Transcript 4.1 The Concepts of Force and Mass

Review: using F=ma
• Force in circular motion
– 𝑎𝑐 and 𝑎𝑡 in curvilinear motion (e.g. nonuniform circular motion)
• Note on other important forces
– Propulsion
– Resistive force (Drag in air/liquid), terminal
velocicty 𝑣𝑇
– Conservative force, spring force Potential
energy
Relevant concepts:
“inertia” and force
a) Inertial reference frame
b) accelerated frame of the car
- fictitious force
Rolling friction
Conical Pendulum (e.g. ชิงช้าหมุน)
•The object is in equilibrium in the
vertical direction.
•It undergoes uniform circular
motion in the horizontal direction.
∑𝐹𝑦 = 0 → 𝑇 cos 𝜃 = 𝑚𝑔
∑𝐹𝑥 = 𝑇 sin 𝜃 = 𝑚 𝑎𝑐
•v is independent of m
Car traveling in a banked curve
•Design the curve with no friction
–in equilibrium in the vertical direction.
–in uniform circular motion in the horizontal direction a component of the normal
force supplies the centripetal force.
•The angle of bank is
ta n  
•Note:
v
2
rg
–The banking angle is independent of the mass of the vehicle.
–If the car rounds the curve at less than the design speed, friction is necessary to
keep it from sliding down the bank.
–If the car rounds the curve at more than the design speed, friction is necessary to
keep it from sliding up the bank.
Car traveling in a horizontal (Flat) Curve
•uniform circular motion in the
horizontal direction.
•in equilibrium in the vertical direction.
•The force of static friction
supplies the centripetal force.
•The maximum speed at which
the car can negotiate the curve
is:
v   s gr
–
Note: this does not depend on the
mass of the car.
Section 6.1
Ferris Wheel
•Uniform circular motion with
constant speed v (controlled by
the motor)
𝑭𝒏𝒆𝒕
𝑭𝒏𝒆𝒕
𝑚𝑣 2
=
(−𝒓)
𝑅
𝑭𝒏𝒆𝒕
•Under gravity, the child feels
apparent weight differently at top
and bottom
Section 6.1
Ferris Wheel (2)
•At the bottom of the loop,
the upward force (the
normal) experienced by
the object is greater than
its weight.
F
 n bot  m g 
mv
•At the top of the circle,
the force exerted on the
object is less than its
weight.
2
F
r

v 
n bot  m g  1 

rg 

2
n to p
Section 6.1
 n to p  m g 
v2

 mg 
 1
 rg

mv
r
2
Non-uniform Circular Motion
If the speed also changes in magnitude
there is non-zero tangential acceleration
Section 6.2
Vertical Circle with Non-Uniform Speed
•Launch by shooting (say, from the
bottom at initial velocity 𝒗𝒃𝒐𝒕 ). The
string restricts motion to a circle of
radius R (1 DOF)
•What is T and
acceleration?
•Tension depends on position 𝜃(𝑡)
d𝜃
and speed 𝑣 𝑡 = 𝑑𝑡 𝑅 ,both varying
with time
•The tension at the bottom is a
maximum.
•The tension at the top is a minimum.
•Requires critical speed 𝑣𝑡𝑜𝑝 ≥
𝑅𝑔 to complete the circle
(𝑇𝑡𝑜𝑝 ≥ 0 at 𝜃 = 𝜋)
Section 6.2
 v2

T  mg 
 co s  
 Rg

Resistive force (or drag force in air/fluid) and
terminal velocity 𝑣𝑇 (this topic is not in the test)
Resistive force 𝑭𝑫 could be of either
form(linear or quadratic to 𝑣), but always
opposite direction to motion (−𝒗)
1. Linear (Stokes law in liquid)
𝑭𝑫 ∝ − 𝒗
𝐹𝐷 = −𝑏𝑣
vT 
mg
b
2. Quadratic (air resistance)
𝑭𝑫 ∝ −𝑣 2
1
𝐹𝐷 = − 𝐶𝐷 𝜌𝐴 𝑣 2
2
𝐶𝐷 is the drag coefficient (empirical, no dimension,
1
typically 𝐶𝐷 ≈ ).
2
𝜌 is the density of air.
𝐴 is the cross-sectional area of the object
𝑣𝑇 ≈
4𝑚𝑔
𝜌𝐴
Lecture 4
Work and Energy
Outline
• Work done by a constant force
– Projection and scalar product of vectors
• Force that results in positive work.
– negative work? forces that do no work?
• Work done by varying force with
displacement
• Work-energy theorem
6.1 Work Done by a Constant Force
W  Fs
1 N  m  1 joule
J 
6.1 Work Done by a Constant Force
Work, cont.
•W = F Dr cos 
– A force does no work on
the object if the force does
not move through a
displacement.
– The work done by a force
on a moving object is zero
when the force applied is
perpendicular to the
displacement of its point of
application.
Section 7.2
6.1 Work Done by a Constant Force
W   F cos  s

