summary lecture 6
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6.4Drag force Terminal velocity 7.1-7.6 Work and Kinetic energy 7.7Power 8.1Potential energy 8.2/3Conservative Forces and Potential energy 8.4Conservation of Mech. Energy 8.5Potential-energy curves 8.7Conservation of Energy. Summary Lecture 6. - PowerPoint PPT PresentationTRANSCRIPT
Summary Lecture 6Summary Lecture 6
6.46.4 Drag forceDrag forceTerminal velocity Terminal velocity
7.1-7.67.1-7.6 Work and Kinetic energyWork and Kinetic energy
7.77.7 PowerPower
8.18.1 Potential energyPotential energy
8.2/38.2/3 Conservative Forces and Potential energyConservative Forces and Potential energy
8.48.4 Conservation of Mech. EnergyConservation of Mech. Energy
8.58.5 Potential-energy curvesPotential-energy curves
8.78.7 Conservation of EnergyConservation of Energy
6.46.4 Drag forceDrag forceTerminal velocity Terminal velocity
7.1-7.67.1-7.6 Work and Kinetic energyWork and Kinetic energy
7.77.7 PowerPower
8.18.1 Potential energyPotential energy
8.2/38.2/3 Conservative Forces and Potential energyConservative Forces and Potential energy
8.48.4 Conservation of Mech. EnergyConservation of Mech. Energy
8.58.5 Potential-energy curvesPotential-energy curves
8.78.7 Conservation of EnergyConservation of Energy
Problems:Chap 6: 32, 33, Chap. 7: 2, 14, 50, 29, 31, Chap. 8 5, 8, 22, 29, 36, 71, 51
Problems:Chap 6: 32, 33, Chap. 7: 2, 14, 50, 29, 31, Chap. 8 5, 8, 22, 29, 36, 71, 51
http://webraft.its.unimelb.edu.au/640141/pub/lectures/mechanics/lecture6.pdfhttp://webraft.its.unimelb.edu.au/640141/pub/lectures/mechanics/lecture6.pdf
VISCOUS DRAG FORCEVISCOUS DRAG FORCEDRAG
VISCOUS DRAG FORCE
Assumptions
low viscosity (like air)
turbulent flow
What is it?
like fluid friction
a force opposing motion as fluid flows past object
Fluid of density
V m
Volume hitting object in 1 sec. =AV
Mass hitting object in 1 sec. = AV
momentum (p) transferred to object in 1 sec. = ( AV)V
Force on object = const AV2
t
pF
Area A
In 1 sec a length of V metres hits the object
Fluid of density
V m
Force on object = const AV2
Area A
V
Air hits object = object moves through air
V
mg
mg
D
mg
D
V
V=0
F = mg - D
F = mg -1/2CAv2
D increases as v2
until F=0
i.e. mg= 1/2CAv2
AC
mg2v
AC
mg2v
term
term2
0mgAv1/2Cdt
dvm 2
F = mg –DD
mg
ma = mg -D
D- mgdt
dvm
2/1Ac
m2
)]e1(Ac
gm2[v
t
2/1]Ac
gm2[v
2/1Ac
m2
)]e1(Ac
gm2[v
t
When entertainment defies reality
D= ½ CAv2
Assume C = 1
v = 700 km h-1
Calculate:
Drag force on presidents wife
Compare with weight force
Could they slide down the wire?
D= ½ CAv2
Assume C = 1
v = 700 km h-1
Calculate:
The angle of the cable relative to horizontal.
Compare this with the angle in the film (~30o)
In working out this problem you will prove the expression for the viscous drag force
2AvC2
1F
Time s
Hei
ght m
Real projectile motion!
Throw a stone up with vel v, what is height as function of time?
Drag force proportional to the square of the velocity
for the ascent, mg and drag force in same direction,
for the descent they are opposite.
VCE PhysicsReal Physics
http://www.colorado.edu/physics/phet/web-pages/simulations-base.html
WORK
You know that if I move a body through a displacement d by applying a constant force F
w = Fd
What if F is NOT in the direction of d?
Work is energy transferred to or from an object by a force acting on the object.
Energy transferred TO the object is positive work, and energy transferred FROM the object is negative work.
F
d
What if the force is NOT constant?
BUT!BUT!
