Dynamics
description
Transcript of Dynamics
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Objectives
● To understand the concept of force and understand its vectorial character
● To establish the relationship between force and the variables that define the movement
● To understand the meaning of Newton's laws
● To identify the different types of motion in terms of applied forces.
● To know the physical meaning of dynamic quantities
● To know and understand the importance of the principles of conservation of energy, angular momentum, and know how to use for solving problems
● Transfer the concepts used for one point system to a system of material points
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Contents
1. Force and Motion: Newton's laws
2. Particle systems
3. Conservation Theorems
4. Work and Energy
BibliographyBibliography
F. Físicos de la Ingeniería II (SPUP-064)F. Físicos de la Ingeniería II (SPUP-064)
Física (Alonso-Finn) Física (Alonso-Finn)
Física (Tipler)Física (Tipler)
Temas Preliminares de Física (CD) (SPUPV-429)Temas Preliminares de Física (CD) (SPUPV-429)
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Relationship with other topics
● Mathematics / Physics
● Mechanics– Rigid body dynamics
– Center of mass calculation
Vectorsd r t dt ∫ dr dt
→ Relationship with various scientific and technological areas (Energy / Principle of conservation)
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Preliminary concepts
● Mass:● Force:● Momentum:● Moment of a force:● Angular momentum:
m
F⃗
p⃗=mv⃗
L⃗= r⃗× p⃗
F⃗
OP1
M⃗ O( F⃗ )= r⃗×F⃗
r⃗
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Newton's Laws
www.schooltube.com/video/2555122616ce8f15ffdb/Force-and-Motion-Newtons-Law-of-Inertia
Δ p⃗=0v⃗=CST
m
v⃗=0
m
First law. Every object
in a state of uniform
motion tends to remain
in that state of motion
unless an external
force is applied to it.
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Newton's Laws
m
m
F⃗F⃗
a⃗
a⃗
Second law. The
acceleration of an object is
in the direction of the net
external force acting on it. It
is proportional to the net
external force and is
inversely proportional to the
mass of the object.
F⃗=m a⃗
a⃗=F⃗m
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Newton's Laws
Third law. Forces
always occur in equal
and opposite pairs. If
object A exerts a force
on object B, an equal but
opposite force is exerted
by object B on object A.
A⃗=−R⃗
A⃗ −R⃗
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Newton's Laws
A⃗=−R⃗
Δ p⃗=0
F⃗=m a⃗
Problems: 1, 4
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Third law: The stubborn horse
According to Newton's third law, whatever force I exert on the cart, the cart will exert an equal and opposite force on me, so the net force will be zero and I will have no chance of accelerating the cart
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Frictional forces
Fluid: ∣F⃗k∣∝ f (v )
∣F⃗k∣=μk∣N⃗∣
∣F⃗ s∣=∣F⃗app∣<μ s∣N⃗∣
∣F⃗ s , k∣=μ∣N⃗∣Solids:
Static friction:
Kinematic friction:
coefficient of static friction
coefficient of kinetic frictionμk
μs
f⃗
F⃗app
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Frictional forces
coefficient of static friction
coefficient of kinetic frictionμk
μs
f⃗
F⃗app
Fluid: ∣F⃗k∣∝ f (v )
∣F⃗k∣=μk∣N⃗∣
∣F⃗ s∣=∣F⃗app∣<μ s∣N⃗∣
∣F⃗ s , k∣=μ∣N⃗∣Solids:
Static friction:
Kinematic friction:
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Applications of Newton's Laws
mg
mg sinθ i⃗(− j⃗)mgcosθ
F⃗ s ,maxN
X
Y
N=mgcos θ
F s ,max=m g sinθ
tanθ=μs
Problems: 6, 7
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Variable mass system
F⃗=d p⃗dt
● Second law. The acceleration of an object is in the direction of the net external force acting on it. It is proportional to the net external force and is inversely proportional to the mass of the object.
