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General Physics T0226A
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General Physics T0226A
Static Equilibrium
~v = 0
∑ ~F
∑ i^ = 0 ~τi = 0 w.r.t. any point If
Fi = 0, and ~τO =
∑
(~rO→i × F~i) = 0,
then ~τP = 0. Proof: ~τ =
∑
(~rP →i × ~Fi)
~rP →i = R~P →O + ~rO→i
~τP = ∑(R~P →O + ~rO→i) × ~Fi
∑
(~rO→i × ~Fi) +
∑
(R~P →O × ~Fi)
= ∑(~rO→i × ~Fi) + ~RP →O × ∑^ F~i
If the net force on an object is zero, and the net torque about some point is zero, then the net torque about any point is also zero.
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General Physics T0226A
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General Physics Y
Center of Gravity
Gravitational torque on a body of arbitrary shape:
~τ = ~rcg × (M~g)
In a unform gravitational field ~g, the center of gravity
of any body is identical to its center of mass.
~τi = ~ri × mi~g
~τ = ∑^ ~τi
= ∑^ mi~ri × ~g
∑ (^) m i~ri
M ×^ M~g
= ~rCM × M~g
~rcg = ~rCM
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General Physics Y
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General Physics K
A leaning ladder
Fe: due to earth FN W : due to wall FN F : due to floor Ff F : frictional force due to floor
minimum value of μs =? ∑ (^) ~
Fi = 0 Ff F = FN W
FN F = Fe ∑ ~τi = 0 HFN W − 12 LFe = 0 FN W = (^) H^1 · 12 LFe = 2 LH Fe
frictional force: Ff F ≤ μsFN F left: ff F = FN W = 2 LH · Fe right: μsFN F = μsFe L 2 H Fe^ ≤^ μsFe
μs ≥ 2 LH
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General Physics K
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General Physics W
A drawbridge m =11,000 kg, L = 14 m, T =?, Fh =?, φ 1 =?
∑ ~F
i = 0, horizontal: Fh cos φ 1 − T cos φ 2 = 0 (1) ∑vertical: Fh^ sin^ φ^1 −^ mg^ −^ T^ sin^ φ^2 = 0^ (2) ~τi = 0, −L 2 mg sin θ 1 + LT sin θ 2 = 0 T = mg 2 sin^ sin θ^ θ 21 = (11000kg)(9.8m/s
(^2) )(sin 120◦) (2)(sin 165◦) =1.80× 105 N From (1), cos φ 1 = T^ cos Fh^ φ^2 From (2), sin φ 1 = mg+T F^ hsin^ φ^2 tan φ 1 = (^) cossin^ φ φ^11 = mg T+ cosT^ sin φ 2 φ^2 = (11000kg)(9.8m/s
(^2) )+(1. 80 × 105 N)(sin 15◦) (1. 80 × 105 N)(cos 15◦) = 0. 887 φ 1 = tan−^1 (0.887) = 41.6◦ From (1), Fh = T cos^ cos φ^ φ 12 = (1.^80 ×^10
(^5) N)(cos 15◦) cos 41. 6 ◦ = 2.3× 105 N
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General Physics Y
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General Physics K
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General Physics K
Elasticity ÄÄÄPPP, Stress TTTæææ, Strain TTT ÄPÿ(;ó) elastic modulus = stress strain Tæstress = f orce area = F A
ùæTæ tensile stress σt = F An
D¹Tæ compressive stress σc = F An
6 æTæ shear stress σs = F Ap Dæ pressure p = F An Tstrain: tensile strain ǫt = ∆ ℓℓ shear strain ǫs = φ ≈ ∆ ℓx Young’s modulus Y = σ ǫtt = F ∆nℓ/ℓ/A
shear modulus S = σ ǫss = F ∆px/ℓ/A
bulk modulus B = − (^) ∆∆V /VP (ÿûYoung’s modulus) (V at P, decrease ∆V when increase ∆P ) [alternate, B = (^) |∆pV V |
]
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General Physics K
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General Physics T0408C
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General Physics T0408C
Fluids At a given point, a fluid exerts the same pressure in all directions.