cos 0  1
cos 90  0

cos 180   1

6.1 Work Done by a Constant Force
Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0-N, the angle is 50.0
degrees, and the displacement is 75.0 m.

W   F cos  s   45 . 0 N  cos 50 . 0
 2170 J

75 .0 m 
6.1 Work Done by a Constant Force
W   F cos 0 s  Fs
W   F cos 180 s   Fs
6.1 Work Done by a Constant Force
Example 3 Accelerating a Crate
The truck is accelerating at
a rate of +1.50 m/s2. The mass
of the crate is 120-kg and it
does not slip. The magnitude of
the displacement is 65 m.
What is the total work done on
the crate by all of the forces
acting on it?
f s  m a  120 kg  1.5 m s
2
  180 N
W  180 N  cos 0 65 m   1 . 2  10 J
4
6.1 Work Done by a Constant Force
The angle between the displacement
and the friction force is 0 degrees.
f s  ma  120 kg 1 . 5 m s
2
  180 N
W  180 N  cos 0 65 m   1 . 2  10 J
4
Scalar Product of Two Vectors
A  B  A B cos 
W  F D r co s   F  D r
Section 7.3
Scalar Product, cont
•The scalar product is commutative.
A B  B  A
•The scalar product obeys the distributive law of
multiplication.


A  B  C  A B  A  C
Section 7.3
Dot Products of Unit Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  1
ˆi  ˆj  ˆi  kˆ  ˆj  kˆ  0
Using component form with vectors:
A  A x ˆi  A y ˆj  A z kˆ
B  B x ˆi  B y ˆj  B z kˆ
A B  Ax B x  Ay B y  Az B z
In the special case where
A  B;
A  A  A x  Ay  Az  A
2
Section 7.3
2
2
2
6.2 The Work-Energy Theorem and Kinetic Energy
Consider a constant net external force acting on an object.
The object is displaced a distance s, in the same direction as
the net force.
F
s
The work is simply
W 
 F s  ma s
6.2 The Work-Energy Theorem and Kinetic Energy
W  m  as   m
1
2
v
v
2
f

2
o
1
2
mv
2
f

1
2
v f  v o  2  ax 
2
2
 ax  
1
2
v
2
f
 vo
2

DEFINITION OF KINETIC ENERGY
The kinetic energy KE of and object with mass m
and speed v is given by
KE 
1
2
mv
2
mv
2
o
6.2 The Work-Energy Theorem and Kinetic Energy
THE WORK-ENERGY THEOREM
When a net external force does work on and object, the kinetic
energy of the object changes according to
W  KE f  KE o 
1
2
mv 
2
f
1
2
mv
2
o
6.2 The Work-Energy Theorem and Kinetic Energy
Example 4 Deep Space 1
The mass of the space probe is 474-kg and its initial velocity
is 275 m/s. If the 56.0-mN force acts on the probe through a
displacement of 2.42×109m, what is its final speed?
6.2 The Work-Energy Theorem and Kinetic Energy
W 
W 
 F  cos  s
1
2
mv 
2
f
1
2
mv
2
o
6.2 The Work-Energy Theorem and Kinetic Energy
 F  cos  s 
5.60  10
-2



N cos 0 2 . 42  10 m 

9
1
2
1
2
mv f 
2
 474
1
2
kg v f 
2
2
mv o
1
2
 474
kg  275 m s 
v f  805 m s
2
6.2 The Work-Energy Theorem and Kinetic Energy
In this case the net force is


