Final Exam Formulae
PHY 121: Physics for Life Sciences
UNIFORM ACCELERATION:
LINEAR MOTION
ω f = ωi + α t
Fc =
1
x = vot + at 2
2
2
2
v f = vo + 2ax
!v +v $
v =# f 0 &
" 2 %
QUADRATIC FORMULA
−b ± b 2 − 4ac
2a
1
θ = ωit + α t 2
2
2
2
ω f = ωi + 2αθ
! ω + ωi $
ω =# f
&
" 2 %
LAW OF UNIVERSAL
GRAVITATION
ROTATIONAL MOTION
Gm1m2
r2
N ⋅ m2
G = 6.67 ×10−11
kg 2
s = θ r v = ω r aT = α r
CENTER OF MASS
LINEAR MOMENTUM & IMPULSE
Fg =
VECTOR COMPONENTS
Ax = A cosθ
Ay = Asin θ
UNIFORM ACCELERATION:
ROTATIONAL MOTION
mv 2
r
v2
ac =
r
2π r
v=
T circumference
= 2π r €
1
T=
f
v f = vo + at
x=
CENTRIPETAL FORCE & MOTION
Spring 2014
€
xCM =
p = mv Δp = mΔv j = FΔt = mΔv
m A x A + mB x B
m A + mB
NEWTON’S SECOND LAW
TORQUE & MOMENT OF INERTIA
WORK & ENERGY


ΣF = ma
Fg = mg
τ = Fr sin θ
τ net = Iα W = Fd cosθ U g = mgh m
g = 9.80 2
s
I = ∑mr 2 1
2
solid disk: I = MR
2
2
2
solid sphere: I = MR 5
1
U s = kx 2 2
1
KEtrans = mv 2 2
W = ΔKE
FORCE OF FRICTION
HOOKE’S LAW & ELASTIC FORCE
ANGULAR MOMENTUM
F f = µFN
FS = −kΔx
# ΔL &
FS = −YA % ( $ L '
WORK & ENERGY UNITS
CONSERVATION OF ENERGY
1 calorie = 4.19 Joules 1 Cal (food) = 1000 cal = 4190 J Emechanical = KE +U g +U s
€
L = Iω 1
KErot = Iω 2
2
POWER
P=
energy F‡d mgh
=
=
= Fsv
t
t
t
THERMODYNAMICS
TEMPERATURE CONVERSIONS
IDEAL GASES
ΔE = W + Q 3
Eth = NK avg = NkBT 2
T (K ) = T (°C) + 273 5
T (°C) = (T (°F) − 32°) 9
9
T (°F) = T (°C) + 32°
5
1u = 1.66 ×10 −27 kg Boltzmann’s Constant: kB = 1.38 ×10
−23
J
K
output
input
W = PΔV e=
PRESSURE & DENSITY
P=
FLUIDS
F
A
1 atm = 101,300 Pa = 14.7 psi m
V
P = Po + ρ gh ρ=
SPECIFIC HEAT &
HEAT OF TRANSFORMATION
Q = mCΔT Q = ±ML f
FB = ρVg A1v1 = A2 v2 ΔV
Q=
= Av
Δt
1
1
P1 + ρ v12 + ρ gh1 = P2 + ρ v22 + ρ gh2 2
2
Lvavg
ΔP = 8πη
A
RATE OF HEAT TRANSFER
Q " kA %
= $ ' ΔT Δt # L &
Q
radiation: = eσ AT 4
Δt
3kBT
m
PV = nRT J
R = 8.31
mol ⋅ K
vrms =
1
T
l
T = 2π
g
f=
T = 2π
f=
1
2π
m
k
mgd
I
THERMAL EXPANSION
ΔL = α Li ΔT ΔV = βVi ΔT SIMPLE HARMONIC MOTION
WAVES
SOUND
x(t) = A cos(2π ft) xmax = A vx (t) = −vmax sin(2π ft) vmax = 2π fA ax (t) = −amax cos ( 2π ft ) v = λ f γ RT
M
P
I=
4π r 2
!I$
β = (10dB)log10 # & " Io %
fbeat = f1 − f2
m
Ts
vstring =
L
µ
! ! x t $$
y(x, t) = A cos # 2π #  && " " λ T %%
8 m
speed of light: c = 3.00 ×10
s
µ=
2
amax = ( 2π f ) A
DOPPLER EFFECT
! v ± vo $
f' =#
& fo
" v  vs %
N A = 6.02 ×10 23 mol −1 PERIOD & FREQUENCY OF
OSCILLATIONS
conduction: Q = ±MLv
Avogadro’s Number: Δf = ±2 fo
vsound =
HARMONICS
vo
v
