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Applications of Force, Momentum & Energy Laws in Phys141: Kepler's Laws, Fluid Dynamics, O, Study notes of Physics

University of MarylandPhysics
Prof. Wolfgang Losert

Various topics from a university physics course, including kepler's laws, fluid dynamics, and oscillations. Concepts such as newton's law of universal gravitation, potential energy, angular momentum conservation, and buoyancy. Students will learn about kepler's laws of planetary motion, the principles of fluid statics and dynamics, and the behavior of oscillating systems.

Typology: Study notes

Pre 2010

Uploaded on 02/13/2009

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Phys141 – Mon 10/31
Applications of Force, Momentum & Energy Laws
- Kepler’s laws (Chapter 13)
- Fluid Dynamics (Chapter 14.1,14.4-6 )
•Wed 11/2: Oscillations: Chapter 15.1-3.
•Fri 11/4 Oscillations: Chapter 15.4-7
•Mon 11/7: Review
•Wed 11/9 Midterm 2 (Wed 11/9): Including Chapter 15
Newtons Law of Universal Gravitation
Force magnitude:
G Universal gravitational constant:
6.673 x 10-11 N⋅m2/ kg2
• This is an example of an inverse square Force law
Potential Energy:
For earth gravity, r is measured relative to the center of the earth.
Therefore if your distance from the center of the earth changes
significantly use this formula, otherwise you can use U=mgh
12
2
g
mm
FG
r
=
12
r
G
Gm m
Ur
dU
Fdr
=−
=
Keplers Laws: Motion due to the
gravitational force (13.4)
[ Kepler’s First Law
– All planets move in elliptical orbits with the Sun at
one focus – NOT DISCUSSED HERE]
Kepler’s Second Law
– The radius vector drawn from the Sun to a planet
sweeps out equal areas in equal time intervals
Kepler’s Third Law
– The square of the orbital period of any planet is
proportional to the cube of the semimajor axis of
the elliptical orbit
Is angular momentum of the earth-sun system
conserved, if we neglect gravitational forces due to other
planets or suns?
No, gravitational force...
No, angular momentu...
Yes
0% 0%0%
1. No, gravitational forces
act between earth and
sun
2. No, angular momentum
changes due to other
forces
3. Yes
4039383736353433323130292827262524232221
2019181716151413121110987654321
pf3
pf4
pf5

Partial preview of the text

Download Applications of Force, Momentum & Energy Laws in Phys141: Kepler's Laws, Fluid Dynamics, O and more Study notes Physics in PDF only on Docsity!

Phys141 – Mon 10/

Applications of Force, Momentum & Energy Laws

- Kepler’s laws (Chapter 13)

- Fluid Dynamics (Chapter 14.1,14.4-6 )

• Wed 11/2: Oscillations: Chapter 15.1-3.

• Fri 11/4 Oscillations: Chapter 15.4-

• Mon 11/7: Review

• Wed 11/9 Midterm 2 (Wed 11/9): Including Chapter 15

Newtons Law of Universal Gravitation

Force magnitude:

G Universal gravitational constant :

6.673 x 10-11^ N⋅m 2 / kg 2

• This is an example of an inverse square Force law

Potential Energy:

For earth gravity, r is measured relative to the center of the earth.Therefore if your distance from the center of the earth changes

significantly use this formula, otherwise you can use U=mgh

Fg G m m^1

= r

r^1 G

U Gm m

r

F dU

dr

Keplers Laws: Motion due to the

gravitational force (13.4)

[ Kepler’s First Law

– All planets move in elliptical orbits with the Sun atone focus – NOT DISCUSSED HERE]

Kepler’s Second Law

– The radius vector drawn from the Sun to a planetsweeps out equal areas in equal time intervals

Kepler’s Third Law

– The square of the orbital period of any planet isproportional to the cube of the semimajor axis of

the elliptical orbit

Is angular momentum of the earth-sun system

conserved, if we neglect gravitational forces due to other

planets or suns?

No, gravitational force...No, angular momentu...

