Note: These formulas only work in RADIANS!
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Topic 6 – Rotation
Chapter 9
III.
Rotational Kinetic Energy.
a. The kinetic energy of a rigid object rotating about a fixed axis
is the sum of the kinetic energies of the individual particles
that collectively constitute the object.
b. The sum in the expression farthest to the right is the object’s
moment of inertia I for the axis of rotation.
c. Example 9 – 2: An object consists of four
point particles, each of mass m, connected
by rigid massless rods to form a rectangle of
edge lengths 2a and 2b, as shown in
Figure 9-3. The system rotates with
angular speed ω about an axis in the plane
of the figure through the center, as shown.
(a) Find the kinetic energy of this object
using Equation 9-11 and Equation 9-12.
(b) Check your result by individually
calculating the kinetic energy of each
particle and then taking their sum.
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Topic 6 – Rotation
Chapter 9
d. Practice Problem 9 – 5: Find the moment of inertia of this
system for rotation about an axis parallel to the first axis but
passing through two of the particles, as shown in Figure 94.
e. Special Notes on Section 9 – 2.
8
Topic 6 – Rotation
Chapter 9
IV.
Calculating the Moment of Inertia.
a. The moment of inertia about an axis is a measure of the
inertial resistance of the object to changes in its rotational
motion about the axis. It is the rotational analog of mass.
b. Unlike the mass of an object, which is a property of the
object itself, the moment of inertia depends on the location of
the axis as well as the mass distribution of the object.
c. For systems consisting of discrete particles, we can compute
the moment of inertia about a given axis.
i. Example 9 – 3: Estimating Moment of Inertia Estimate the moment of inertia
of a thin uniform rod of length L
and mass M about an axis
perpendicular to the rod and
through one end. Execute this
estimation by modeling the rod
as three point masses, each
point mass representing onethird of the rod.
ii. Practice Problem 9 – 6: The contribution to the
moment of inertia of the third of the rod farthest from
the axis is many times greater than is the contribution
of the third closest to the axis. About how many times
greater is it?
d. To calculate the moment of inertia for continuous objects, we
imagine the object to consist of a continuum of very small
mass elements.
9
Topic 6 – Rotation
Chapter 9
where r is the radial distance from the axis to mass element
dm. To evaluate this integral, we first express dm as a
density times an element of length, area, or volume.
e. We can calculate I for continuous objects of various shapes.
Example 9 – 4: Moment of Inertia of a Thin Uniform Rod -Find the moment of inertia of a thin uniform rod
of length L and mass M about an axis perpendicular to the rod and through one end.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
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Topic 6 – Rotation
Chapter 9
Hoop about a perpendicular axis through its center Assume that a hoop has mass M and radius R
(Figure 9-7). The axis of rotation is the symmetry axis of the hoop, which is perpendicular to the plane of
the hoop and through its center.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
Uniform disk about a perpendicular axis through its center
The uniform disk is made of many hoops. The
area of the disk has a constant R, but the radius
of each hoop varies and is differential. The
hoop radius = r.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
Uniform solid cylinder about its axis We consider a cylinder to be a set of disks. The radius of each
disk is constant and is equal to R.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
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Topic 6 – Rotation
Chapter 9
Uniform hollow cylinder shell about its axis We consider a hollow cylinder shell to be a set of hoops.
Each hoop has a constant radius, R. All the mass is the same distance from the center.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
Uniform solid cylinder about its axis. We consider a sd cylinder to be a set of cylinder shells. The
radius of each shell varies and is differential.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
Uniform washer about its axis. We consider the washer to be made of smaller hoops, each with a
different radius. Therefore the radius is differential.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
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Topic 6 – Rotation
Chapter 9
Uniform hollow cylinder about its axis. We consider the hollow cylinder to be made of washers, each
with a constant radius.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia:
Uniform Hollow Spherical Shell about its axis. We can think of the hollow spherical shell to be made of
many differential rings. Each ring has a different radius. The differential area of this ring is its
circumference (2r) times its differential width (Rd).
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia
13
Topic 6 – Rotation
Chapter 9
Uniform Solid Sphere about its axis. We can think of the solid sphere as being made of many spherical
shells, each with a different radius. The radius is differential.
