Transcript Ch 8

Physics
CHAPTER 8
ROTATIONAL MOTION
The Radian

The radian is a unit of
angular measure

The radian can be
defined as the arc
length s along a
circle divided by the
radius r

s
 
r
57.3
More About Radians

Comparing degrees and radians
360 
1 rad 
 57 . 3 
2

Converting from degrees to radians

 [rad ] 
 [deg rees ]
180 
Angular Displacement

Axis of rotation is the
center of the disk

Need a fixed
reference line

During time t, the
reference line moves
through angle θ
Angular Displacement,
cont.


The angular displacement is defined
as the angle the object rotates
through during some time interval
   f   i

The unit of angular displacement is the
radian

Each point on the object undergoes
the same angular displacement
Average Angular Speed

The average angular
speed, ω, of a
rotating rigid object is
the ratio of the
angular
displacement to the
time interval
 av
 f   i 


tf  ti
t
Angular Speed, cont.


The instantaneous angular speed is
defined as the limit of the average
speed as the time interval
approaches zero
Units of angular speed are radians/sec



rad/s
Speed will be positive if θ is increasing
(counterclockwise)
Speed will be negative if θ is
decreasing (clockwise)
Average Angular
Acceleration


The average angular acceleration,
, 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:
 av
 f   i 


tf  ti
t
Angular Acceleration,
cont



Units of angular acceleration are rad/s²
Positive angular accelerations are in the
counterclockwise direction and negative
accelerations are in the clockwise
direction
When a rigid object rotates about a fixed
axis, every portion of the object has the
same angular speed and the same
angular acceleration
Angular Acceleration,
final

The sign of the acceleration does not
have to be the same as the sign of the
angular speed

The instantaneous angular acceleration is
defined as the limit of the average
acceleration as the time interval
approaches zero
Angular Acceleration
Angular acceleration α
measures how rapidly the
angular velocity is changing:
Slide 7-17
Linear and Circular Motion Compared
Slide 7-18
Linear and Circular Kinematics Compared
Slide 7-19
Sign of the Angular Acceleration
Slide 7-20
Relationship Between Angular
and Linear Quantities

Displacements

Speeds
s  r
vt   r

Accelerations
at   r

Every point on the
rotating object has
the same angular
motion

Every point on the
rotating object does
not have the same
linear motion
Centripetal Acceleration
and Angular Velocity

The angular velocity and the linear velocity are
related (v = ωr)

The centripetal acceleration can also be related
to the angular velocity
aC   r
2
Vector Nature of Angular
Quantities

Angular displacement,
velocity and acceleration
are all vector quantities

Direction can be more
completely defined by
using the right hand rule

Grasp the axis of rotation
with your right hand

Wrap your fingers in the
direction of rotation

Your thumb points in the
direction of ω
Velocity Directions, Example


In a, the disk rotates
clockwise, the
velocity is into the
page
In b, the disk rotates
counterclockwise,
the velocity is out of
the page
Righty-tighty
Lefty-loosey
Centripetal and Tangential Acceleration
Slide 7-22
Force vs. Torque
 Forces
cause accelerations
 Torques
cause angular
accelerations
 Force
and torque are related
Torque
The door is free to rotate about an axis through O
 There are three factors that determine the effectiveness
of the force in opening the door:




The magnitude of the force
The position of the application of the force
The angle at which the force is applied
Torque, cont
Torque, t, is the tendency of a force
to rotate an object about some
axis
 t = Fr
t is the torque
F is the force
symbol is the Greek tau
r is the length of the position
vector
 SI unit is N.m

Interpreting Torque
Torque is due to the component of the force perpendicular to the
radial line.
Torque is a vector quantity
The direction is
perpendicular to the
plane determined by
the position vector and
the force
t  rF  rF sin 
Slide 7-25
A Second Interpretation of Torque
Fsin
t  r F  rF sin 
Slide 7-26
Signs and Strengths of the Torque
If the turning tendency of the force is counterclockwise, the
torque will be positive
If the turning tendency is clockwise, the torque will be negative
Slide 7-27
Multiple Torques

When two or more torques are
acting on an object, the torques
are added
 As

vectors
If the net torque is zero, the object’s
rate of rotation doesn’t change
General Definition of
Torque

The applied force is not always perpendicular to
the position vector

The component of the force perpendicular to the
object will cause it to rotate

When the force is parallel to the position vector,
no rotation occurs

When the force is at some angle, the
perpendicular component causes the rotation
General Definition of
Torque, final

Taking the angle into account leads to a
more general definition of torque:
t
 Fr sin 
F
is the force
r
is the position vector
is the angle between the force
and the position vector

Lever Arm

The lever arm, d, is the perpendicular distance from
the axis of rotation to a line drawn along the
direction of the force

d = r sin 
Net Torque

The net torque is the sum of all the
torques produced by all the forces
 Remember
to account for the
direction of the tendency for rotation
 Counterclockwise
positive
 Clockwise
torques are
torques are negative
Checking Understanding
The four forces shown have the same strength. Which force
would be most effective in opening the door?
A.
B.
C.
D.
E.
Force F1
Force F2
Force F3
Force F4
Either F1 or F3
Slide 7-23
Answer
The four forces shown have the same strength. Which force
would be most effective in opening the door?
A.
B.
C.
D.
E.
Force F1
Force F2
Force F3
Force F4
Either F1 or F3
Slide 7-24
Moment of Inertia

The moment of inertia, I, of a point mass is
equal to mass of the object times the
square of the distance from the object’s
axis of rotation.
I  m r
 SI
2
units are kg m2
 Applying Newton’s 2nd Law results in 
 t/I
Moment of Inertia

The moment of inertia of an object is the
rotational equivalent to the mass of the object in
a linear motion.