Yes

1. No, gravitational forces

act between earth and

sun

2. No, angular momentum

changes due to other

forces

3. Yes

Angular momentum conservation: Kepler 2nd

The force produces no torque, so

angular momentum is conserved

L = r x p = MP r x v = const

In a time sweeps out the area dt , the radius vector dA , which is r

half the area of

| r x d r |

Magnitude of L constant:

-> Area swept by radius vector perunit time is constant

L = M P ^ r^^ × dt dr = M P dAdt = const

The earth does not fall into the sun because

211 222 233 244 255 266 277 288 299 3010 3111 It’s pulled by sun’s gra... 3212 The net force on it is ... 3313 It is beyond the main p... 3414 It is being pulled by ot... 3515 3616 3717 3818 3919 4020

All of the aboveNone of the above

  1. It’s pulled by sun’s gravity.
  2. The net force on it is zero.
  3. It is beyond the main pull of sun’s gravity.
  4. It is being pulled by other planets as well as by the sun.
  5. All of the above
  6. None of the above

Force balance: Kepler 3 rd

(Assume a circular orbit of radius r and period T )

Gravitational force = centripetal force

Note: Ks is depends only on mass of sun

->Tsame sun^2 proportional to r 3 for any planet rotating around the

Sun Planet Planet^2

2 π

=

GM M M v r r v r T

Sun

= ⎜⎛^4 π ⎞⎟

= S

T GM r

T K r

Chapter 14: Forces and Flows of Fluids

  • Forces in Fluid (Statics)
    • Definition of pressure
    • Pascal’s law - Hydraulics
    • Archimedes Principle - Buoyancy
  • Fluid flow (dynamics)
    • Conservation of fluid
    • Bernulli equation (energy conservation for fluid)

Derivation of buoyancy - Archimedes’s Principle

Consider fluid object of volume V

and fluid densitymass ρ ρfluid (and hence

fluid V)

Take away surrounding fluid, fluidobject will fall down due to

gravitational force of the fluid

F g = - ρfluidV g

-> Surrounding fluid exerts upward

buoyancy force B = ρfluid V g

Archimedes’s Principle

Buoyancy

Place wood sphere in place of fluid

sphere (i.e. wood sphere in water)

Take away surrounding fluid, wood willfall down

F g = - ρWoodV g

Surrounding fluid still exerts upward

buoyancy force B = ρfluid V g

Net force on wood:

F= ρfluidV g – ρWoodV g = ( ρfluid – ρWood)V g

Buoyancy in partially submerged bodies

V ice is the total volume of the ice

V water is the volume of the water

displaced

– Equal to the volume of the submerged fraction of the iceberg

(89% of the ice is below water)

V water =0.89* V ice

Buoyancy force: ρwater V water g

= Weight of iceberg: ρiceV ice g

ρwater V water g = ρiceV ice g

ρwater* 0.89 V ice = ρice V ice

ρwater *0.89 = ρice

Ideal Fluid Flow

Simplifying assumptions

(1) The fluid is nonviscous – internal friction is

neglected

(2) The flow is steady – the velocity of each point

remains constant

(3) The fluid is incompressible – the density

remains constant

(4) The flow is irrotational – the fluid has no

angular momentum about any point

Equation of Continuity

• fluid moving through a

pipe of nonuniform

diameter

• The mass that crosses

A 1 in some time interval

is the same as the mass

that crosses A 2 in that

same time interval

Equation of Continuity, cont

• m 1 = m 2 →ρ A 1 v 1 = ρ A 2 v 2

• Since the fluid is incompressible, ρ is a

constant

• A 1 v 1 = A 2 v 2

– Equation of continuity for fluids

• The product, Av , is called the volume

flux or the flow rate

A blood platelet drifts along with the flow of blood through an

artery that is partially blocked by deposits. As the platelet moves

from the narrow region to the wider region, its speed

increases. remains the same.

decreases

1. increases.

2. remains the

same.

3. decreases

Energy conservation in a fluid

Bernoulli’s equation: P + ½ ρ v^2 + ρ gy = constant

Multiply by volume V = Adx

PV + ½ ρ Vv^2 + ρ Vgy = constant

PAdx + ½ ρ Vv^2 + ρ Vgy = constant

Fdx + ½ m v^2 + m gy = constant

Work + Kinetic Energy + Gravitational energy= constant