PHSX 114, PHSX 114, Monday, October 20, 2003Monday, October 20, 2003

• Reading for today: Reading for today: Chapter 10 (10-1 -- 10-4)Chapter 10 (10-1 -- 10-4)

• Reading for next lecture (Wed.): Reading for next lecture (Wed.): Chapter 10 Chapter 10 (10-5 -- 10-8)(10-5 -- 10-8)

• Homework for today's lecture: Homework for today's lecture: Chapter 10, Chapter 10, problems 3, 5, 9, 14, 61problems 3, 5, 9, 14, 61

Conservation of angular Conservation of angular momentummomentum

• If the torque is zero, then ΔL= 0 If the torque is zero, then ΔL= 0 (no change in L)(no change in L)

• The total angular momentum of a system The total angular momentum of a system remains constant if there is no net external remains constant if there is no net external torque acting on the system.torque acting on the system.

• Ice skater exampleIce skater example

Your turnYour turn

A 1-kg ball of radius 0.1 m is spinning initially A 1-kg ball of radius 0.1 m is spinning initially with ωwith ω11= 2 rad/s. The ball’s radius increases to = 2 rad/s. The ball’s radius increases to

0.2 m. Find the final angular velocity.0.2 m. Find the final angular velocity.

Answer: L=Iω is constant, so IAnswer: L=Iω is constant, so I11ωω11=I=I22ωω22. .

II11=2MR=2MR1122/5, I/5, I22=2MR=2MR22

22/5 => 2MR/5 => 2MR1122ωω11/5 = /5 =

2MR2MR2222ωω22/5 => R/5 => R11

22ωω11 = R = R2222ωω22 => ω => ω22= =

RR1122ωω11/R/R22

22= (0.1)= (0.1)22(2 rad/s)/(0.2)(2 rad/s)/(0.2)22 = 0.5 rad/s = 0.5 rad/s

Kepler’s 2Kepler’s 2ndnd law and angular law and angular momentum conservationmomentum conservation

• Kepler’s 2Kepler’s 2ndnd law: law: An imaginary line joining any An imaginary line joining any planet to the sun sweeps out equal areas in planet to the sun sweeps out equal areas in equal time periods.equal time periods.

• For area A, ΔA/Δt is constantFor area A, ΔA/Δt is constant• Angular momentum is conserved because there Angular momentum is conserved because there

is no torque on the planet (about an axis through is no torque on the planet (about an axis through the sun) (F and r are anti-parallel).the sun) (F and r are anti-parallel).

• L=Iω=(L=Iω=(mmrr22)(v/r)=)(v/r)=mmrv is constantrv is constant• Area swept out: ΔA=½rvΔt => Area swept out: ΔA=½rvΔt =>

ΔA/Δt=½rv=½ΔA/Δt=½rv=½mmrv/rv/mm=½L/=½L/mm => => constant!

Chapter 9 conceptsChapter 9 concepts

• No net force means no translational accelerationNo net force means no translational acceleration

• No net torque means no rotational accelerationNo net torque means no rotational acceleration

• Object is said to be in Object is said to be in equilibriumequilibrium if no net if no net force and no net torqueforce and no net torque

• Study of forces on objects at rest is called Study of forces on objects at rest is called staticsstatics

• Most solids are elastic (for a range of forces) Most solids are elastic (for a range of forces) and obey Hooke's law (F= -kx) and obey Hooke's law (F= -kx) (more on this in Chapter 11)(more on this in Chapter 11)

Fluid dynamicsFluid dynamics

• When apply a force to a solid it accelerates, When apply a force to a solid it accelerates, doesn't deform (much)doesn't deform (much)

• When apply a force to a liquid or gas, it When apply a force to a liquid or gas, it flows, does not keep a fixed shapeflows, does not keep a fixed shape

• In physics-speak liquids and gasses are In physics-speak liquids and gasses are known as known as fluidsfluids

• Fluid dynamics is the subject of what Fluid dynamics is the subject of what happens to a fluid when you apply a forcehappens to a fluid when you apply a force

DensityDensity

• An object's density is its mass per unit An object's density is its mass per unit volume: volume: ρ=m/Vρ=m/V

• Specific gravity (SG) is the density divided Specific gravity (SG) is the density divided by the density of waterby the density of water

• ρρwaterwater= 1000 kg/m= 1000 kg/m33 = 1 g/cm = 1 g/cm33

• ExampleExample

Your turnYour turn

What is the mass of a small gold brick What is the mass of a small gold brick 3 cm x 4 cm x 5 cm? (3 cm x 4 cm x 5 cm? (ρρgoldgold= 19,300 kg/m= 19,300 kg/m33))

Answer: Answer: m=ρVm=ρV= = (19,300 kg/m(19,300 kg/m33)(0.03 m)(0.04 m)(0.05 m) = )(0.03 m)(0.04 m)(0.05 m) = 1.16 kg1.16 kg

PressurePressure

• Pressure is force per unit area:Pressure is force per unit area: P P=F/=F/AA

• Units: 1 N/mUnits: 1 N/m22= 1 Pa (Pascal)= 1 Pa (Pascal)

• Named after Blaise Pascal (1623-1662)Named after Blaise Pascal (1623-1662)

• English unit: pounds per square inch (psi): English unit: pounds per square inch (psi): 1 lb/in1 lb/in22 = 6900 Pa = 6900 Pa