Hydrostatic Equilibrium with Gravity top: P , bottom: P + dP fluid element: dFpress = (P + dP )A − P A = AdP gravity: dFg = −g dm = −g · ρAdh equilibrium: AdP − g · ρAdh = 0, dP dh = ρg That is P = P 0 + ρgh where P 0 is the pressure at the liquid surface. Pascal’s Principle A pressure applied to a confined fluid at rest is trans- mitted throughout that fluid. Pin = Pout Fin Ain =^
Fout Aout Atmospheric pressure 1 atm = 1. 013 × 105 Pa = 101.3 kPa = 760 mmHg
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General Physics T0408C
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General Physics T0408D
Archimedes’ Principle The buoyant force on a submerged object is a result of the increase in pressure with depth.
top: F 1 = P 1 A bottom: F 2 = P 2 A Fbuoyant = F 2 − F 1 = P 2 A − P 1 A = (P 0 + ρgh 2 )A − (P 0 + ρgh 1 )A = ρgA(h 2 − h 1 ) = (ρV )g = mfluidg Archimedes’ principle The buoyant force acting on a submerged volume is equal to the weight of an equivalent volume of the fluid displaced.
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General Physics T0409A
Fluid Dynamics nøLaminar flow (øa) BøTurbulent flow §øP²: (1) incompressible (2) nonviscous (3) steady (4) irrotational Streamline: øaÎ>ºa, øü. Continuity Equation mass flow rate = ∆ ∆mt = const.
left: ∆ ∆mt = ρ^1 A ∆^1 vt^1 ∆t= ρ 1 A 1 v 1
right: ∆ ∆mt = ρ^2 A ∆^2 vt^2 ∆t= ρ 2 A 2 v 2
continuity (no sinks, no sources): ρ 1 A 1 v 1 = ρ 2 A 2 v 2
Since it is incompressible, that is ρ 1 = ρ 2 , we have A 1 v 1 = A 2 v 2
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General Physics T0409A
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General Physics T0409B
Bernoulli’s Effect
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General Physics T0409B
Bernoulli’s Equation v: fluid speed at a given position. Work-energy theorem: W = ∆K W 1 = F 1 ∆x 1 = P 1 A 1 ∆x 1
W 2 = −F 2 ∆x 2 = −P 2 A 2 ∆x 2 (F 2 to left, ∆x 2 to right, dot product)
W 3 = −mg(h 2 − h 1 ) (mg downward)
W = W 1 + W 2 + W 3 = P 1 A 1 ∆x 1 − P 2 A 2 ∆x 2 − mg(h 2 − h 1 ) = ∆K = 12 mv^22 − 12 mv^21 Therefore, P 1 A 1 ∆x 1 + mgh 1 + 12 mv 12 = P 2 A 2 ∆x 2 + mgh 2 + 12 mv 22
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General Physics T0409B
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General Physics T0409C
Bernoulli’s Equation P 1 A 1 ∆x 1 + mgh 1 + 12 mv 12 = P 2 A 2 ∆x 2 + mgh 2 + 12 mv 22 Divided by equal volume: A 1 ∆x 1 = A 2 ∆x 2
P 1 + (^) A 1 m∆x 1 gh 1 + (^12) A 1 m∆x 1 v^21 = P 2 + (^) A 2 m∆x 2 gh 2 + (^12) A 2 m∆x 2 v^22
equal mass: m = ρA 1 ∆x 1 = ρA 2 ∆x 2 that is, (^) A 1 m∆x 1 = (^) A 2 m∆x 2 = ρ we have P 1 + ρgh 1 + 12 ρv^21 = P 2 + ρgh 2 + 12 ρv^22 Bernoulli’s equation: P + 12 ρv^2 + ρgh = const.
At same h, P + 12 ρv^2 =const.
Bernoulli’s effect means the higher flow speed and therefore lower pressure.
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General Physics W
Venturi meter Bernoulli’s eq.: P 1 + 12 ρv^21 = P 2 + 12 ρv 22
Continuity eq.: v 1 A 1 = v 2 A 2 v 2 = v^1 AA 21
P 1 − P 2 = 12 ρv^22 − 12 ρv 12
= 12 ρ(v^1 AA 21 )^2 − 12 ρv^21
= 12 ρv^21 [(A A^12 )^2 − 1]
v^21 = (^) ρ[(2(A 1 P/A^1 − 2 P)^22 )−1]
v 1 =
√ 2(P 1 −P 2 ) ρ[(A 1 /A 2 )^2 −1]
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General Physics W
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General Physics H
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General Physics T0229A
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General Physics T0229A
Oscillation The motion repeats itself, back and forth, over the same path.