Length
Area
Volume
Differential L, A, or V
Density:
Differential Mass:
Moment of Inertia
f. We can often simplify the calculation of
moments of inertia for various objects
by using the parallel-axis theorem,
which relates the moment of inertia
about an axis through the center of
mass to the moment of inertia about a
second, parallel axis (Figure 9-10).
i. Let I be the moment of inertia, and let I cm be the
moment of inertia about a parallel axis through the
center of mass.
ii. In addition, let M be the total mass of the object and
let h be the distance between the two axes.
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Topic 6 – Rotation
Chapter 9
iii. Example 9 – 5: Applying the Parallel Axis Theorem - A
thin uniform rod of mass M and length L on the x axis
(Figure 9-11) has one end at the origin. Using the
parallel-axis theorem, find the moment of inertia about
the y’ axis, which is parallel to the y axis, and through
the center of the rod.
iv. Practice Problem 9 – 7: Using the parallel-axis
theorem, show that when comparing the moments of
inertia of an object about two parallel axes, the
moment of inertia is less about the axis that is nearer
to the center of mass.
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Topic 6 – Rotation
Chapter 9
v. Example 9 – 6: A Flywheel – Powered Car - You are
driving an experimental hybrid vehicle that is designed
for use in stop-and-go traffic. In a car with
conventional brakes, each time you brake to a stop the
kinetic energy is dissipated as heat. In this hybrid
vehicle, the braking mechanism transforms the
translational kinetic energy of the vehicle’s motion into
the rotational kinetic energy of a massive flywheel. As
the car returns to cruising speed, this energy is
transferred back into the translational kinetic energy of
the car. The 100-kg flywheel is a hollow cylinder with
an inner radius R1 of 25.0 cm, an outer radius R2 of
40.0 cm, and a maximum angular speed of 30,000
rev/min. On a dark and dreary night, the car runs out
of gas 15.0 mi from home with the flywheel spinning at
maximum speed. Is there sufficient energy stored in
the flywheel for you and your nervous grandmother to
make it home? (When driving at the minimum highway
speed of 40.0 mi/h, air drag and rolling friction
dissipate energy at 10.0 kW.)
16
Topic 6 – Rotation
Chapter 9
vi. Example 9 – 7: The Pivoted Rod A uniform thin rod of length L and
mass M, pivoted at one end as
shown in Figure 9-13, is held
horizontal and then released from
rest. Assuming that effects due to
friction and air resistance are
negligible, find (a) the angular
speed of the rod as it sweeps
through the vertical position, and
(b) the force exerted on the rod by the pivot at this
instant. (c) What initial angular speed would be needed
for the rod to just reach a vertical position at the top of
its swing?
17
Topic 6 – Rotation
Chapter 9
vii. Example 9 – 8: A Winch
and a Bucket - A winch is
at the top of a deep well.
The drum of the winch has
mass mw and radius R.
Virtually all its mass is
concentrated a distance R
from the axis. A cable
wound around the drum
suspends a bucket of
water of mass mb. The
entire cable has mass mc
and length L. Just when
you have the bucket at the highest point, your hand
slips and the bucket falls back down the well,
unwinding the winch cable as it goes. How fast is the
bucket moving after it has fallen a distance d, where d
is less than L? Assume that effects due to friction and
air resistance are negligible.
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Topic 6 – Rotation
Chapter 9
V.
Newton’s Second Law for Rotation.
a. To set a top spinning, you twist it. In Figure 9-17, a disk is
set spinning by the forces
and
exerted at the edges of
the disk in the tangential direction.
b. The directions of these forces and their points of application
are important. If the same forces are applied at the same
points but in the radial direction (Figure 9-18a), the disk will
not start to spin. In addition, if the same forces are applied in
the tangential direction, but at points closer to the center of
the disk (Figure 9-18b), the disk will not gain angular speed
as quickly.
c. Figure 9-19 shows a particle of mass m attached to one end
of a massless rigid rod of length
r. The rod is free to rotate about
a fixed axis perpendicular to the
rod and passing through the end
of the rod at A. Consequently,
the particle is constrained to
move in a circle of radius r. A
single force is applied to the
particle as shown.
