Transcript P1710_MWFsummary

Physics 1710
Section 004
Mechanics and Thermodynamics
Final Review
Physics 1710
MWF Session 1 Introduction
The “Structure” of this course:
Dynamics
Fluid Mechanics
Gravitation
Elasticity
Applications
Statics
Kinematics
Oscillations
Waves
Thermodynamics
Physics 1710—Chapter 1 Measurement
Summary
•Fundamental Dimensions and Units
Time, measured in seconds;
Length, measured in meters;
Mass, measured in kilograms.
• Prefixes scale units to convenient size.
k =1000, M = 1 000 000
c = 1/100, m = 1/1000, μ =1/1 000 000
• Density is mass per unit volume.
ρ = m/V [kg/m3 ]
• Avogadro’s number is the number of atoms
in a mole of an element. 6.022 x1023 atom/mole
Physics 1710
Chapter 2 Motion in One Dimension—II
Summary:
• The change in the instantaneous velocity is equal to the
(constant) acceleration multiplied by its duration. ∆v = at
•The displacement is equal to the displacement at constant
velocity plus one half of the product of the acceleration and
the square of its duration.
∆x = vinitial t + ½ at 2
•The change in the square of the velocity is equal to two
times the acceleration multiplied by the distance traveled
during acceleration.
∆v 2 = 2a ∆x
•The acceleration of falling bodies is 9.8 m/s/s downward.
a = - g = - 9.8 m/s/s
Physics 1710 Chapter 3 Vectors
Summary:
•To add vectors, simply add the components separately.
•Use the Pythagorean theorem for the magnitude.
•Use trigonometry to get the angle.
•The vector sum will always be equal or less than the
arithmetic sum of the magnitudes of the vectors.
Physics 1710 Chapter 4: 2-D Motion—II
Summary:
• Kinematics in two (or more) dimensions obeys the
same 1- D equations in each component
independently.
– rfinal = rinitial + vinitial t + ½ a t 2
– vfinal = vinitial + a t
– vx,final2 = vx,initial 2+ 2 ax /∆x
– vy,final2 = vy,initial 2+ 2 ay /∆y
• Projectiles follow a parabola [y(x) = A + Bx +Cx2]
Physics 1710 Chapter 4: 2-D Motion—II
Summary:
• In a moving or accelerating Frame of Reference
• v ′ = v – vframe of reference
• a ′ = a – aframe of reference
• The Centripetal acceleration is
• a = - ω2 r
or |a| = v 2/ |r|, toward the center.
Physics 1710 Chapter 5: Laws of Motion—II
Summary:
• Newton’s Laws of Motion are:
(1) Acceleration (or deceleration) occurs if and only
if there is a net external force.
(2) a = F/m [Note this is a vector eqn.]
(3) The force exerted by a first object on a second is
always equal and opposite the the force exerted by
the second on the first. F12 = - F21
Physics 1710 Chapter 5: Laws of Motion—II
Summary (cont’d.) :
• Weight is the force of gravity equal to g times the
mass of the object.
• g =9.80 N/kg
• The force of friction is opposed to the motion of a
body and proportional to the normal force.
• Free body diagrams are sketches of all the forces
acting on a body.
Physics 1710 Chapter 6—Circular Motion
Summary
• The net force on a body executing circular motion is
equal to the mass times the centripetal acceleration of the
body.
•acentripedal = v 2/ R [toward the center]
• The “centrifugal” force is a fictitious force due to a noninertial frame of reference.
Physics 1710 Chapter 7—Work
Summary
• Work is defined to be the distance traveled
multiplied by the distance over which the force
acts.
• W = ∫ F•d r
•[Joules] = [N ][m]
Physics 1710 Chapter 7&8—Power & Energy
Summary:
•The Potential Energy is equal to the negative of
the work done on the system to put it in its
present state.
U = -∫ F•d r
• The sum of all energy, potential and kinetic, of a
system is conserved, in the absence of
dissipation.
E=U+K–W
• F = - ∇U
•P = dE/dt
Physics 1710—Chapter 1 Measurement
Summary
• F = - ∇U = negative gradient of U.
• The Potential Energy graph is a complete
description of the dynamics of a system.
Physics 1710—Chapter 10 Rotating Bodies
Summary:
•Angular displacement is the angle through
which a body has rotated.
•Instantaneous angular speed is the time rate
of angular displacement.
•Instantaneous angular acceleration is the time
rate of change in angular speed.
Physics 1710—Chapter 10 Rotating Bodies
Summary (cont’d):
•The moment of inertia is the measure of the
(inertial) resistance to angular acceleration and
equal to the second moment of the mass
distribution.
•Torque (“twist”) is the vector product of a force
and the “moment” arm.
Physics 1710—Chapter 10 Rotating Bodies
Summary:
•The moment of inertia I is the measure of the
(inertial) resistance to angular acceleration and
equal to the second moment of the mass
distribution about an axis.