displacement x amplitude A period T frequency f angular frequency ω = 2πf equilibrium position x = 0 restoring force F = −kx
Simple Harmonic Motion simple harmonic oscillator “simple”: F = −kx Since F = ma, we have −kx = md
(^2) x dt^2. Equation of motion: d^2 x dt^2 +^
k mx^ = 0 d^2 x dt^2 +^ ω
(^2) x = 0
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General Physics T0229A
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General Physics T0229B
Solution of Simple Harmonic Motion sinusoidal function: x = a cos ωt + b sin ωt substituting x into equation of motion, d^2 x dt^2 +^
k m dxx^ = 0 dt =^ −aω^ sin^ ωt^ +^ bω^ cos^ ωt d^2 x dt^2 =^ −aω
(^2) cos ωt − bω (^2) sin ωt = −ω^2 (a cos ωt + b sin ωt) Therefore, −ω^2 (a cos ωt + b sin ωt) + (^) mk(a cos ωt + b sin ωt) = 0 ( (^) mk − ω^2 )(a cos ωt + b sin ωt) = 0
If (^) mk = ω^2 , then the sinusoidal function x = a cos ωt+ b sin ωt is the solution of equation.
ω =
√ k m : A simple harmonic oscillator with mass (m) and spring(k) has angular frequency ω =
√ (^) k m or frequency f = (^21) π
√ (^) k m
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General Physics T0304A
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General Physics T0304A
Solution of Simple Harmonic Motion x = a cos ωt + b sin ωt Since cos(ωt + φ) = cos ωt cos φ − sin ωt sin φ we have A cos(ωt + φ) = A cos ωt cos φ − A sin ωt sin φ = a cos ωt + b sin ωt where a = A cos φ b = −A sin φ Solution can be rewritten as x = A cos(ωt + φ) φ: phase angle, determined by initial conditions.
Kinematics of Simple Harmonic Motion x = A cos(ωt + φ) v = dx dt = −ωA sin(ωt + φ)
a = d dt^22 x = −ω^2 A cos(ωt + φ)
Initial conditions (t = 0) x 0 = x(t = 0) = A cos φ v 0 = v(t = 0) = −ωA sin φ = −vmax sin φ a 0 = a(t = 0) = −ω^2 A cos φ = −amax cos φ
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General Physics T0304A
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General Physics C
The Physical Pendulum restoring torque: τ = −mgl sin θ τ = Iα −mgl sin θ = Iα α = −(mgl I ) sin θ ≈ −(mgl I )θ ⇐ θ is small. α = d
(^2) θ dt^2 =^ −(
mgl I )θ
d^2 θ dt^2 + (
mgl I )θ^ = 0
θ(t) = θmax cos(ωt + φ) ω =
√mgl I f = (^) T^1 = 2 ωπ = (^21) π
√mgl I (simple pendulum: I = ml^2 )
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General Physics C
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General Physics C
The Torsion Pendulum restoring torque: τ = −κθ κ: torsional constant τ = Iα (F = ma) −κθ = Iα = I d dt^22 θ d^2 θ dt^2 +^
κ I θ^ = 0 d^2 θ dt^2 +^ ω
(^2) θ = 0 ω^2 = κ I ω =
√κ I T = 2π
√ (^) I κ
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General Physics C
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General Physics T0311A
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General Physics T0311A
Damped Harmonic Motion restoring force: F = −kx damping force:∑ F d = −bv = −bdx dt F = ma ∑ (^) F = −kx − bdx dt = md dt^22 x md
(^2) x dt^2 +^ b
dx dt +^ kx^ = 0
d^2 x dt^2 +^
b m
dx dt +^
k mx^ = 0 Solution of Damped Harmonic Motion:
x = Ae−(^ 2 bm)t^ cos(ω′t + φ)
where ω′^ =
√ (^) k m −^ (^
b 2 m)
2
√ ω 02 − α^2
ω 0 =
√ (^) k m (natural frequency) α = 2 bm
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General Physics T0311A
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General Physics T0311B
Damped Harmonic Motion Equation: d
(^2) x dt^2 + 2α
dx dt +^ ω^0
(^2) x = 0 Solution: x = Ae−αt^ cos(ω′t + φ) dx dt =^ −αx^ −^ ω
′Ae−αt (^) sin(ω′t + φ)
d^2 x dt^2 =^ −α
dx dt +^ ω
′αAe−αt (^) sin(ω′t + φ) − ω′ (^2) x
Substituting into eq., d^2 x dt^2 =^ α
(^2) x + 2ω′αAe−αt (^) sin(ω′t + φ) − ω′ (^2) x
2 αdx dt = − 2 α^2 x − 2 αω′Ae−αt^ sin(ω′t + φ)