F  mg sin 25  f k
6.2 The Work-Energy Theorem and Kinetic Energy
Conceptual Example 6 Work and Kinetic Energy
A satellite is moving about the earth
in a circular orbit and an elliptical orbit.
For these two orbits, determine whether
the kinetic energy of the satellite
changes during the motion.
W>0
W<0
W>0
W<0
6.3 Gravitational Potential Energy
W   F cos  s
W gravity  mg h o  h f

6.3 Gravitational Potential Energy
W gravity  mg h o  h f
Work due to gravity is
independent of path!
We call this phenomenon
“conservative force”
 We can define “potential
energy” based on this force

6.3 Gravitational Potential Energy
Example 7 A Gymnast on a Trampoline
The gymnast leaves the trampoline at an initial height of 1.20 m
and reaches a maximum height of 4.80 m before falling back
down. What was the initial speed of the gymnast?
6.3 Gravitational Potential Energy
W gravity  mgh o  mgh
f
DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy PE is the energy that an
object of mass m has by virtue of its position relative to the
surface of the earth. That position is measured by the height
h of the object relative to an arbitrary zero level:
PE  mgh
1 N  m  1 joule
J 
6.4 Conservative Versus Nonconservative Forces
DEFINITION OF A CONSERVATIVE FORCE
Version 1 A force is conservative when the work it does
on a moving object is independent of the path between the
object’s initial and final positions.
Version 2 A force is conservative when it does no work
on an object moving around a closed path, starting and
finishing at the same point.
6.4 Conservative Versus Nonconservative Forces
6.4 Conservative Versus Nonconservative Forces
Version 1 A force is conservative when the work it does
on a moving object is independent of the path between the
object’s initial and final positions.
W gravity  mg h o  h f

6.4 Conservative Versus Nonconservative Forces
Version 2 A force is conservative when it does no work
on an object moving around a closed path, starting and
finishing at the same point.
W gravity  mg h o  h f

ho  h f
6.4 Conservative Versus Nonconservative Forces
An example of a nonconservative force is the kinetic
frictional force.
W   F cos  s  f k cos 180 s   f k s

The work done by the kinetic frictional force is always negative.
Thus, it is impossible for the work it does on an object that
moves around a closed path to be zero.
The concept of potential energy is not defined for a
nonconservative force.
6.4 Conservative Versus Nonconservative Forces
In normal situations both conservative and nonconservative
forces act simultaneously on an object, so the work done by
the net external force can be written as
W  W c  W nc
W  KE f  KE o  D KE
W c  W gravity  mgh o  mgh
f
 PE o  PE f   D PE
Spring force and Spring (Elastic) Potential Energy
6.4 Conservative Versus Nonconservative Forces
Work done by conservative force
W  W c  W nc
Work done by non-conservative force
D KE   D PE  W nc
THE WORK-ENERGY THEOREM
W nc  D KE  D PE
6.5 The Conservation of Mechanical Energy
W nc  D KE  D PE  KE f  KE o    PE f  PE o 
W nc  E f  E o
If the net work on an object by nonconservative forces
is zero, then its energy does not change:
Ef  Eo
6.5 The Conservation of Mechanical Energy
THE PRINCIPLE OF CONSERVATION OF
MECHANICAL ENERGY
The total mechanical energy (E = KE + PE) of an object
remains constant as the object moves, provided that the net
work done by external nononservative forces is zero.
6.5 The Conservation of Mechanical Energy
6.5 The Conservation of Mechanical Energy
Example 8 A Daredevil Motorcyclist
A motorcyclist is trying to leap across the canyon by driving
horizontally off a cliff 38.0 m/s. Ignoring air resistance, find
the speed with which the cycle strikes the ground on the other
side.
6.5 The Conservation of Mechanical Energy
Ef  Eo
mgh
f

1
2
mv
gh f 
1
2
2
f
 mgh o 
v f  gh o 
2
1
2
1
2
mv
2
vo
2
o
6.5 The Conservation of Mechanical Energy
gh f 
vf 
vf 

