Download Midterm 3 - overview

Midterm 3 - overview
=I
(compare to F=ma)
Moment of inertia I: I=(miri2)
: angular acceleration
I depends on the choice of rotation axis!!
Rotational Kin. Energy KEr=½I2
Conservation of energy for rotating object:
[PE+KEt+KEr]initial= [PE+KEt+KEr]final
[mgh+0.5mv2+0.5I2]I= [mgh+0.5mv2+0.5I2]F
=v/r I=xMr2 with x: depending on the object
Rolling of a slope:
[mgh]top= [0.5mv2+0.5I2]bottom
[mgh]top= [mgh+0.5mv2+0.5xmv2]bottom
The smaller I (and thus x), the larger the linear
speed at the bottom.
Conservation of angular momentum
If the net torque equals zero, the
angular momentum L does not change
Li=Lf
Iii=Iff
Rotational Kin. Energy KEr=½I2=½L
Solids:
General:
FL0
F/A
Y

L / L0 AL
Young’s modulus
F/A
Fh
Shear modulus
S

x / h Ax
F / A
P
Bulk modulus
B

V / V0
V / V0 Also fluids
P  pressure
P=F/A (N/m2=Pa)
=M/V (kg/m3)
Fpressure-difference=PA
Pascal’s principle: a change in pressure applied
to a fluid that is enclosed is transmitted to the whole
fluid and all the walls of the container that hold the fluid.
Bouyant Force B:
weight of the water in the volume displaced by the object:
B=mwater,displacedg
= waterVdisplacedg
If object is fully submerged: Vdisplaced=Vobject
If floating: Vdisplaced=Vpart of object under water
Gravitational force acting on object in/under water:
Fg=mobjectg= objectVobjectg
If floating: B=Fg so waterVdisplacedg= objectVobjectg
P0
Pressure at depth h
P
=
P0+ fluidgh
h: distance between liquid surface
and the point where you measure P
h
P
Buoyant force for submerged object
B = fluidVobjectg = Mfluidg = wfluid
The buoyant force equals the weight of the
amount of water that can be put in the
volume taken by the object.
If object is not moving: B=wobject object= fluid
Buoyant force for floating object
The buoyant force equals the weight of the amount
of water that can be put in the part of the volume
of the object that is under water.
objectVobject= waterVdisplaced h= objectVobject/(waterA)
B h
w
Bernoulli’s equation
P1+½v12+gy1= P2+½v22+gy2
P+½v2+gy=constant
The sum of the pressure (P), the kinetic energy per unit
volume (½v2) and the potential energy per unit volume (gy)
is constant at all points along a path of flow.
Note that for an incompressible fluid:
A1v1=A2v2
This is called the equation of
continuity.
Contact surface A
moving
Viscous flow
F=Av/d
=coefficient of viscosity
unit: Ns/m2
or poise=0.1 Ns/m2
Poiseuille’s Law
Rate of flow Q= v/t=
R4(P1-P2)
8L
(unit: m3/s)
Temperature scales
Conversions
Tcelsius=Tkelvin-273.5
Tfahrenheit=9/5*Tcelcius+32
We will use Tkelvin.
If Tkelvin=0, the atoms/molecules
have no kinetic energy and every
substance is a solid; it is called the
Absolute zero-point.
Celsius
Kelvin
Fahrenheit
Thermal expansion
length
L
L=LoT
surface
A=AoT =2
volume
V=VoT =3 
L0
: coefficient of linear expansion
different for each material
T=T0
T=T0+T
Boyle & Charles & Gay-Lussac
IDEAL GAS LAW
PV/T = nR
n: number of particles in the gas (mol)
R: universal gas constant 8.31 J/mol·K
If no molecules are extracted from or added to a system:
PV
 constant
T
P1V1 P2V2

T1
T2
2 1 2
PV  N  mv  Microscopic
3 2

Macroscopic
PV  Nk B T
2 1 2
T
( mv )
3k B 2
Temperature ~ average molecular
kinetic energy
1 2 3
mv  k B T Average molecular kinetic energy
2
2
3
3
E kin  Nk B T  nRT Total kinetic energy
2
2
3k bT
3RT rms speed of a molecule
2
v rms  v 

M=Molar mass (kg/mol)
m
M
Calorimetry
If we connect two objects with different temperature
energy will transferred from the hotter to the cooler
one until their temperatures are the same.
If the system is isolated:
Qcold=-Qhot
mcoldccold(Tfinal-Tcold)=-mhotchot(Tfinal-Thot)
the final temperature is: Tfinal=
mcoldccoldTcold+mhotchotThot
mcoldccold+mhotchot
Phase Change
GAS(high T)
Q=cgasmT
Gas 
liquid
Q=mLv
Q=csolidmT Solid (low T)
liquid (medium T)
Q=cliquidmT
liquid 
solid
Q=mLf
Heat transfer via conduction
Rate of energy transfer P
P=Q/t (unit Watt)
P=kA(Th-Tc)/x=kAT/x
k: thermal conductivity
Unit:J/(msoC)
multiple layers:
Q A(Th  Tc )
P

t  ( Li / ki )
i
Li=thickness of layer i
ki=thermal conductivity of layer i
Radiation
P=AeT4 : Stefan’s law (J/s)
=5.6696x10-8 W/m2K4
A: surface area
e: object dependent constant emissivity (0-1)
T: temperature (K)
P: energy radiated per second.
P=Ae(T4-T04) where
T: temperature of object
T0: temperature of
surroundings.
Isobaric compression
Let’s assume that the pressure does not
change while lowering the piston (isobaric
compression).
W=-Fy=-PAy
(P=F/A)
W=-PV=-P(Vf-Vi) (in Joule)
W: work done on the gas
+ if V<0
- if V>0
This corresponds to the area under
the curve in a P-V diagram
Work done on gas: signs.
If the arrow goes from right to left, positive work is
done on the gas. If the arrow goes from left to right,
negative work is done on the gas (the gas has done
positive work on the piston) Not mentioned in the book!
First Law of thermodynamics
U=Uf-Ui=Q+W
U=change in internal energy
Q=energy transfer through heat (+ if heat is
transferred to the system)
W=energy transfer through work (+ if work is
done on the system)
This law is a general rule for conservation of energy
Types of processes
A: Isovolumetric V=0
B: Adiabatic Q=0
C: Isothermal T=0
D: Isobaric P=0
PV/T=constant