ω 02 x = ω^20 x ︷︸︸︷ 0 =
︷ ︸︸ ︷ (ω 02 − ω′^2 − α^2 )x This requires ω′^2 = ω 02 − α^2 = (^) mk − ( 2 bm)^2
that is ω′^ =
√ k m −^ (^
b 2 m)
2 in solution x = Ae−(^ 2 bm)t^ cos(ω′t + φ)
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General Physics T0312A
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General Physics T0312A
Wave Properties
- Wave forms
- pulse
- continuous wave
- wave train
- Parameters
- amplitude A
- wave length λ
- frequency f
- period T
- wave speed v, v = λ T = λf
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General Physics T0312A
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General Physics T0312A
- 1-dimensional wave
- longitudinal wave
- transverse wave
- 2-dimensional wave
- 3-dimensional wave
- sound wave
- seismic wave (earthquake)
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General Physics T0312A
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General Physics T0312A
- Reflection and Transmission
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General Physics T0312B
Traveling Harmonic Wave At t = 0, y(x) = A sin[^2 λπ x] At t = ∆t, y(x) = A sin[^2 λπ (x − v∆t)] original y(x) by replacing x with x − v∆t
Wave function for a harmonic wave is: y(x, t) = A sin[^2 λπ (x − vt)] = A sin[^2 πx λ − 2 πvt λ ] = A sin[^2 πx λ − 2 πλf t λ ] = A sin[^2 πx λ − 2 πf t] = A sin[kx − ωt] k: wave number
Traveling Harmonic Wave: to right: y(x, t) = A sin(kx − ωt + φ) to left: y(x, t) = A sin(kx + ωt + φ) wave number: k = (^2) λπ phase of wave: (kx − ωt + φ) (argument of the sine function) phase angle: φ phase velocity: v = λf = (^2 kπ )( 2 ωπ ) = ω k (Note: group velocity)
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General Physics T0312B
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General Physics W
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General Physics T0319A
< cos^2 (kx − ωt) > = (^) T^1
∫ (^) t+T t cos^2 (kx^ −^ ωt′)dt′
Let kx − ωt′^ = α, then
< cos^2 (kx − ωt) > = (^) −^1 ωT
∫ cos^2 αdα
= (^) −^1 ωT
∫ (^) 1+cos 2α 2 dα
= (^) −^1 ωT [α 2 + sin 2 4 α]kx kx−−ωωt(t+T^ )
= (^) −^1 ωT [kx−ω 2 (t +T^ )− kx− 2 ωt]
= (^) −− 2 ωTωT
= (^12)
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General Physics T0319B
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General Physics T0319B
Standing Waves Two harmonic waves of equal frequency traveling in opposite direction can set up standing waves.
y 1 = A sin(kx − ωt) y 2 = A sin(kx + ωt) y(x, t) = y 1 + y 2 = A sin(kx − ωt) + A sin(kx + ωt) Since sin α + sin β = 2 sin(α+ 2 β) cos(α− 2 β ) we have y(x, t) = 2A cos(ωt) sin(kx) The points of permanent zero displacement are called nodes; the points of maximum displacement are called antinodes.
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General Physics T0319B
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General Physics T0319C
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General Physics T0319C
Resonant Standing Waves Standing waves with constraints, imposed by boundary condi- tions, on the frequencies or wave lengths. y(x, t) = 2A cos(ωt) sin(kx) boundary conditions: y(0, t) = 0; y(L, t) = 0 At x = 0, y = 2A cos ωt sin 0 = 0 At x = L, y = 2A cos ωt sin kL = 0 Thus, kL = π, 2 π, 3 π, ..., nπ, ... kL = nπ, n = 1, 2 , 3 , ... 2 π λ L^ =^ nπ λ = (^2) nL normal mode: each resonant standing wave pattern λn = (^2) nL fn = nv 2 L n = 1, fundamental frequency or first harmonic resonant (normal mode) frequencies of a string:
fn = 2 nL
√ F μ
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General Physics T0319C
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