• Average atmospheric pressure at sea level Average atmospheric pressure at sea level provides another unit: 1 atm = 1.013 x 10provides another unit: 1 atm = 1.013 x 105 5 Pa = Pa = 14.7 lb/in14.7 lb/in22

• ExampleExample

Gauge PressureGauge Pressure

• Most pressure gauges (like tire gauges) Most pressure gauges (like tire gauges) measure the pressure relative to that of the measure the pressure relative to that of the atmosphereatmosphere

• Absolute pressure = Absolute pressure = atmospheric pressure + gauge pressure: atmospheric pressure + gauge pressure: PP==PPAA+ + PPGG

• Example: If your tire gauge reads 32.1 psi, Example: If your tire gauge reads 32.1 psi, then the absolute pressure inside your tire is then the absolute pressure inside your tire is PP=14.7 psi + 32.1 psi = 46.8 psi=14.7 psi + 32.1 psi = 46.8 psi

Your turnYour turnSuppose your tire gauge reads 30 psi and the Suppose your tire gauge reads 30 psi and the

inside surface area of your tire is 900 square inside surface area of your tire is 900 square inches. What is the total force on the inside inches. What is the total force on the inside walls of your tire?walls of your tire?

Answer: Answer: PP=14.7 psi + 30 psi = 44.7 psi; =14.7 psi + 30 psi = 44.7 psi; F=F=PAPA=(44.7 lb/in=(44.7 lb/in22)(900 in)(900 in22)= 40200 lb)= 40200 lb

Note: the force on the outside of the tire will Note: the force on the outside of the tire will be be PP=14.7 psi; =14.7 psi; F=F=PAPA=(14.7 lb/in=(14.7 lb/in22)(900 in)(900 in22)= 13200 lb)= 13200 lb

Fluid staticsFluid statics

• Study of fluids at rest (not Study of fluids at rest (not flowing)flowing)

• Consider a small volume of Consider a small volume of water inside of a beakerwater inside of a beaker

• Forces must balance in all Forces must balance in all directions or else the fluid directions or else the fluid would flowwould flow

Two key ideas of fluid staticsTwo key ideas of fluid statics

• Fluids exert pressure in all directionsFluids exert pressure in all directions

• The force a fluid exerts on a surface is The force a fluid exerts on a surface is directed perpendicular to the surface directed perpendicular to the surface

• (If the force had a component along the (If the force had a component along the surface, from Newton's 3surface, from Newton's 3rdrd law, the fluid law, the fluid would flow along the surface and not be at would flow along the surface and not be at rest.)rest.)

Variation of pressure with Variation of pressure with depthdepth

• Assume fluid is Assume fluid is incompressibleincompressible (density of (density of fluid does not change with pressure) fluid does not change with pressure) (Note: (Note: notnot a good assumption for gases.) a good assumption for gases.)

• In order to be at rest, the pressure must In order to be at rest, the pressure must balance the weight of the fluidbalance the weight of the fluid

Variation of pressure Variation of pressure with depthwith depth

• Top pressure force plus weight equals bottom Top pressure force plus weight equals bottom (upward) pressure force: (upward) pressure force: PP00AA + + mmg = g = PP11AA

• Relate mass to density: Relate mass to density: m=ρV= ρAhm=ρV= ρAh

• PP00A + (ρAh)A + (ρAh)gg = P = P11A => PA => P00 + ρ + ρggh = Ph = P11

• PP11 - P - P00 = ρ = ρggh => h => ΔΔPP= = ρρgΔgΔyy

• Pressures at equal depths Pressures at equal depths hh are equal: are equal: PP = = ρρgghh

• ExampleExample

PP11

PP00

mmgg h

Your turnYour turn

What is the gauge pressure 5 cm below the What is the gauge pressure 5 cm below the surface?surface?

Answer: Answer: PPGG = =P-PP-PAA=P-P=P-P00= ρ= ρgghh= =

(1000 kg/m(1000 kg/m33)(9.8 m/s)(9.8 m/s22)(0.05 m)= )(0.05 m)= 490 Pa490 Pa

Pascal's principlePascal's principle

• Pressure applied to a confined fluid Pressure applied to a confined fluid increases the pressure throughout by the increases the pressure throughout by the same amountsame amount

• External pressure acting on a fluid is External pressure acting on a fluid is transmitted to all points of the fluidtransmitted to all points of the fluid

• Principle behind the hydraulic liftPrinciple behind the hydraulic lift

Hydraulic lift exampleHydraulic lift example• Suppose the car in the figure weighs 10,000 N Suppose the car in the figure weighs 10,000 N

and the area of the lift where the car sits is 10 and the area of the lift where the car sits is 10 mm22. If the area where the force F. If the area where the force F inin is applied is is applied is

0.5 m0.5 m22, find the force F, find the force Finin needed to lift the car. needed to lift the car.

• PPoutout= F= Foutout//AAoutout = (10,000 N)/(10 m = (10,000 N)/(10 m22) = 1000 ) = 1000

N/mN/m22; ; PPinin= = PPoutout => F => Finin//AAinin =1000 N/m =1000 N/m22 => =>

FFinin=(1000 N/m=(1000 N/m22))AAinin =(1000 N/m =(1000 N/m22)(0.5 m)(0.5 m22)= 500 )= 500

NN• A 500 N force can lift a 10,000 N car using A 500 N force can lift a 10,000 N car using

pressure in an incompressible fluid.pressure in an incompressible fluid.