Physics 1710—Chapter 11 Rotating Bodies
Summary:
•The total Kinetic energy of a rotating system is the sum
of the rotational energy about the Center of Mass and
the translational KE of the CM.
K = ½ ICM ⍵ 2 + ½ MR 2 ⍵ 2
τ=rxF
Physics 1710—Chapter 11 Rotating Bodies
Summary:
•Angular momentum L is the vector product of the
moment arm and the linear momentum.
L=rxp
• The net externally applied torque is equal to the time
rate of change in the angular momentum.
∑ τz = d Lz /dt = Iz ⍺
Physics 1710—Chapters 6-10
Summary
• Rotary (circular) motion obeys laws that are analogous
to those of translational motion.
• Linear Momentum is conserved in absence of external
forces.
• F = d p/dt
• Energy is related to the work done or stored
• Work is the cumulative force times distance moved.
• Power is the rate of expenditure of work or energy.
• Force is the negative of the gradient of the potential.
Physics 1710—Chapter 11 Rotating Bodies
Summary:
• Angular momentum about an axis z is equal to the
product of the moment of inertia of the body about that axis
and the angular velocity about z.
L=I⍵
Lz = Iz ⍵
• In the absence of torques, the angular momentum is
conserved.
• In the presence of torques the angular moment will
change with time.
Physics 1710—Chapter 11 App: E & E
Summary
•
Static equilibrium implies that all forces and
torques balance.
•
The center of mass is often the center of
gravity.
•
The moduli of elasticity characterizes the
stress-strain relation:
•
stress= modulus x strain
Stress = modulus x strain
σ = F/A = Y ε
Physics 1710—Chapter 13 Apps: Gravity
Summary:
• The force of attraction between two bodies with mass
M and m respectively is proportional to the product of
their masses and inversely proportional to the distance
between their centers squared.
F = - G M m/ r 2
• The proportionality constant in the Universal Law of
Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 .
Physics 1710—Chapter 13 Apps: Gravity
Summary:
•The gravitational force constant g is equal to
G M/(R+h) 2, R is the radius of the planet.
• Kepler’s Laws
–The orbits of the planets are ellipses.
–The areal velocity of a planet is constant.
–The cube of the radius of a planet’s orbit
is proportional to the square of the period.
• The gravitation field is the force divided by the mass.
g = Fg / m
Physics 1710—Chapter 13 Apps: Gravity
Summary:
• The force of attraction between two bodies with mass
M and m respectively is proportional to the product of
their masses and inversely proportional to the distance
between their centers squared.
F = - G M m/ r 2
• The proportionality constant in the Universal Law of
Gravitation G is equal to 6.673 x 10 –11 N m2 /kg2 .
Physics 1710—Chapter 13 Apps: Gravity
Summary:
• The gravitation potential energy for a point mass is
proportional to the product of the masses and inversely
proportional to the distance between their centers:
U = GMm / r
• The escape velocity is the minimum speed a projectile must
have at the surface of a planet to escape the gravitational
field.
vescape = √[ 2GM/R]
• Total Energy E is conserved for two body geavitational
problem; bodies are bound for E ≤ 0
E = L2/2mr 2 – GMm/r
Physics 1710—Chapter 14 Fluid Dynamics
Summary:
•Pressure is the force per unit area. P =F/A
• Unit of pressure [Pacal] = [N]/[m2]
•The hydrostatic pressure is P = Po + ρgh
• Archimedes’ Principle: Fbouyant = ρfluid g V
• Equation of Continuity: A1v1 = A2v2
• Bernoulli’s Equation: P + ½ ρv2 + ρgy =
constant.
Physics 1710—Chapter 15 SHO
Summary :
• Simple Harmonic Motion is sinusoidal.
x = Xo cos(ωt +φ)
• The period is the reciprocal of the frequency.
• For a mass m on a spring of spring constant k,
the period
T = 1/ f
T = 2π√(m/k)
• For Damped SHO, the frequency is decreased
and the amplitude decays exponentially.
x = Xo e – ½ (b/m)t cos(ωt +φ)
with ω = √[k/m – ½ b/m]
Physics 1710—Chapter 15 SHO
Summary :
• For a driven SHO the amplitude is a maximum when the
drive frequency is equal to the natural frequency; a condition
known as “resonance.”
• A simple pendulum oscillates at a frequency of
f = (1/2π) √(g/L)
•A physical pendulum oscillates at a frequency of
f = (1/2π) √(mgL/I)
Physics 1710—Chapter 16 Waves
Summary :
• A traveling wave has the form
y(x,t) = Y sin(kx – ωt),
with k = 2π/λ, k : wave number, λ : wavelength
& ω= 2π f = 2π/ T as previously defined
• d 2y/dx 2 = (1/v 2) d 2y/dt 2 is the linear wave equation.
