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Chapter 10
Rotation of a Rigid Object about a Fixed Axis
+
Rigid Object
A rigid


object is one that is non-deformable.
The relative locations of all particles making up the object
remain constant.
All real objects are deformable to some extent, but the rigid
object model is very useful in many situations where the
deformation is negligible.
Introduction
Angular Position
is located at (r, q) where r is the distance from the origin to P and q is the
measured counterclockwise from the reference line.
P
the particle moves, the only coordinate that changes is q, and as the
particle moves through q, it moves though an arc length s.
As
The

arc length and r are related:
s=qr
+
Radian
This
can also be expressed as:
q=
s
r
Comparing
1 rad =
degrees and radians
360°
= 57.3°
2p
Converting
q (rad ) =
from degrees to radians
p
180°
q (degrees )
Section 10.1
+
Angular Position, final
can associate the angle q with the entire rigid object as well
as with an individual particle.
We

Remember every particle on the object rotates through the same
angle.
angular position of the rigid object is the angle q between
the reference line on the object and the fixed reference line in
space.
The

The fixed reference line in space is often the x-axis.
angle θ plays the same role in rotational motion that the
position x does in translational motion.
The
Section 10.1
Angular Displacement
The
angular displacement is
defined as the angle the object
rotates through during some time
interval.
Dq = q f - qi
This
is the angle that the
reference line of length r sweeps
out.
Section 10.1
+ Average/Instantaneous Angular
Speed
average angular speed, ωavg, of a rotating rigid object is the
ratio of the angular displacement to the time interval.
The
wavg =
qf - q i
tf - t i
=
Dq
Dt
The
instantaneous angular speed is defined as the limit of
the average speed as the time interval approaches zero.
wº
lim
Dt ®0
Dq dq
=
Dt
dt
Section 10.1
+
Angular Acceleration
average angular acceleration, a avg, of an object is defined
as the ratio of the change in the angular speed to the time it takes
for the object to undergo the change.
The
aavg =
wf - wi
tf - t i
=
Dw
Dt
The
instantaneous angular acceleration is defined as the limit of
the average angular acceleration as the time goes to 0.
aº
lim
Dt ®0
Dw dw
=
Dt
dt
Section 10.1
Directions, details
Strictly
speaking, the speed and
acceleration (w, a) are the
magnitudes of the velocity and
acceleration vectors.
The
directions are actually given
by the right-hand rule.
Section 10.1
+
Rotational Kinematic Equations
The
kinematic expression for the rigid object under
constant angular acceleration are of the same
mathematical form as those for a particle under
constant acceleration.
Substitutions

x→θ

v→ω

a→α
from translational to rotational are
+

Example
A wheel rotates with a constant angular acceleration of 3.50
rad/s2.

If the angular speed of the wheel is 2.00 rad/s at ti = 0, through what
angular displacement does the wheel rotate in 2.00 s?

Through how many revolutions has the wheel turned during this
time period?

What is the angular speed of the wheel at t = 2.00s?
Relationship Between Angular &
Translational Motion
The
linear velocity is always tangent
to the circular path.

Called the tangential velocity
The
magnitude is defined by the
tangential speed.
ds
dq
v=
=r
= rw
dt
dt
Since
r is not the same for all points
on the object, the tangential speed of
every point is not the same.
The
tangential speed increases as
one moves outward from the center of
rotation.
Section 10.3
Relationship Between Angular &
Translational Motion
The
tangential acceleration is the
derivative of the tangential velocity.
at =
dv
dw
=r
= ra
dt
dt
An
object traveling in a circle, even
though it moves with a constant
speed, will have an acceleration.

Therefore, each point on a rotating
rigid object will experience a
centripetal acceleration.
v2
aC =
= rw2
r
Section 10.3
+
Resultant Acceleration
The
tangential component of the acceleration is due to changing
speed.
The
centripetal component of the acceleration is due to
changing direction.
Total
acceleration can be found from these components:
a = at2 + ar2 = r 2a 2 + r 2w 4 = r a 2 + w 4
Section 10.3
+
Rotational Motion Example
Find the angular speed of the
disc in revolutions per minute
when information is being read
from the innermost first track (r =
23 mm) and the outermost final
track (r = 58 mm)

The
maximum playing time of a
standard music disc is 74 min and
33 s. How many revolutions does
the disc make during that time?
What
is the angular acceleration
of the compact disc over the 4473
s time interval?
Section 10.3
+
Torque
Torque,
t, is the tendency of a force to
rotate an object about some axis.



Torque is a vector, but we will deal with its
magnitude here:
t = r F sin f = F d
 F is the force
 f is the angle the force makes with the
horizontal
 d is the moment arm (or lever arm) of
the force
There is no unique value of the torque on
an object.
 Its value depends on the choice of a
rotational axis.
Section 10.6
Net Torque
force F1 will tend to cause a
counterclockwise rotation about
O.
The
force F2 will tend to cause a
clockwise rotation about O.
The
 St
= t1 + t2 = F1d1 – F2d2
Section 10.6
+
Example


What is the net torque acting
on the cylinder about the
rotation axis?
Suppose T1 = 5.0 N R1 = 1.0 m,
T2 = 15 N, and R2 = 0.50 m.
What is the net torque about
the rotation axis, and which
way does the cylinder rotate
starting from rest?