open-­‐ended pipe:
! v $
! v $
fm = m # & closed-­‐ended pipe: fm = m # &
" 2L %
" 4L %
Formulae PHY121 Final Exam Spring 2014
Vectors:
A = Axi + Ayj with: i (j)  the unit vector along the x (y) axis
Trigonometry: sinθ ≡ b/c; cosθ ≡ a/c; tanθ ≡ b/a = “slope” of c
sin30°= cos60°= ½
c
thus: b=c sinθ, a=c cosθ, b=a tanθ,
b sin60°= cos30°= ½ √3
a2 + b2 = c2, hence: sin2θ + sin2θ = 1
sin45°= cos45°= ½ √2,
θ
sinθ = cos(90°– θ), sinθ = –sin(–θ), cosθ = cos(–θ)
tan45°=1
a
Ax=A cosθ; Ay=A sinθ;
A = (Ax,Ay)
Components: (if θ  angle with the +x-axis!)
B
Scalar Product (“Dot” Product):
A·B = AxBx + AyBy = AB cosθA,B
A
= AB// = A//B
(θA,B  angle between the vectors A and B)
θ B//
2
2
3
Circle: circumference= 2πR; Area=πR ; π≡3.14159
Sphere: surface= 4πR ; volume=4πR /3
2
Quadratic Equation: ax + bx + c=0; constants a, b, and c
Solutions: x+, x–=[ –b ≤ √(b2 – 4ac) ] /(2a)
Differentiation: d/dx[Axn]=Anxn–1; d/dt[cos(ωt)]= –ωsin(ωt); d/dt[sin(ωt)]= ωcos(ωt); d/dx[lnx]= x–1
f=10–15, p=10–12, n=10–9, μ=10–6, m=10–3, k=103, M=106, G=109, T=1012, P=1015
Prefixes:
1 mile=1609 m, 1 L=10–3m3, g=9.80 m/s2; ME=5.98×1024 kg; RE=6.37×106 m; NA=6.022×1023mol–1
Constants:
Uncertainty:
S  AB S  A2  B 2 ; S  AB S S 
 A A    B B 
2
2
; S  An S S  n A A
Kinematics:
the “motion”: position s as function of time t
position: s = s(t) =(x,y)
velocity v; acceleration a:
v  ds /dt =(vx,vy); speed v|v|; a  dv/dt=(ax,ay)
Linear motion with constant a:
v = v0 + at,
s = s0 +v0t + ½at2;
2
eliminating t:
v = v02 + 2a·(s – s0)
rotation angle  (radians; rotation radius R):
  s(=arc length)/R = (t) (angular position)
angular velocity  :
  d /dt = v/R
(v  speed along the circle)
angular acceleration  :
  d /dt = a///R
(//  parallel to circle)
Circular motion (radius R) with constant :
 = 0 + t ;  = 0 +0t + ½t2
eliminating t:
2 = 02 + 2( –0)
2
circular motion – centripetal acceleration ac:
ac = v /R (radially inwards)
Center-of-Gravity of system (mass M):
rcgΣimiri/Σimi=Σimiri/M; xcg=Σimixi/M, ycg=Σimiyi/M
Moment of Inertia I:
I  Σmiri2; ri = distance between rotation axis and cg of mi
axis through cg: I = ML2/12
thin uniform rod or slab (M, L), axis  rod:
axis through end: I= ML2/3
hollow cylinder (M, Rin, Rout), axis is cylinder axis:
I = ½ M(Rin2+Rout2) (= ½ MR2 solid disk!)
uniform solid sphere (M, R), axis through center:
I = 2MR2/5
Momentum p
p  Σimivi = Mvcg
where M=Σimi the total mass
Angular Momentum L
L  Σiripisinθp,v= Iω L depends on the axis of rotation!