Ex. For linear motion, the heavier the mass the
more difficult it is to get it to move. In
rotational motion, a high I, moment of inertia,
means that it is difficult to get the object to
rotate on an axis.

The size of the moment of inertia, depends on
the radius of rotation from the center axis and
the distribution of mass around the axis of
rotation.
Moment of Inertia

If the radius length is
large, it will be more
difficult to get the mass
to rotate which
indicates a higher
moment of inertia. So if
we apply a torque
closer to the axis of
rotation, it will be easier
to cause the rotation to
occur.

If the mass is distributed
closer to the axis of
rotation, the rotation will
be easier to start and
the Moment of Inertia
will be less
I1 > I2
Moments of Inertia for Various
Objects
Object
Location of Axis
Thin hoop of radius Through central
r
diameter
Solid uniform
cylinder of radius r
Through center
Uniform Sphere of
radius r
Through center
Long uniform rod
of length l
Through center
Long uniform rod
of length l
Through end
Thin rectangular
plane of length l
and width w
Through center
Diagram
Moment of Inertia
Equation
Newton’s Second Law for
a Rotating Object
 t  I

The angular acceleration is directly
proportional to the net torque

The angular acceleration is
inversely proportional to the
moment of inertia of the object
More About Moment of
Inertia

There is a major difference between
moment of inertia and mass: the
moment of inertia depends on the
quantity of matter and its distribution
in the rigid object.

The moment of inertia also depends
upon the location of the axis of
rotation
Moment of Inertia of a Uniform
Ring

Image the hoop is
divided into a
number of small
segments, m1 …

These segments are
equidistant from the
axis
I   m i ri2  M R 2
Other Moments of Inertia
Example, Newton’s Second
Law for Rotation

Draw free body diagrams
of each object

Only the cylinder is
rotating, so apply t = I 

The bucket is falling, but
not rotating, so apply F
=ma

Remember that a =  r
and solve the resulting
equations
Torque and Equilibrium

First Condition of Equilibrium
 The
net external force must be zero
 F  0 or
 Fx  0 and  Fy  0
This is a necessary, but not sufficient,
condition to ensure that an object is in
complete mechanical equilibrium
 This is a statement of translational
equilibrium

Torque and Equilibrium,
cont

To ensure mechanical equilibrium, you
need to ensure rotational equilibrium as
well as translational

The Second Condition of Equilibrium
states
 The
net external torque must be zero
t  0
Equilibrium Example

The woman, mass m, sits
on the left end of the seesaw

The man, mass M, sits
where the see-saw will be
balanced

Apply the Second
Condition of Equilibrium
and solve for the
unknown distance, x
Axis of Rotation

If the object is in equilibrium, it does
not matter where you put the axis of
rotation for calculating the net torque
The location of the axis of rotation is
completely arbitrary
 Often the nature of the problem will
suggest a convenient location for the axis
 When solving a problem, you must specify
an axis of rotation

 Once
you have chosen an axis, you must
maintain that choice consistently
throughout the problem
Notes About Equilibrium

A zero net torque does not mean the
absence of rotational motion
 An
object that rotates at uniform
angular velocity can be under the
influence of a zero net torque
 This
is analogous to the translational
situation where a zero net force
does not mean the object is not in
motion
Solving Equilibrium
Problems

Draw a diagram of the system


Include coordinates and choose a
rotation axis
Isolate the object being analyzed and
draw a free body diagram showing all
the external forces acting on the
object

For systems containing more than one
object, draw a separate free body
diagram for each object
Problem Solving, cont.

Apply the Second Condition of Equilibrium


Apply the First Condition of Equilibrium


This will yield a single equation, often with one
unknown which can be solved immediately
This will give you two more equations
Solve the resulting simultaneous equations for
all of the unknowns

Solving by substitution is generally easiest
Example of a Free Body
Diagram (Forearm)

Isolate the object to be analyzed

Draw the free body diagram for that object

Include all the external forces acting on the object
Example of a Free Body
Diagram (Beam)

The free body
diagram includes the
directions of the
forces

The weights act
through the centers
of gravity of their
objects
Fig 8.12, p.228
Slide 17
Example of a Free Body
Diagram (Ladder)


The free body diagram shows the normal force and
the force of static friction acting on the ladder at
the ground
The last diagram shows the lever arms for the forces
Center of Gravity

The force of gravity acting on an
object must be considered

In finding the torque produced by
the force of gravity, all of the
weight of the object can be
considered to be concentrated at
a single point
Calculating the Center of
Gravity

The object is
divided up into a
large number of
very small particles
of weight (mg)

Each particle will
have a set of
coordinates
indicating its
location (x,y)
Calculating the Center of
Gravity, cont.

We assume the object is free to
rotate about its center

The torque produced by each
particle about the axis of rotation is
equal to its weight times its lever
arm
 For
example, t1  m1 g x1
Calculating the Center of
Gravity, cont.

We wish to locate the point of
application of the single force
whose magnitude is equal to the
weight of the object, and whose
effect on the rotation is the same
as all the individual particles.

This point is called the center of
gravity of the object
Coordinates of the Center
of Gravity

The coordinates of the center of gravity
can be found from the sum of the
torques acting on the individual particles
being set equal to the torque produced
by the weight of the object
x cg
 mi x i
 mi y i

and y cg 
 mi
 mi
Center of Gravity of a
Uniform Object

The center of gravity of a homogenous,
symmetric body must lie on the axis of
symmetry.

Often, the center of gravity of such an
object is the geometric center of the
object.