19
Topic 6 – Rotation
Chapter 9
d. Applying Newton’s second law to the particle and taking
components in the tangential direction gives
e. The product rFt is the torque
associated with the force.
about the rotation axis
f. Torque about a point is defined as a vector quantity in
Chapter 10. What we refer to as “torque about an axis” is
the component of the torque vector parallel with the axis.
g. In Chapter 8, we saw that the net force acting on a system
of particles is equal to the net external force acting on the
system because the internal forces (those exerted by the
particles within the system on one another) cancel in pairs.
The treatment of internal torques exerted by the particles
within a system on one another leads to a similar result, that
is, the net torque acting on a system equals the net external
torque acting on the system.
h. Figure 9-20 shows a force acting on
an object constrained to rotate about a
fixed axis A, not shown, which passes
through O and is perpendicular to the
page.
20
Topic 6 – Rotation
Chapter 9
i. The positive tangential direction is shown at the point
of application of the force, and r is the radial distance
of this point of application from axis A.
ii. The torque due to this force about axis A is
.
In principle, the expression Ftr is all that is needed to
calculate torques.
iii. However, in practice, calculations are often simpler if
alternative expressions for torque are used.
iv. The line of action of a force is the line
through the point of application of the
force that is parallel to the force.
v. The moment arm is the
perpendicular distance between A and
the line of action. (The moment arm is
also called the lever arm.)
vi. The torque of a force about an axis is
also called the moment of the force
about the axis.
i. The net gravitational torque can be calculated by considering
the total gravitational force (the sum of the microscopic
gravitational forces) to act at a single point-the center of
gravity.
i. The torque due to a uniform gravitational field is
calculated as if the entire gravitational force is applied
at the center of gravity.
21
Topic 6 – Rotation
Chapter 9
ii. For any object in a uniform gravitational field, the center
of gravity and the center of mass coincide.
VI.
Applications of Newton’s Second Law for Rotation.
a. Applying Newton’s second law for Rotation
i. PICTURE Angular accelerations for rigid objects can be
found by using free-body diagrams and Newton’s
second law for rotation, which is
. If
is constant, then the constant angular
acceleration equations apply. Time intervals and
angular positions, velocities, and angular accelerations
can then be determined using these equations.
ii. SOLVE
1. Draw a free-body diagram with the object shown
as a likeness of the object (not just as a dot).
2. Draw each force vector along the line of action of
that force.
3. On the diagram indicate the positive direction
(clockwise or counterclockwise) for rotations.
iii. CHECK Make sure that the signs of your results are
consistent with your choice for the positive directions of
rotation.
b. Problems.
i. Example 9 – 9: A Stationary Bike To get some exercise without going
anywhere, you set your bike on a
stand so that the rear wheel is free
to turn. As you pedal, the chain
applies a force of 18 N to the rear
sprocket wheel at a distance of rs =
7.0 cm from the rotation axis of
the wheel. Consider the wheel to
2
be a hoop (I = MR ) of radius R =
35 cm and mass M = 2.4 kg. What
is the angular velocity of the wheel
after 5.0 s?
22
Topic 6 – Rotation
Chapter 9
ii. Example 9 – 10: A Uniform Rod, Pivoted at One End A uniform thin rod of length L
and mass M is pivoted at one
end. It is held horizontal and
released. Effects due to
friction and air resistance are
negligible. Find (a) the
angular acceleration of the
rod immediately following its
release, and (b) the
magnitude of the force FA
exerted on the rod by the pivot at this instant.
iii. Practice Problem 9 – 8: A small pebble of mass
is placed on top of the rod at its center. Just after the
rod is released find (a) the acceleration of the pebble,
and (b) the force it exerts on the rod.
c. Nonslip Conditions.
23
Topic 6 – Rotation
Chapter 9
i. In physics courses, there are many situations in which
a taut string is in contact with a rotating pulley wheel.
For the string not to slip on the pulley wheel, the parts
of the string and the wheel that are in direct contact
with each other must share the same tangential
velocity.
where is the tangential velocity of the string and Rω is the tangential
velocity of the perimeter of the pulley wheel. The wheel has radius R and
is rotating with angular velocity ω.
ii. Differentiating both sides of the nonslip condition with
respect to time gives
where at is the tangential acceleration of the string and α is the angular
acceleration of the wheel.
iii. Example 9 – 11: Tension in a String An object of mass m is suspended
from a light string that is wound
around the rim of a pulley wheel that
has moment of inertia I and radius R.