Kinetic Energy K [J≡Nm]:
1
2
Moving axis: K tot  mvcg2 
1
2
1
2
I cgcg2 ; Fixed axis A: K rot ,A  IA A2
Forces [N≡kgm/s2] and consequences:
FNet=iFi = ma; FA on B = –FB on A
Force of Gravity between M and m,
FG = GMm/r2 (–i) (attractive! G=6.67×10–11 Nm2/kg2);
at center-to-center distance r
near sea level: FG = mg(–j) (downwards); g=9.80 m/s2
Force of a Spring (spring constant k):
FS = –k ΔL (opposes compression/stretch ΔL)
Force of an Elastic solid bar of length L and
FS = –Y A(ΔL/L)
(opposes compression/stretch ΔL)
cross sectional area A:
Y ≡ the material’s Young’s modulus (units: N/m2)
Normal Force n: a reaction force
Direction: perpendicular to (^) surfaces
Force of Friction
Direction: // to surface and opposing the motion
Note: static friction is a reaction force:
Static Friction: fs ≤ sn
( = friction coefficient; n = normal force)
Kinetic friction: fk = kn
Torque by F:
τ ≡ RFsinθR,F =RF=RF; τNet= iτi = Iα=dL /dt; R points from the rotation axis to the point
in which F acts; sign of τ: Counter Clock Wise=+, Clock Wise= –
Equilibrium
if iFi = dptot /dt = 0  Δptot = 0  p0=pf ; if iτi = dLtot /dt = 0  ΔLtot = 0  L0=Lf
& Collisions:
(In)elastic: K is (not) conserved; Completely Inelastic: objects stick afterwards
Impulse J by a force F over a time interval Δt:
JF  Fdt = FavgΔt = Δp
Work W done by a force F over a trajectory Δx:
WF  Fdx = FavgΔx= FavgΔx cosθF,Δx
Units: Work-Energy: [J≡Nm]
1 cal=4.19 J, 1 Cal(“food calorie”)=4190 J
Power P [W≡J/s; 1 hp=746 W]:
PF  dWF/dt (≈ WF/Δt) = F·v = Fv cosθF,v
Work-Kinetic energy relationship (from Fj= ma):
Mechanical work: WNet=iWi=K  Kf – K0
Potential Energy UF of a Conservative force F:
ΔUF = –WF ; e.g. UG = mgy, US = ½ kx2
Net work by Non-Conservative forces (friction, drag)
Emech  K + ΣiUi
changes the Mechanical Energy:
WNC=Emech=Ef – E0  Kf +ΣiUfi – (K0+ΣiU0i)
Etotal = Emech + Eth + Echem + …
When taking into account all forms of energy, we find that energy in any
process is conserved: work can transfer energy from one system to another,
but the total energy Etotal remains unchanged.
density ρ≡ m/Volume, pressure p ≡F/Area [Pa]
Fluids (Buoyant force = –ρVg of displaced fluid)
Fluid flow:
1 Pa ≡ 1 N/m2; 1 atm ≡ 1.013×105 Pa
continuity equation (incompressible fluids):
Volume flow rate: Q=ΔV/Δt=Av = constant throughout fluid
Bernoulli’s equation (incompressible+frictionless): p + ρgy + ½ ρv2 = constant throughout fluid
Poiseuille’s Equation (pipe radius R, cross section A, length L; pressure drop Δp; viscosity η):
Av=πR4Δp/(8ηL)
9
Temperature scales (Kelvin K, Celsius °C, °F)
T = TK = 273.15 + TC; TF = /5 TC + 32 °F; TC = 5/9(TF – 32) °C
expansion of solids and liquids:
LT = L0(1+ αΔT); VT = V0(1+ βΔT), β ≈ 3α ; ΔT ≡ T – T0
heat flow rate H≡dQ/dt : conduction: H=kAΔT/L; radiation: H=eσAT4, σ =5.67×10–8 J/m2/K4, e=emissivity
–23
2
Kinetic-molecular model ( K tr the average translational kinetic
K tr  12 mv 2  12 mvrms
 32 kBT ; kB=1.38×10 J/K
energy of the molecules, vrms the root-mean-square speed):
ideal gas Law (number of moles n, T in kelvin!)
