summary lecture 6

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