F⃗=m a⃗
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Variable mass system
v⃗v⃗+ d v⃗
m m+dm
dmv⃗ p
F⃗ F⃗
p⃗ i=m v⃗+dm v⃗ p p⃗ f=(m+dm)( v⃗+d v⃗ )
d p⃗= p⃗ f− p⃗i=md v⃗−( v⃗ p− v⃗ )dm=md v⃗−u⃗ dm
F⃗=d p⃗dt
=md v⃗dt
− u⃗dmdt
the velocity of the impacting material relative to the object
u⃗
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Variable mass system
Thrust: Rocket propulsion
p⃗i=m v⃗
t t+ d t
v⃗+ d v⃗v⃗
m m−dm
dm
v⃗+ u⃗
The gas is exhausted at a speed u relative to the rocket
v=v0+ u lnm0
m
F⃗=d p⃗dt
=md v⃗dt
− u⃗dmdt
For F⃗=0
Problems: 3,
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Conservation of Momentum
If the net external force on a system is zero, the total linear momentum of the system remains constant
R⃗=d p⃗dt
R⃗=0 ⇒ d p⃗=0
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Conservation of Angular Momentum
If the net external torque acting on a system is zero, the total angular momentum of the system is constant
d L⃗Adt
=ddt
( r⃗ A× p⃗ )=( d r⃗ Adt × p⃗+ r⃗ A×d p⃗dt )=
=r⃗ A×d p⃗dt
=r⃗ A×F⃗=M⃗ A
M⃗ A=0 ⇒ d L⃗A=0
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Systems of particles
Point particles
Extended objects
mi
mi
mj
mk m=∑i=1
n
mi
Forces and momentum
{ R⃗ , M⃗ 0}f⃗ ijf⃗ ji
F⃗ iF⃗ j
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The Centre of Mass
r⃗ cm=∑ mi r⃗im
r⃗ cm=∫ r⃗ dmm
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G
Centre of gravity
mi g⃗im j g⃗ j
Centre of the parallel sliding vector system
r⃗G=∑i
(migi r⃗ i )
∑i
migi
r⃗ cm=∑i
(mi r⃗ i )
∑i
mi
r⃗ i
r⃗ j
O⃗C=
∑i
(v i*O⃗Pi )
∑i
v i* =
∑i
(v iO⃗Pi)
∑i
v i
Problems: 8, 9, 10
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Centre of mass theoremP1O
0
F → External forces
f → Internal forces
f⃗ ijf⃗ ji
F⃗ i
F⃗ jr⃗ j
r⃗ i
F⃗ i+∑j
f⃗ ij=mi a⃗ i
∑i
F⃗ i+∑i , j
f⃗ ij=∑i
mi a⃗ i
R⃗=∑i
mid 2 r⃗idt 2
=
d 2
(∑i mi r⃗ i)dt 2
=d 2
(m r⃗ cm )
dt 2=m
d 2 r⃗ cmdt 2
R⃗=md 2 r⃗ cmdt 2
R⃗=m a⃗cm
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Centre of mass theorem
CM
m1m2
m3
mi
...
≡ CM
M
R
R⃗=md 2 r⃗ cmdt 2
R⃗=m a⃗cm
F⃗1 F⃗ 2
F⃗3
F⃗ i
The motion of any object or system of particles can be described in terms of the motion of the centre of mass plus the motion of individual particles in the system relative to the centre of mass
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The Centre of MassThe motion of any object or system of particles can be described in terms of the motion of the centre of mass plus the motion of individual particles in the system relative to the centre of mass
R⃗=md 2 r⃗ cmdt 2
R⃗=m a⃗cmhammer
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Conservation of Linear Momentum
R⃗=d∑ p⃗idt
=d p⃗cmdt
=d p⃗dt
R⃗=0 → Δ p⃗=0
If the net external force on a system is zero, the total momentum of thesystem remains constant.
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Conservation of angular momentum
d L⃗Adt
=ddt
∑i
( r⃗ Ai× p⃗i)=∑i
( d r⃗ Aidt× p⃗ i+ r⃗ Ai×
d p⃗idt )=
=∑i
( r⃗ Ai×d p⃗idt )=∑i
r⃗ i×( F⃗ i+ f⃗ i)=∑i
r⃗ i×F⃗ i=M⃗ A(*)
M⃗ A=d∑i
L⃗Aidt
=d L⃗Adt
M⃗ A=0 → Δ L⃗A=0
If the net external torque acting on a system is zero, the total angular momentum of the system is constant.
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Conservation of angular momentum
∑i
r⃗ i× f⃗ i=∑i ( r⃗ i×∑
j
f⃗ ij)=0
r⃗ i× f⃗ ij+ r⃗ j× f⃗ ji=
O
=( r⃗ i− r⃗ j)× f⃗ ij=
= r⃗ ij× f⃗ ij=0
(*)
f⃗ ij
f⃗ jir⃗ ij
r⃗ j
r⃗ i
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Conservation theorems
Linear momentum R⃗=d∑ p⃗idt
=d p⃗cmdt
=d p⃗dt
R⃗=0 → Δ p⃗=0
Angular momentum M⃗ A=d∑i
L⃗Aidt
=d L⃗Adt
M⃗ A=0 → Δ L⃗A=0
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Conservation theorems: applications
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Conservation theorems: applications
p⃗i p⃗f
F⃗L
F⃗
p⃗f
p⃗iΔ p⃗
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Work and energy
Work
W AB=∫
A
B
F⃗⋅d r⃗=12m (vB
2−v A
2 )=K B−K A
Power P=dWdt
Kinetic energy: kinetic energy theorem
W AB=∫
A
B
F⃗⋅d r⃗=∫A
B
M⃗⋅d θ⃗
Kinetic energy in a system of particles:
K=∑i
12mi v i
2=
12 (∑i mi)vcm
2+
12∑i
mi v ' i2
Ec≡K
Problem: 13
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Potential energy
Potential energy is the energy stored in a body or in a system due to its position in a force field or due to its configuration.
U=mgh
h
U=12k x2
Problem: 2, 14, 15
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Potential energy
F (x)=−dUdx
∫A
B
F (x)dx=−ΔU=U A−UB
Ep≡U
A force is conservative if the total work it does on a particle is zero when the particle moves around any closed path returning to its initial position.
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Conservation of mechanical energy
W AB=W T=W c+W nc=Δ K
W T=−Δ U+W nc=Δ K
W nc=Δ K+ Δ U
W nc=0 ⇒ Δ K+ Δ U=0
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Conservation of mechanical energy
Potential energy HIGHKinetic energy low
Potential energy lowKinetic energy HIGH
W nc=0 ⇒ Δ K+ Δ U=0
Roller coaster
Problems: 5, 11, 12, 16