Figure 10.9
Torque and Angular Acceleration
The
tangential force provides a tangential
acceleration:

Ft = mat
The
magnitude of the torque produced
byå Ft on a particle about an axis
through the center of the circle is

St = SFt r = (mat) r
The
tangential acceleration is related to
the angular acceleration.

St = (mat) r = (mra) r = (mr 2) a
mr 2 is the moment of inertia of
the particle,
Since


St = Ia **
The torque is directly proportional to the
angular acceleration and the constant of
proportionality is the moment of inertia.
+
Moment of Inertia
The
definition of moment of inertia is
I = å ri 2mi
i
dimensions of moment of inertia are ML2
and its SI units are kg.m2.
The
We
can calculate the moment of inertia of an
object more easily by assuming it is divided into
many small volume elements, each of mass Dmi.
Mass
is an inherent property of an object, but
the moment of inertia depends on the choice of
rotational axis.
Moment
of inertia is a measure of the resistance
of an object to changes in its rotational motion,
similar to mass being a measure of an object’s
resistance to changes in its translational motion.
 The moment of inertia depends on the mass
and how the mass is distributed around the
rotational axis.

Figure 10.11
+
+
Example

A uniform rod of length L and
mass M is attached to one end
of a frictionless pivot and is
free to rotate about the pivot in
the vertical plane. The rod is
released from rest in the
horizontal position. What are
the initial angular acceleration
of the rod and the initial
translational acceleration of its
right end?

Figure 10.12
Falling Smokestack Example
When
a tall smokestack falls
over, it often breaks somewhere
along its length before it hits the
ground.
Each
higher portion of the
smokestack has a larger
tangential acceleration than the
points below it.
The
shear force due to the
tangential acceleration is greater
than the smokestack can
withstand.
The
smokestack breaks.
Section 10.7
+ Torque and Angular Acceleration,
Extended
Consider
the object consists of an
infinite number of mass elements dm
of infinitesimal size.
Each
mass element rotates in a
circle about the origin, O.
Each
mass element has a tangential
acceleration.
From

Newton’s Second Law
dFt = (dm) at
The
torque associated with the force
and using the angular acceleration
gives
 dt ext = r dFt = atr dm = ar 2 dm
Section 10.7
+ Torque and Angular Acceleration,
Extended cont.
Finding


åt
ext
the net torque
= ò a r 2dm = a ò r 2dm
This becomes St = Ia
This
is the same relationship that applied to a particle.
This
is the mathematic representation of the analysis model of a
rigid body under a net torque.
The
result also applies when the forces have radial components.

The line of action of the radial component must pass through the axis of
rotation.

These components will produce zero torque about the axis.
Example

A wheel of radius, R, mass M,
and moment of inertia I is
mounted on a frictionless,
horizontal axis as shown to the
right. A light cord wrapped
around the wheel supports an
object of mass m. When the
wheel is released, the object
accelerates downward, the cord
unwraps off the wheel, and the
wheel rotates with an angular
acceleration. Find the
expressions for the angular
acceleration of the wheel, the
translational acceleration of the
object, and the tension in the
cord.
+
Moment of Inertia, cont
The
moment of inertia of a system of discrete particles can be
calculated by applying the definition for I.
For
a continuous rigid object, imagine the object to be divided
into many small elements, each having a mass of Δmi.
We
can rewrite the expression for I in terms of Dm.
I = Dmi ®lim0 å ri 2 Dmi = ò r 2dm
i
With
the small volume segment assumption,
I = ò r r 2dV
If
r is constant, the integral can be evaluated with known
geometry, otherwise its variation with position must be known.
Section 10.5
+
Notes on Various Densities
Volumetric
Mass Density → mass per unit volume: r = m / V
Surface
Mass Density → mass per unit thickness of a sheet of
uniform thickness, t : s = r t
Linear
Mass Density → mass per unit length of a rod of uniform
cross-sectional area: l = m / L = r A
Section 10.5
+ Moment of Inertia of a Uniform Rigid
Rod
The

shaded area has a mass
dm = l dx
Then
the moment of inertia is
I y = ò r dm = ò
2
L/2
-L / 2
I=
x2
M
dx
L
1
ML2
12
Section 10.5
+ Moment of Inertia of a Uniform Solid
Cylinder

A uniform solid cylinder has a
radius R, mass M, and length L.
Calculate its moment of inertia
about its central axis.
Section 10.5
+ Parallel-Axis Theorem
In
the previous examples, the axis of rotation coincided with the axis of
symmetry of the object.
For
an arbitrary axis, the parallel-axis theorem often simplifies calculations.
theorem states I = ICM + MD 2
I is about any axis parallel to the axis through the center of mass of the
object.
ICM is about the axis through the center of mass.
D is the distance from the center of mass axis to the arbitrary axis.
The




