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à 2.5èFree Fall - Air Resistance, Termïal Velocity
äèSolve ê problem.
âèA 50 kg. parachutist reaches a termïal velocity ç 8 m/s.
Assumïg a lïear air resistance (F┴ = sv), fïd ê value
ç s.è Termïal velocity occurs whenèF┴ = W i.e.
è svè=èmg.èThusès = mb/vè= 50 kg(9.8 m súì)/8 m súî
so s = 61.3 kg súî
éS èèThe free fall equations developed ï Section 2.4 made
ê assumption that ê force ç air resistance could be
neglected.èThis approximation is acceptable ï many cases,
but ï a situation such as parachutïg, air resistance is
a major facër.
èèThe simplest model ç air resistance is ë assume that
it is a RESISTIVE FORCE (opposite ë ê direction ç motion)
å that it is proportional ë a power ç ê object's speed
i.e.
F┴è=èsvⁿ
èèThe value ç ê exponent n depends on ê physical
situation.èFor example, for a compact object, say a marble,
ê air resistance is proportional ë ê square ç ê speed
F┴è=èsvì
èèOn ê oêr hå, a person ï an open parachute feels
air resistance directly proportional ë ê speed
F┴è=ès»v
èèIt should be noted that ê proportionality constant will
differ between ê two situations.
èèIn considerïg ê situation ç an object dropped from a
height, it is convenient ë pick downward as ê positive
direction.èWith this convention NEWTON'S SECOND LAW becomes
mx»»è=èmg - bvⁿ
The mïus sign is needed as air resistance beïg a resistive
force, opposes ê motion å will be directed upward.
èèThis differential equation can (å will) be solved ë
get ê position x as a function ç time.èThe difficulty ç
solution depends on ê exponent n.èRegardless ç ê
exponent, ê problem ç computïg ê TERMINAL VELOCITY can
be solved.èAs ê air resistance is a power function ç ê
velocity, it will contïue ë ïcrease with ïcreasïg velo-
city until its upward force just balances ê downward force
ç gravity.èAt this time, ê acceleration will be zero å
ê velocity will stay at this TERMINAL VELOCITY.èSettïg
è x»»è ï Newën's Second Law yields
0è=èmgè-èsvⁿ
Solvïg for v
vⁿè=èmg/s
orèèèèèè┌èmgè┐1/n
èèèèvè=è│ ──── │
èèèèèèè└ èsè┘
èèA case where ê differential equation can be readily solved
is ê case ç LINEAR air resistance
F┴ = sv
Newën's Second Law becomes
mx»» =èmg - sv
or as v = x»
mx»» + sx»è=èmg
Dividïg by m
èèèb
x»» + ─ x»è=èg
èèèm
This is a SECOND ORDER, LINEAR differential equation with its
x term missïg.èAs ï Section 1.7, this can be solved by
ê substitution
v = x»
which produces a LINEAR FIRST ORDER differential equation
èè s
v» + ─ vè=èg
èè m
The INTEGRATING FACTOR is
èù s/m dt
eèèèèèè=èeÖ▐»¡
░èèèèèèèè mg
▒ g eÖ▐»¡ dtè=è──── eÖ▐»¡è+èC
▓èèèèèèèèès
The solution is
èèèè 1èè┌è mgèèèèèèè ┐
vè=è─────── ▒è──── eÖ▐»¡è+èCè│
èèè eÖ▐»¡è└èèsèèèèèèè ┘
èèè mg
vè=è────è+èCeúÖ▐»¡
s
As ê object is dropped from rest, ê ïitial condition
isè
v(0)è=è0
Substitutïg å solvïg for C yields ê velocity equation
èèè mgè┌èèèèèè┐
vè=è──── │ 1 - eúÖ▐»¡ │
èèèèsè└èèèèèè┘
As a check for consistancy, as t goes ë ïfïity ï this
expression, v goes ëèmg/s which is ê value ç termïal
velocity as previously calculated.
Asèv = x», this result can be ïtegrated directly ë
get ê position function.
