07 LectureSlides 20171018

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Gravity Obeys an Inverse-Square Law
•  Gravity is a universal
force that affects all
objects in the universe.
•  Newton proposed that
the force of gravity
has the following
properties:
1.  The force is inversely proportional to the square of the distance
between the objects.
2.  The force is directly proportional to the product of the masses of the
two objects.
G = 6.67 x 10-11 N.m2/kg2
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Slide 7-1
Gravity on a Grand Scale
•  No matter how far apart two objects may be, there is a
gravitational attraction between them.
•  Galaxies are held together by gravity.
•  All of the stars in a galaxy are different distances from the
galaxy’s center, and so orbit with different periods.
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Slide 7-2
Lecture
Presentation
Chapter 7
Rotational Motion
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Chapter 7 Rotational Motion
Chapter Goal: To understand the physics of rotating
objects.
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Slide 7-4
Chapter 7 Preview
Looking Ahead: Rotational Kinematics
•  The spinning roulette wheel isn’t going anywhere, but it is
moving. This is rotational motion.
•  You’ll learn about angular velocity and other quantities we use
to describe rotational motion.
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Slide 7-5
Chapter 7 Preview
Looking Ahead: Torque
•  To start something moving, apply a force. To start something
rotating, apply a torque, as the sailor is doing to the wheel.
•  You’ll see that torque depends on how hard you push and also
on where you push. A push far from the axle gives a large
torque.
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Slide 7-6
Chapter 7 Preview
Looking Ahead: Rotational Dynamics
•  The girl pushes on the outside edge of the merry-go-round,
gradually increasing its rotation rate.
•  You’ll learn a version of Newton’s second law for rotational
motion and use it to solve problems.
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Slide 7-7
Section 7.3 Torque
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A question
•  A force is required for motion.
•  I: Force causes an acceleration
•  II: Acceleration changes velocity
•  III: Velocity changes position
•  This is true for any translational, circular motions etc.
•  What about rotational motion? (when the whole object or
a system of objects rotates about an axis)
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Slide 7-9
Torque
•  Forces with equal strength
will have different effects
on a swinging door.
•  The ability of a force to
cause rotation depends on
•  The magnitude F of the force.
•  The distance r from the pivot—the axis about which the
object can rotate—to the point at which force is applied.
•  The angle at which force is applied.
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Slide 7-10
Torque
•  Torque (τ) is the rotational equivalent of force.
Greek letter
Tau
•  Torque units are newton-meters, abbreviated N ⋅ m.
•  A force is required to push or pull an object. A torque is
required to “turn” an object, i.e., to give a rotational
motion to the object.
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Slide 7-11
Torque
•  The radial line is the line
starting at the pivot and
extending through the point
where force is applied.
•  The angle ϕ (Greek letter: Phi)
is measured from the
radial line to the
direction of the force.
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Slide 7-12
Torque
•  Torque is dependent on
the perpendicular component
of the force being applied.
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Slide 7-13
Torque
•  The equivalent expression for
torque is
•  For both methods for
calculating torque, the
resulting expression is
the same:
•  The greater the torque, more effective it would be in
rotating the object.
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Slide 7-14
How to increase torque
•  Depends on :
•  r
•  F
•  ϕ
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Slide 7-15
QuickCheck 7.10
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
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Slide 7-16
QuickCheck 7.10
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
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Your intuition likely led you to choose F1.
The reason is that F1 exerts the largest torque
about the hinge.
Slide 7-17
Example 7.9 Torque in opening a door
Ryan is trying to open a stuck door.
He pushes it at a point 0.75 m from
the hinges with a 240 N force
directed 20° away from being
perpendicular to the door.
There’s a natural pivot point, the
hinges. What torque does Ryan exert?
How could he exert more torque?
F sin 70° = 226 N. The distance from the hinge to the point at which
the force is applied is r = 0.75 m.
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Slide 7-18
Example 7.9 Torque in opening a door
Ryan is trying to open a stuck door.
He pushes it at a point 0.75 m from
the hinges with a 240 N force
directed 20° away from being
perpendicular to the door.
There’s a natural pivot point, the
hinges. What torque does Ryan exert?
How could he exert more torque?
In FIGURE 7.20 the radial line is shown drawn from the
pivot—the hinge—through the point at which the force is applied.
We see that the component of that is perpendicular to the radial line
is F⊥ = F cos 20° = 226 N which is also = F sin 70° = 226 N. The
distance from the hinge to the point at which the force is applied is r =
0.75 m.