Figure 10.17
Section 10.5
+ Rotational Kinetic Energy
object rotating about some axis with an angular speed, ω, has
rotational kinetic energy even though it may not have any
translational kinetic energy.
An
Each

particle has a kinetic energy of
Ki = ½ mivi2
Since
the tangential velocity depends on the distance, r, from the
axis of rotation, we can substitute vi = wi r.
The
total rotational kinetic energy of the rigid object is the sum
of the energies of all its particles.
1
K R = å K i = å mi ri 2w 2
i
i 2
KR =
There
1æ
1 2
2ö 2
m
r
w
=
Iw
å i i ÷ø
2 çè i
2
is an analogy between the kinetic energies associated
with linear motion (K = ½ mv 2) and the kinetic energy
associated with rotational motion (KR= ½ Iw2).
Work and Power in Rotational Motion
the work done by F on the
object as it rotates through an
infinitesimal distance ds = r dq
Find
dW = F ids
= (F sin f )r dq
The
radial component of the force
does no work because it is
perpendicular to the displacement.
The
rate at which work is being
done in a time interval dt is
Power = P =
dW
dq
=t
= tw
dt
dt
Section 10.8
+
Work-Kinetic Energy Theorem in
Rotational Motion
The
work-kinetic energy theorem for rotational motion states
that the net work done by external forces in rotating a
symmetrical rigid object about a fixed axis equals the change in
the object’s rotational kinetic energy.
The
rotational form can be combined with the linear form which
indicates the net work done by external forces on an object is the
change in its total kinetic energy, which is the sum of the
translational and rotational kinetic energies.
Section 10.8
+
Summary of Useful Equations
Section 10.8
+
Example

A uniform rod of length L and
mass M is free to rotate on a
frictionless pin passing
through one end. The rod is
released from rest in the
horizontal position.

What is its angular speed
when the rod reaches its
lowest position?

Determine the tangential
speed of the center of mass
and the tangential speed of the
lowest point on the rod when it
is in the vertical position.

Figure 10.21
+
Energy in an Atwood Machine,
Example
Two
blocks having different
masses m1 and m2 are
connected by a string pulley
system. The pulley has a radius R
and moment of inertia I about its
axis of rotation. The string does
not slip on the pulley, and the
system is released from rest. Find
the translational speeds of the
blocks after block 2 descends
through a distance h and find the
angular speed of the pulley at this
time.
Section 10.8
Rolling Object
The

This path is called a cycloid.
The
In
red curve shows the path moved by a point on the rim of the object.
green line shows the path of the center of mass of the object.
pure rolling motion, an object rolls without slipping.
In
such a case, there is a simple relationship between its rotational and
translational motions.
Section 10.9
+
Pure Rolling Motion, Object’s
Center of Mass
The
translational speed of the
center of mass is
v CM
ds
dq
=
=R
= Rw
dt
dt
The
linear acceleration of the
center of mass is
aCM
dvCM
dw
=
=R
= Ra
dt
dt
Section 10.9
+
Rolling Motion Cont.
Rolling
motion can be modeled as a combination of pure translational motion
and pure rotational motion.
The
contact point between the surface and the cylinder has a translational
speed of zero (c).
Section 10.9
+
Total Kinetic Energy of a Rolling
Object
The
total kinetic energy of a rolling object is the sum of the
translational energy of its center of mass and the rotational
kinetic energy about its center of mass.

K = ½ ICM w2 + ½ MvCM2


The ½ ICMw2 represents the rotational kinetic energy of the
cylinder about its center of mass.
The ½ Mv2 represents the translational kinetic energy of the
cylinder about its center of mass.
Section 10.9
+
Total Kinetic Energy, Example
Accelerated
rolling motion is
possible only if friction is present
between the sphere and the incline.
 The friction produces the net
torque required for rotation.
 No loss of mechanical energy
occurs because the contact
point is at rest relative to the
surface at any instant.
 In reality, rolling friction causes
mechanical energy to transform
to internal energy.

Rolling friction is due to
deformations of the surface and
the rolling object.
Section 10.9
+
Total Kinetic Energy, Example cont.
Apply
Conservation of Mechanical Energy:

Let U = 0 at the bottom of the plane

Kf + U f = Ki + Ui

Kf = ½ (ICM / R2) vCM2 + ½ MvCM

Ui = Mgh

Uf = Ki = 0
Solving
v cm
for v
é
ù
ê
ú
2gh
=ê
ú
I
æ
ö
ê1 + ç CM
ú
2÷
MR ø úû
êë è
=
2
1 æ ICM
ö 2
+
M
vCM
ç
÷
2
2èR
ø
1
2
Section 10.9
+ Sphere Rolling Down an Incline,
Example
For the solid sphere shown to
the right, calculate the
translational speed of the center
of mass at the bottom of the
incline and the magnitude of the
translational acceleration of the
center of mass.

Section 10.9