If the Force is not in the direction of displacementIf the Force is not in the
direction of displacement
F
w = F . d (Scalar product)
F = iFx +jFy
Remember for a scalar product
i.i = 1 j.j = 1 i.j=0 j.i=0
w = Fxdx + 0 + 0 + Fydy
here: dx= d dy= 0
W = Fcos d +0
Fy=Fsin
Fx=Fcosd = idx + jdy
Thus w = (iFx +jFy) . (idx + jdy)
vectorsscalary
x
F
d
=i.iFxdx + i.jFxdy + j.iFydx + j.jFydy
F
0 d
w = F . d
=Fcos |d |
If = 0 w = Fd
if = 90 w = 0
component of F parallel to d,
multiplied by magnitude of d
Work is a SCALAR: the product of 2 vectors
The unit of work is JOULE
xfxi
What if the force is NOT constant?i.e F depends on x: F(x)
Move a distance x
w = F(x). x
f
i
x
xxxFw ).(
In the limit as x 0
f
i
x
x
dx).x(Fw
x
F(x) F
x
How much work is done by F in moving object from xi to xf?
or the area under the F-x curve
F(x)
x
Frest = -kx
Work done BY the spring
f
i
x
x
dxxFw )(
f
i
x
x
dxkxw
2
2
10 fi kxwthenxif
Work done BY the spring
The Spring Force
f
i
f
i
x
x
x
x
xkdxxkw 221/
xf
x
+ve
Work = area of this triangle!
Frest
PowerPowerPower is the rate of doing work
If we do work w in a time t
dt
dw
Δt
Δwp
0Δt
limitinst
Δt
Δx F.
Δt
Δwpinst
= F cos |v |F cos
v
F
Power is a scalar: the product of F and v
Unit of power is J s-1= watt
1kw = 1000 w 1 HP = 746 w
Δt
Δwpav
F.v
Kinetic Energy
Work-Kinetic Energy Theorem
Change in KE work done by all forces
K w
F
xxi
xf
f
i
f
i
vv
vv vmdvvm ]/[. 221
= 1/2mvf2 – 1/2mvi
2
= Kf - Ki
= KK
Work done by net force
= change in KE
f
i
xx dxFw .
f
i
xx dxma .
f
i
f
i
xx
xx dv
dtdx
mdxdtdv
m ..
Work-Kinetic Energy Theorem
vect
or s
um o
f al
l for
ces
F
xxi
xf
f
i
f
i
vv
vv vmdvvm ]/[. 221
= 1/2mvf2 – 1/2mvi
2
= Kf - Ki
= KK
Work done by net force
= change in KE
f
i
xx dxFw .
f
i
xx dxma .
f
i
f
i
xx
xx dv
dtdx
mdxdtdv
m ..
Work-Kinetic Energy Theorem
vect
or s
um o
f al
l for
ces
mg
F
h
Lift mass m with constant velocity
Work done by me (take down as +ve)
= F.(-h) = -mg(-h) = mghWork done by gravity
= mg.(-h) = -mgh ________
Total work by ALL forces (W) = 0
What happens if I let go?
=K
Gravitation and work
Compressing a spring
Compress a spring by an amount x
Work done by me Fdx = kxdx = 1/2kx2
Work done by spring -kxdx =-1/2kx2
Total work done (W) =
0=K
What happens if I let go?
x
F -kx
Ff
dWork done by me = F.d
Work done by friction = -f.d = -F.d
Total work done = 0What happens if I let go? NOTHING!!
Gravity and spring forces are Conservative
Friction is NOT!!
Moving a block against friction at constant velocity
Conservative Forces
A force is conservative if the work it does on a particle that moves through a round trip is zero; otherwise the force is non-conservative
work done for round trip:On way up: work done by gravity = -mgh
On way down: work done by gravity = mgh
Total work done = 0
Sometimes written as 0ds.F
h-g
Consider throwing a mass up a height h
Conservative Forces
A force is conservative if the work done by it on a particle that moves between two points is the same for all paths connecting these points: otherwise the force is non-conservative.
-g
Each step height=h
= -mg(h1+h2+h3 +……)
= -mgh
Same as direct path (-mgh)
Work done by gravity
w = -mgh1+ -mgh2+-mgh3+…
h