The wheel bearing is frictionless and
the string does not slip on the rim.
The wheel is released from rest. It
starts to rotate as the object
descends and the string unwinds.
Find the tension in the string and the
acceleration of the object.
24
Topic 6 – Rotation
Chapter 9
iv. Example 9 – 12: Two Block and a Pulley I
- The system shown in Figure 9-28 is
released from rest. The mass of the pulley
wheel is not negligible, but the friction in
the bearing is negligible. The string does
not slip on the pulley wheel. Given that m1
> m2 what can be determined about the
tensions T1 and T2?
25
Topic 6 – Rotation
Chapter 9
v. Example 9 – 13: Two Blocks and a Pulley II - Two
blocks are connected by a string that passes over a
pulley of radius R and moment of inertia I. The block of
mass m1 slides on a
frictionless, horizontal surface;
the block of mass m2 is
suspended from the string
(Figure 9-29). Find the
acceleration a of the blocks
and the tensions T1 and T2. The
string does not slip on the
pulley.
vi. Power.
1. When you spin up an object, you do work on it,
increasing its kinetic energy.
a. Consider the force acting on a rotating
object.
b. As the object rotates through an angle dθ,
the point of application of the force moves
a distance ds = r dθ, and the force does
work.
26
Topic 6 – Rotation
Chapter 9
c. The rate at which the torque does work is
the power input of the torque.
d. Equations in b and c are analogous to
2. Example 9 – 14: Torque Exerted by an
Automobile Engine - The maximum torque
produced by the 5.4-L V8 engine of a 2005 Ford
GT is 678 N · m of torque at 4500 rev/min. Find
the power output of the engine operating at
these maximum torque conditions.
3. Practice Problem 9 – 9: The maximum power
produced by the Ford GT engine is 500 hp at
6000 rev/min. What is the torque when the
engine is operating at maximum horsepower?
27
Topic 6 – Rotation
Chapter 9
4. Example 9 – 15: Stopping the Wheel - The
specifications for the London Eye include that it
be able to brake to a stop so that the passenger
compartments move no more than 10 m during
braking. The operating speed of the 135-mdiameter 1600-tonne wheel is 2.0 rev/h. (One
tonne equals 1000 kg.) A picture of the wheel
can be found at the beginning of this chapter. (a)
Estimate the torque that is required to stop the
wheel so the rim travels 10 m during the
braking. (b) Assuming that the braking force is
applied at the rim, what is the magnitude of the
breaking force?
28
Topic 6 – Rotation
Chapter 9
VII.
Rolling Objects.
a. Rolling Without Slipping.
i. When a spool rolls without
slipping down an incline
(Figure 9-31), the points of the
spool in contact with the incline
are instantaneously at rest and
the spool rotates about a rotation
axis through the contact points.
ii. Point P on the wheel moves as shown where r is the
radial distance from the rotation axis to point P.
29
Topic 6 – Rotation
Chapter 9
iii. For a point on the very top of the wheel, r = 2R, so the
top of the wheel is moving at twice the speed of the
center of the wheel.
iv. A wheel of radius R is rolling without
slipping along a straight path. As the
wheel rotates through angle
(Figure 9-33), the point of contact
between the wheel and the surface
moves a distance s that is related to
.
v. If the wheel is rolling on a flat surface, the wheel’s
center of mass remains directly over the point of
contact, so it also moves through distance R .
vi. In Chapter 8 we saw (Equation 8-7) that the kinetic
energy of a system can be written as the sum of its
translational kinetic energy
and the kinetic
energy relative to the center of mass Krel. For an object
that is rotating, the kinetic energy relative to an inertial
frame moving with center of mass
.
vii. Remember, a rolling object has both translational and
rotational kinetic energy.
30
Topic 6 – Rotation
Chapter 9
viii. Example 9 – 16: A Bowling Ball - A bowling ball that
has an 11-cm radius and a 7.2-kg mass is rolling
without slipping at 2.0 m/s on a horizontal ball return.
It continues to roll without
slipping up a hill to a height
h before momentarily
coming to rest and then
rolling back down the hill.