2 9 .8 m s
2
1
2
v f  gh o 
2
2 g h
o
 hf
1
2
2
vo
 v
2
o
35 .0 m   38 .0 m s 
2
 46 . 2 m s
6.5 The Conservation of Mechanical Energy
Conceptual Example 9 The Favorite Swimming Hole
The person starts from rest, with the rope
held in the horizontal position,
swings downward, and then lets
go of the rope. Three forces
act on him: his weight, the
tension in the rope, and the
force of air resistance.
Can the principle of
conservation of energy
be used to calculate his
final speed?
6.7 Power
DEFINITION OF AVERAGE POWER
Average power is the rate at which work is done, and it
is obtained by dividing the work by the time required to
perform the work.
P 
Work
Time
joule

W
t
s  watt (W)
6.7 Power
P 
Change in energy
Time
1 horsepower
 550 foot  pounds
P  Fv
second  745.7 watts
6.7 Power
6.8 Other Forms of Energy and the Conservation of Energy
THE PRINCIPLE OF CONSERVATION OF ENERGY
Energy can neither be created not destroyed, but can
only be converted from one form to another.
• Where does work due to
nonconservative force (Wnc) go
to, other than kinetic energy?
• Thermal energy (heat) as
internal energy of materials
• This means energy of the
universe (isolated system) is
always conserved.
– When we say “not energy
conserving” we mean just in the subsystem.
Lecture 5
Impulse and Momentum
7.1 The Impulse-Momentum Theorem
There are many situations when the
force on an object is not constant.
7.1 The Impulse-Momentum Theorem
DEFINITION OF IMPULSE
The impulse of a force is the product of the average
force and the time interval during which the force acts:


J  F Dt
Impulse is a vector quantity and has the same direction
as the average force.
newton  seconds (N  s)
7.1 The Impulse-Momentum Theorem


J  F Dt
7.1 The Impulse-Momentum Theorem
DEFINITION OF LINEAR MOMENTUM
The linear momentum of an object is the product
of the object’s mass times its velocity:


p  mv
Linear momentum is a vector quantity and has the same
direction as the velocity.
kilogram
 meter/seco nd (kg  m/s)
7.1 The Impulse-Momentum Theorem

a 



vf  vo


F  ma

 m v  m v
f
o
F 
Dt
 



F Dt  mv f  mv o
Dt
7.1 The Impulse-Momentum Theorem
IMPULSE-MOMENTUM THEOREM
When a net force acts on an object, the impulse of
this force is equal to the change in the momentum
of the object
impulse
 



F Dt  mv f  mv o
final momentum
initial momentum
7.1 The Impulse-Momentum Theorem
Example 2 A Rain Storm
Rain comes down with a velocity of -15 m/s and hits the
roof of a car. The mass of rain per second that strikes
the roof of the car is 0.060 kg/s. Assuming that rain comes
to rest upon striking the car, find the average force
exerted by the rain on the roof.
 



F Dt  mv f  mv o
7.1 The Impulse-Momentum Theorem
Neglecting the weight of
the raindrops, the net force
on a raindrop is simply the
force on the raindrop due to
the roof.



F Dt  mv f  mv o

 m 
F  
vo
 Dt 

F   0 . 060 kg s  15 m s    0 . 90 N
7.1 The Impulse-Momentum Theorem
Conceptual Example 3 Hailstones Versus Raindrops
Instead of rain, suppose hail is falling. Unlike rain, hail usually
bounces off the roof of the car.
If hail fell instead of rain, would the force be smaller than,
equal to, or greater than that calculated in Example 2?
7.2 The Principle of Conservation of Linear Momentum
sum
of average external


forces D t  P f  P o
If the sum of the external forces is zero, then


0  Pf  Po


Pf  Po
PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM
The total linear momentum of an isolated system is constant
(conserved). An isolated system is one for which the sum of
the average external forces acting on the system is zero.
7.2 The Principle of Conservation of Linear Momentum
Conceptual Example 4 Is the Total Momentum Conserved?
Imagine two balls colliding on a billiard
table that is friction-free. Use the momentum
conservation principle in answering the
following questions. (a) Is the total momentum
of the two-ball system the same before
and after the collision? (b) Answer
part (a) for a system that contains only
one of the two colliding
balls.
7.2 The Principle of Conservation of Linear Momentum
Example 6 Ice Skaters
Starting from rest, two skaters
push off against each other on
ice where friction is negligible.
One is a 54-kg woman and
one is a 88-kg man. The woman
moves away with a speed of
+2.5 m/s. Find the recoil velocity
of the man.
7.2 The Principle of Conservation of Linear Momentum