PREPARE
Problem-solving tips: (1) identify the pivot, (2) identify the radial line (start at the pivot and go to
the point where force is applied), (3) determine the angle ϕ. Remember that the angle is measured.
© 2015 Pearson Education, Inc.
Slide 7-19
Example 7.9 Torque in opening a door (cont.)
SOLVE We
can find the torque
on the door from Equation 7.10:
The torque depends on how hard Ryan pushes, where he
pushes, and at what angle. If he wants to exert more torque, he
could push at a point a bit farther out from the hinge, or he could
push exactly perpendicular to the door. Or he could simply push
harder!
As you’ll see by doing more problems, 170 N ⋅ m is a
significant torque, but this makes sense if you are trying to free a
stuck door.
ASSESS
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Slide 7-20
Torque
•  A torque that tends to rotate the object in a counterclockwise direction is positive, while a torque that
tends to rotate the object in a clockwise direction is
negative.
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Slide 7-21
Net Torque
•  The net torque is the sum
of the torques due to the
applied forces:
[Insert Figure 7.23]
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Slide 7-22
Section 7.4 Gravitational Torque
and the Center of Gravity
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Gravitational Torque and the Center of Gravity
•  Gravity pulls downward on
every particle that makes up
an object (like the gymnast).
•  Each particle experiences a
torque due to the force of
gravity.
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Slide 7-24
Gravitational Torque and the Center of Gravity
•  The gravitational torque can
be calculated by assuming
that the net force of gravity
(the object’s weight) acts as
a single point.
•  That single point is called the
center of gravity.
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Slide 7-25
Example 7.12 The torque on a flagpole
A 3.2 kg flagpole extends
from a wall at an angle of
25° from the horizontal.
Its center of gravity is
1.6 m from the point
where the pole is attached
to the wall. What is the gravitational torque on the flagpole
about the point of attachment?
FIGURE 7.26 shows the situation. For the purpose of
calculating torque, we can consider the entire weight of the pole
as acting at the center of gravity. Because the moment arm r⊥ is
simple to visualize here, we’ll use Equation 7.11 for the torque.
PREPARE
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Slide 7-26
Example 7.12 The torque on a flagpole (cont.)
From Figure 7.26,
we see that the moment arm
is r⊥ = (1.6 m) cos 25° =
1.45 m. Thus the gravitational
torque on the flagpole, about
the point where it attaches to
the wall, is
SOLVE
We inserted the minus sign because the torque tries to rotate the pole
in a clockwise direction.
If the pole were attached to the wall by a hinge, the
gravitational torque would cause the pole to fall. However, the actual
rigid connection provides a counteracting (positive) torque to the pole
that prevents this. The net torque is zero.
ASSESS
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Slide 7-27
Examples in everyday life
•  You use torque every day without realizing it. You apply
torque three times when you simply open a locked door.
Turing the key, turning the doorknob, and pushing the
door open so it swings on its hinges are all methods of
applying a torque.
•  Athletes
•  Physical Therapy!!!
•  Engineering
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Slide 7-28
Example 7.16 Angular acceleration of a falling
pole
In the caber toss, a contest of
strength and skill that is part
of Scottish games, contestants
toss a heavy uniform pole,
landing it on its end. A
5.9-m-tall pole with a mass
of 79 kg has just landed on
its end. It is tipped by 25°
from the vertical and is
starting to rotate about the
end that touches the ground.
Estimate the angular
acceleration.
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Slide 7-29
Example 7.16 Angular acceleration of a falling
pole (cont.)
PREPARE The
situation is shown
in FIGURE 7.37, where we
define our symbols and list the
known information. Two forces
are acting on the pole: the pole’s
weight which acts at the
center of gravity, and the force
of the ground on the pole (not shown). This second force exerts
no torque because it acts at the axis of rotation. The torque on
the pole is thus due only to gravity. From the figure we see that
this torque tends to rotate the pole in a counterclockwise
direction, so the torque is positive.
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Slide 7-30
Example 7.16 Angular acceleration of a falling
pole (cont.)
ASSESS The
final result for the
angular acceleration did not
depend on the mass, as we
might expect given the analogy
with free-fall problems. And
the final value for the angular
acceleration is quite modest.
This is reasonable: You can see that the angular acceleration is
inversely proportional to the length of the pole, and it’s a long
pole. The modest value of angular acceleration is fortunate—the
caber is pretty heavy, and folks need some time to get out of the
way when it topples!