• λ f = v, the phase velocity.
• For a longitudinal wave on a string v = √(T/μ).
T = tension, μ = dm/dx = linear mass density
• The time averaged power transmitted on a string is
₧ = ½ μ ω2A2v
Physics 1710—Chapter 17 Sound
Summary :
• Sound is a longitudinal pressure/displacement
•V = √B/ρ, the phase velocity is equal to the square root of
the ratio of the bulk modulus to the density.
• The Doppler effect is a shift in frequency due to the relative
motion of the source and observer of a sound.
Physics 1710—Chapter 16 Waves
Summary :
• A traveling wave has the form
y(x,t) = Y sin(kx – ωt),
with k = 2π/λ, k : wave number, λ : wavelength
& ω= 2π f = 2π/ T as previously defined
• d 2y/dx 2 = (1/v 2) d 2y/dt 2 is the linear wave equation.
• λ f = v, the phase velocity.
• For a longitudinal wave on a string v = √(T/μ).
T = tension, μ = dm/dx = linear mass density
• The time averaged power transmitted on a string is
₧ = ½ μ ω2A2v
Physics 1710—Chapter 18 Chapter 18
Superposition and Standing Waves
Summary:
The propagation of waves is characterized by
Reflection — the rebound of the wave.
Refraction — the bending of a wave’s
direction due to a velocity gradient
Diffraction — the bending of a wave
around obstacles.
Interference — the combination of two
or more waves in space.
Beats — the combination of two waves
in time.
Physics 1710—Chapter 18 Chapter 18
Superposition and Standing Waves
Summary:
•
•
•
•
•
Angle of incidence = angle of reflection;
θi=θr
sin θ 1 /v1 = sin θ 2 / v2
fave = ( f1 + f2 )/2; fbeat = ( f1 - f2 )
fn = n /(2L) √(T/μ)
A = (Fext /m)/ [ω0 2 - ω2]
Physics 1710
Chapter 19 Temperature
Summary:
•Temperature is a measure of the average kinetic
energy of a system of particles.
• Thermal Equilibrium means that two bodies are at the
same temperature.
• The “Zeroth Law of Thermodynamics” states that if
system A and B are n thermal equilibrium with system
C, then A and B are in thermal Equilibrium with each
other.
Physics 1710
Chapter 19 Temperature
•Kelvin is a unit of temperature where one
degree K is 1/279.16 of the temperature of the
triple point of water (near freezing).
TC = (100/180) (TF – 32 ⁰F)
TF = (180/100) TC + 32 ⁰F
• ∆L/L = α∆T
• PV = n R T = N kT
Physics 1710 Chapter 20 Heat & 1st Law of Thermo
Summary
•The internal energy is the total average energy of the
atoms of an object.
• Heat is the change in internal energy.
• The change in temperature is proportional to the
change in internal energy (heat flow) when there is no
change of phase and the system does no work.
• The first law of thermodynamics states
∆E = ∆Q - W
Physics 1710 Chapter 21 Kinetic theory of Gases
Summary:
• The Ideal Gas Law results from the cumulative action of
atoms or molecules.
• The average kinetic energy of the atoms or molecules
of an ideal gas is equal to 3/2 kT.
½ m<v2> = 3/2 kT
• Energy average distributes equally (is equipartitioned)
into all available states.
•Each degree of freedom contributes 1/2 kT to the
energy of a system.
Physics 1710 Chapter 21 Kinetic theory of Gases
Summary (cont’d.)
γ = CP / CV
PV γ = constant
B=γP
• The
distribution of particles among available energy
states obeys the Boltzmann distribution law.
nV = no e –E/kT
Physics 1710
Chapter 22 Heat Engines etc
Summary:
•The work done by a heat engine is equal to
the difference in the heat absorbed at the high
temperature and expelled at the low.
∆W = ∆Qh – ∆Qc
• The thermal efficiency is the work done
divided by the heat absorbed.
e = 1 - ∆Qc / ∆Qh
Physics 1710
Chapter 22 Heat Engines etc
Summary:
• Kelvin-Planck form of 2 nd Law of Thermo: It is
impossible to construct a heat engine that,
operating in a cycle, produces no effect other
than the absorption of energy from a reservoir
and the performance of an equal amount of
work.
•Clausius Form of 2 nd Law of Thermo: It is
impossible to construct a cyclical machine
whose sole effect is the continuous transfer of
energy from one object to another at a higher
temperature without the input of work.
Physics 1710
Chapter 22 Heat Engines etc
Summary:
•The maximum efficiency is obtained via a
Carnot cycle and is equal to the temperature
difference divided by the high temperature.
eCarnot = 1 - Tc / Th
• Entropy
S is a measure of the disorder of a
system.
• ∆S = ∫dQ/T
• S≡ k ln N