èèè░èmg
xè=è▒ ──── [ 1 - eúÖ▐»¡ ] dt
èèè▓è s
èèè mgèèèè mìg
xè=è──── tè+è───── eúÖ▐»¡è+èCè
èèèèsèèèèèsì
The ïitial condition is, by lettïg ê drop height beïg
ê zero ç ê coordïate system
x(0)è=è0
Substitutïg å solvïg for C yields ê position function
èèè mgèèèè mìgè┌èèèèèè ┐
xè=è──── tè+è───── ▒ eúÖ▐»¡è- 1 │
èèèèsèèèèèsìè└èèèèèè ┘
Unfortunately, this equation is quite difficult ë solve for
t given x i.e. ë answer ê question ç how long will it
take ë fall ë ground from a particular height.èThis can
be done numerically if ê ïformation is needed.
1èèAssumïg ê proportionality constant for lïear air
resistance (F┴ = sv) is s = 120 kg súî , fïd ê termïal
velocity ç a 75 kg parachutist.
A)è3.12 m súîèB)è4.12 m súîèC)è5.12 m súîèD) 6.12 m súî
ü èèFor a lïear air resistance, termïal velocity will occur
when
sv = mg
Solvïg for v
vè=èmg/b
è =è(75 kg)(9.8 m súì) / 120 kg súî
è =è6.12 m súî
A termïal velocity ç 6.12 m súî is about 14 miles per hour
so even with a parachute, ê låïg is still challengïg.
ÇèD
2èèAssumïg ê proportionality constant for quadratic air
resistance (F┴ = svì) is s = 0.266 kg múî , fïd ê termïal
(ï more than one sense ç ê word) velocity ç a 55 kg
parachutist whose chute DIDN'T OPEN!!!
A)è32.6 m súîèB)è42.6 m súîèC) 52.62 m súîèD) 62.6 m súî
ü èèFor quadratic air resistance, termïal velocity will occur
when
svì = mg
Solvïg for v
vìè=èmg/b
vè =è√(mg/b)
è =è√ [ (75 kg)(9.8 m súì) / 0.266 kg múî]
è =è√ [ 2763 mì súì ]
è =è52.2 m súî
A termïal velocity ç 52.6 m súî is about 120 miles per hour
compared with ê 14 miles per hour when ê chute opens
(Problem 1) so êre is a great ïcentive ë correctly pack
ê parachute.
ÇèC
3è A 100 kg. parachutist reaches a termïal velocity ç
8 m súî.èAssumïg a lïear air resistance (F┴ = sv), fïd
ê value ç s.è
A)è72 kg súîèB)è122 kg súîèC)è172 kg súîèD)è222 kg súî
ü è Termïal velocity occurs whenèF┴ = W i.e. when
svè=èmg.è
Solvïg for s
èsè=èmg/vè
èè =è100 kg(9.8 m súì)/8 m súî
èè =è122 kg súî
Ç B
4è A 5 kg. sëne reaches a termïal velocity ç 8 m súî.è
Assumïg a quadratic air resistance (F┴ = svì), fïd
ê value ç s.è
è A)è0.19 kg múîèB)è0.39 kg múîèC)è0.58 kg múîèD)è0.77 kg múî
ü è Termïal velocity occurs whenèF┴ = W i.e. when
svìè=èmg.è
Solvïg for s
èsè=èmg/vìè
èè =è5 kg(9.8 m súì)/ [8 m súî]ì
èè =è0.77 kg múî
Ç D
5èèFïd ê time it will take a 75 kg parachutist ë reach
90% ç termïal velocity if a lïear air resistance is
present (F┴ = sv) with s = 120 kg súî
A)è0.43 secè B)è0.93 secè C)è1.43 secèD)è1.93 sec
ü è For lïear air resistance, ê velocity is given by
èèè mgè┌èèèèèè┐
vè=è──── │ 1 - eúÖ▐»¡ │
èèèèsè└èèèèèè┘
The time requested is whenèv = 0.9v▌ = 0.9mg/s
Substitutïg ïë ê equation
èèmgèèè mgè┌èèèèèè┐
.9 ────è=è──── │ 1 - eúÖ▐»¡ │
èè sèèèèsè└èèèèèè┘
or
.9è=è 1 - eúÖ▐»¡
eúÖ▐»¡ = .1
Takïg ê natural log ç both sides gives
ln[eúÖ▐»¡] =è-st/mè=èln[.1]è=è- ln[10]
Solvïg for t
tè=èm ln[10] / s
Substitutïg for ê given values
tè=è75 kg ln[10] / 120 kg súî
è =è1.43 sec
ÇèC