Model the bowling ball as a
uniform sphere and find h.
ix. PRACTICE PROBLEM 9-10 Find the initial kinetic
energy of the ball.
x. Example 9 – 17: Playing Pool - A cue stick strikes a cue
ball horizontally at a point a distance d above the
center of the ball (Figure 9-35). Find the value of d
for which the cue ball will roll without slipping from the
beginning. Express your answer in terms of the radius
R of the ball.
31
Topic 6 – Rotation
Chapter 9
xi. When an object rolls down an incline, its center of
mass accelerates. The analysis of such a problem is
simplified by an important theorem concerning the
center of mass:
xii. Equation 9-31 is the same as Equation 9-19, except
that in Equation 9-31 the torques and the moment of
inertia are computed from a reference frame moving
with the center of mass.
1. When the center of mass is accelerating (a ball
rolling down an incline, for example), the centerof-mass reference frame is a noninertial one,
where we would not expect our equations for
Newton’s second law for rotation to be valid.
2. Nevertheless, they are valid for this special case.
xiii. Example 9 – 19: Acceleration of a Ball
That Is Rolling Without Slipping – A
uniform solid ball of mass m and radius
R rolls without slipping down a plane
inclined at an angle above the
horizontal. Find the frictional force and
the acceleration of the center of mass.
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Topic 6 – Rotation
Chapter 9
1. The results of steps 5 and 6 in Example 9-18
apply to any round object with the center of
mass at the geometric center that is rolling
without slipping.
2. For such objects,
for a
solid sphere, for a rolling solid cylinder, 1 for a
thin hollow cylinder, and so forth. (These values
of β are obtained from the expressions for I
found in Table 9-1.)
xiv. The linear acceleration of any object rolling without
slipping down an incline is less than g sin because of
the frictional force directed up the incline.
1. Note that these accelerations are independent of
the mass and the radius of the objects.
2. That is, all uniform solid spheres will roll without
slipping down an incline with the same
acceleration.
3. However, if we release a sphere, a cylinder, and
a hoop at the top of an incline, and if they all roll
without slipping, the sphere will reach the
bottom first because it has the greatest
acceleration.
xv. Static frictional forces do no work on the rolling
objects, and if there is no slipping there is no
dissipation of energy.
1. Therefore, we use the conservation of
mechanical energy to find the final speed of an
object released from rest that is rolling without
slipping down an incline.
33
Topic 6 – Rotation
Chapter 9
2. At the top of the incline, the total energy is the
potential energy mgh. At the bottom, the total
energy is kinetic energy.
3. For a cylinder, with
. Note that this
speed is independent of both the mass and the
radius of the cylinder.
4. For an object rolling without slipping down an
incline, the frictional force fs is less than or equal
to its maximum value.
5. Practice Problem 9 – 11: A uniform cylinder rolls
down a plane inclined at = 50°. What is the
minimum value of the coefficient of static friction
for which the cylinder will roll without slipping?
34
Topic 6 – Rotation
Chapter 9
6. Practice Problem 9 – 12: For a uniform hoop of
mass m that is rolling without slipping down an
incline, (a) what is the force of friction, and (b)
what is the maximum value of tan for which the
hoop will roll without slipping?
b. Rolling With Slipping.
i. When an object slips (skids) as it rolls, the nonslip
condition
does not hold.
ii. The kinetic frictional force will both
reduce its linear speed
(Figure 939) and increase its angular speed ω
until the nonslip condition
reached, after which the ball rolls
without slipping.
is
iii. Another example of rolling with slipping is a ball with
topspin, such as a cue ball struck at a point higher than
above the center (see
Example 9-17) so that
. Then the kinetic
frictional force both increases
and decreases ω until the
nonslip condition
reached (Figure 9-40).
is
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Topic 6 – Rotation
Chapter 9
iv. Example 9 – 19: A Skidding Bowling Ball - A bowling
ball of mass M and radius R is released at floor level so
that at release it is moving
horizontally with speed
and is not rotating. The coefficient of
kinetic friction between the ball and
the floor is μk = 0.080. Find (a) the
time the ball skids along the floor
(after which it begins rolling without
slipping), and (b) the distance the
ball skids.
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