Pf  Po
m1v f 1  m 2 v f 2  0
vf2  
vf2  
54
m1v f 1
m2
kg   2 . 5 m s 
88 kg
  1 .5 m s
7.2 The Principle of Conservation of Linear Momentum
Applying the Principle of Conservation of Linear Momentum
1. Decide which objects are included in the system.
2. Relative to the system, identify the internal and external forces.
3. Verify that the system is isolated.
4. Set the final momentum of the system equal to its initial momentum.
Remember that momentum is a vector.
การชนกัน (collision)
- ในการชนส่ วนใหญ่ เราไม่ทราบแรงชน จึงไม่สามารถวิเคราะห์
การเคลื่อนที่โดยใช้กฎของนิวตันได้
- แต่หากทราบลักษณะการเคลื่อนที่ (ความเร็ ว) การวิเคราะห์การเคลื่อนที่
สาหรับการชนสามารถใช้หลักอนุรักษ์ โมเมนตัมได้
- กรณี ที่เป็ นการชนแบบยืดหยุ่น เป็ นการชนแบบไมสู
่ ญเสี ย
พลังงานจลน์
สามารถใช้หลักอนุรักษ์ พลังงานจลน์ ซ่ ึง
หมายถึง พลังงานจลน์รวมของระบบก่อนและหลังชนมีค่าเท่ากัน
- กรณี ที่เป็ นการชนแบบไม่ ยดื หยุ่น พลังงานจลนของระบบ
์
จะลดลง
เนื่องจากพลังงานจลน์เปลี่ยนรู ปเป็ น เสี ยง ความร้อน เป็ นต้น
80
เชน หลังชนวัตถุตด
ิ กัน
7.3 Collisions in One Dimension
The total linear momentum is conserved when two objects
collide, provided they constitute an isolated system.
Elastic collision -- One in which the total kinetic
energy of the system after the collision is equal to
the total kinetic energy before the collision.
Inelastic collision -- One in which the total kinetic
energy of the system after the collision is not equal
to the total kinetic energy before the collision; if the
objects stick together after colliding, the collision is
said to be completely inelastic.
Example of elastic collision in 1D
7.3 Collisions in One Dimension
Example 8 A Ballistic Pendulum
The mass of the block of wood
is 2.50-kg and the mass of the
bullet is 0.0100-kg. The block
swings to a maximum height of
0.650 m above the initial position.
Find the initial speed of the
bullet.
7.3 Collisions in One Dimension
Apply conservation of momentum
to the collision:
m 1v f 1  m 2 v f 2  m 1v o1  m 2 v o 2
m 1  m 2 v f
 m1v o1
Inelastic collision
v o1 
m 1  m 2 v f
m1
7.3 Collisions in One Dimension
Applying conservation of energy
to the swinging motion:
mgh 
 m 1  m 2  gh f
1
2
mv

1
2
gh f 
vf 
2 gh f 
2
 m 1  m 2 v 2f
1
2
2
vf

2 9 . 80 m s
2
0 .650 m 
7.3 Collisions in One Dimension
vf 
v o1 
v o1
2 9 . 80 m s
2
0 . 650 m 
m 1  m 2 v f
m1
 0 . 0100 kg  2 . 50 kg 
2
 2 9 . 80 m s 0 . 650 m    896 m s
 
0.0100 kg




7.5 Center of Mass
The center of mass is a point that represents the average location for
the total mass of a system.
x cm 
m 1 x1  m 2 x 2
m1  m 2
7.5 Center of Mass
D x cm 
m 1 D x1  m 2 D x 2
m1  m 2
v cm 
m 1 v1  m 2 v 2
m1  m 2
7.5 Center of Mass
v cm 
m 1 v1  m 2 v 2
m1  m 2
In an isolated system, the total linear momentum does not change,
therefore the velocity of the center of mass does not change.
7.5 Center of Mass
BEFORE
v cm 
m 1 v1  m 2 v 2
m1  m 2
0
AFTER
v cm 
88 kg  1 . 5 m s   54
kg   2 . 5 m s 
88 kg  54 kg
 0 . 002  0