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Slide 7-31
Example 7.18 Starting an airplane engine
The engine in a small air-plane is
specified to have a torque of
500 N ⋅ m. This engine drives a
2.0-m-long, 40 kg single-blade
propeller. On start-up, how long
does it take the propeller to reach
2000 rpm?
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Slide 7-32
Example 7.18 Starting an airplane engine
(cont.)
PREPARE The
propeller can be
modeled as a rod that rotates
about its center. The engine
exerts a torque on the propeller.
FIGURE 7.38 shows the propeller
and the rotation axis.
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Slide 7-33
Example 7.18 Starting an airplane engine
(cont.)
ASSESS We’ve
assumed a constant
angular acceleration, which is
reasonable for the first few seconds
while the propeller is still turning
slowly. Eventually, air resistance
and friction will cause opposing
torques and the angular acceleration
will decrease. At full speed, the
negative torque due to air resistance
and friction cancels the torque of the engine. Then
and the propeller turns at constant angular velocity with no
angular acceleration.
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Slide 7-34
Why Athletes Need to Understand the Concept
of Torque
•  The concept of torque is a foundation of human
movement and is a core principle in physical therapy,
personal training, and weightlifting. All movement
generates torque to varying degrees and, in reality, it’s
what makes the world of biomechanics tick.
•  You push and pull countless objects in all directions every
day in the gym, and the human musculoskeletal system is
nothing more than an intricate system of levers and
pulleys. While we must generate torque to operate
these levers and pulleys, we need to gain awareness of
how to minimize torque to avoid injury.
Credit: http://breakingmuscle.com/strength-conditioning/why-athletes-need-to-understand-the-concept-of-torque
© 2015 Pearson Education, Inc.
Slide 7-35
•  The resolution of forces occurs when we can:
• 
•  Visualize the potential effect of a force on our body.
•  Determine how much torque we should generate.
•  Make our muscles and joints act in proportion to this
awareness.
Credit: http://breakingmuscle.com/strength-conditioning/why-athletes-need-to-understand-the-concept-of-torque
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Slide 7-36
•  So how can you get your torque on? The answer is simpler
than the explanation of torque may lead you to believe:
use your muscles. Often in the gym, I see people move
into a position without regard for the joint position or joint
support. They place ligaments and fascia in charge of
holding a joint together (wrong), rather than engaging
some of the 642 muscles they have at their disposal
(right). Muscles need to do the work of supporting the
joint under the strain of torque.
Credit: http://breakingmuscle.com/strength-conditioning/why-athletes-need-to-understand-the-concept-of-torque
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Slide 7-37
Example: The Snatch
A successful snatch is all about torque and how quickly you
can move your body around the bar, not the bar around your
body.
There is a moment at the apex of extension when
the sum of all up and down forces are zero. At that
point in time, you are exerting no torque, which is
precisely how humans are able to lift the amount of
weight that we do and avoid the pain and suffering of
an annihilating injury
Coaches must present the concept of torque to
athletes in a consistent and digestible fashion.
The health of your joints and your ability
to move mass will reflect your
understanding of the ability to generate
torque.
Credit: http://breakingmuscle.com/strength-conditioning/why-athletes-need-to-understand-the-concept-of-torque
© 2015 Pearson Education, Inc.
Slide 7-38
Everyday
See-saw
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Lifting an object
Slide 7-39
Torque in Automobiles!!
•  Torque is one of the terms
commonly thrown around
to describe how powerful a
car is, but what exactly
does it mean? In a car,
torque is applied by the
pistons on the crankshaft,
causing it and the wheels to
turn.
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http://study.com/academy/lesson/what-is-torque-definition-equation-calculation.html
Slide 7-40
Torque in automobiles
•  Initial energy that moves a car forward: from
thermodynamics. (Example: Internal Combustion engine).
•  This forces a piston (one or more pistons) down in a
straight line, which pushes on a connecting rod and turns
the engine's crankshaft.
•  It's this turning crankshaft where the twisting force of
torque initiates. From there the force is carried through a
flywheel, transmission, driveshaft, axle(s) and wheel(s)
before moving the car.
Credit: http://www.edmunds.com/car-technology/the-twist-on-torque.html
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Slide 7-41
Summary: General Principles
Text: p. 217
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Slide 7-42
Summary: Important Concepts
Text: p. 217
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Slide 7-43
Summary: Important Concepts
Text: p. 217
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Slide 7-44
Summary: Important Concepts
Text: p. 217
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Slide 7-45
Summary
Text: p. 217
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Slide 7-46
Summary
Text: p. 217
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Slide 7-47
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