pV = nRT; R =NAkB=8.314 J/K/mol, NA= 6.022×1023/mol
Q1→2= mc(T2 – T1)
Heat Q [J] added to fluid/solid (specific heat c, latent heat L) in
(no phase change)
same phase, and for a phase change f, v:
Q1→2= mLf,v (change to fluid or vapor)
cliquid water=4190 J/kg/K, cice=2090 J/kg/K, Lice→water=333 kJ/kg
Heat Q [J] added to a gas at constant p or V, (Cp or V is the molar heat capacity):
Q1→2= nCp or V(T2 – T1)
3
5
CV = /2 R for ideal monatomic gas; CV = /2 R for ideal di-atomic gas
Cp= CV + R; γ ≡ Cp /CV
1st Law: Q = ΔEth + Wg (heat Q is positive if added to the gas; change in thermal
ΔEth = nCVΔT, Wg = ∫pdV
energy ΔEth; work Wg is positive if done by the gas; all have unit [J]):
Thermodynamic Processes: isochoric: V constant; isobaric: p constant; isothermic: T constant  pV constant;
adiabatic Q = 0  pV γ constant. Note: Q and Wg depend on the detailed path p as a function of V !
State variable: variable determined uniquely by the state of the system, not the path: n, p, V, T, Eth, S
cycle in a pV-diagram (efficiency ε):
ε ≡ WNet/Qinput ≤ εCarnot
Carnot cycle (between two isotherms (TH, TC) and two adiabats):
εCarnot= 1+ QC/QH = 1– TC/TH
x(t) = A cos(ωt+φ), ω ≡ 2π/T, f ≡ 1/T
Periodic Motion (period T, frequency f, ang.freq. ω) for restoring
ω = √(k/m)
forces. E.g. spring force F = –kx and mass m:
ω = √(g/L); ω = √(mgL/I)
pendulum (distance L between cg and rotation point):
with damping force F = –bv:
x(t)=Ae–(b/(2m))tcos(ωdt)=Ae–t/τcos(ωdt), ωd ≡ √[k/m – (b/(2m))2]
with driving force F = Fmaxcos(ωdt) and damping:
A = Fmax/√[(k – mωd2)2 + (ωdb)2]
y(t) = A cos(kx – ωt + φ),
Traveling Waves (T, f, ω; wave length λ and wave number k;
ω ≡ 2π/T, k ≡ 2π/λ
propagation velocity v ≡ λ/T = λf in the +x direction); e.g.:
vstring= √(TS/μ), Pav= ½ ω2A2√(μTS)
waves in a string of linear mass μ ≡ m/L under tension TS:
waves in fluids/gases (bulk modulus B, γ≡Cp /CV, molar mass M): vfluid= √(B/ρ), vgas= √(γRT/M),
vair = 343 m/s, I ≡ Pav/Area;
sound waves in air (T = 19 °C, p=1 atm) (intensity I):
–12
2
sound level β (I0≡10 W/m ):
β ≡ (10 dB) log10(I/I0)
Doppler effect (velocity of source vS, listener vL ; +ve if they are approaching):
fL = fS (v + vL)/(v – vS)
superposition principle : ‘displacements add’  interference!
ytot(t) = y1(t) + y2(t)
Slight frequency difference: Beat frequency:
fbeat = |f1 – f2|
y(t) = A sin(kx) cos(ωt)
Standing Waves (interference of wave with its reflections  nodes
λm= 2L/m, m=1,2,3,...; fm= mv/(2L)
and anti-nodes. E.g. string or open-ended pipe of length L:
Open-closed (one end open, one end closed) pipe of length L:
λm= 4L/m, m=1,3,5,...; fm= mv/(4L)