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PHSX 114, Monday, October 20, 2003

PHSX 114, Monday, October 20, 2003. Reading for today: Chapter 10 (10-1 -- 10-4) Reading for next lecture (Wed.): Chapter 10 (10-5 -- 10-8) Homework for today's lecture: Chapter 10, problems 3, 5, 9, 14, 61. Conservation of angular momentum.

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PHSX 114, Monday, October 20, 2003

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  1. PHSX 114, Monday, October 20, 2003 • Reading for today: Chapter 10 (10-1 -- 10-4) • Reading for next lecture (Wed.): Chapter 10 (10-5 -- 10-8) • Homework for today's lecture: Chapter 10, problems 3, 5, 9, 14, 61

  2. Conservation of angular momentum • If the torque is zero, then ΔL= 0 (no change in L) • The total angular momentum of a system remains constant if there is no net external torque acting on the system. • Ice skater example

  3. Your turn A 1-kg ball of radius 0.1 m is spinning initially with ω1= 2 rad/s. The ball’s radius increases to 0.2 m. Find the final angular velocity. Answer: L=Iω is constant, so I1ω1=I2ω2. I1=2MR12/5, I2=2MR22/5 => 2MR12ω1/5 = 2MR22ω2/5 => R12ω1 = R22ω2 => ω2= R12ω1/R22= (0.1)2(2 rad/s)/(0.2)2 = 0.5 rad/s

  4. Kepler’s 2nd law and angular momentum conservation • Kepler’s 2nd law: An imaginary line joining any planet to the sun sweeps out equal areas in equal time periods. • For area A, ΔA/Δt is constant • Angular momentum is conserved because there is no torque on the planet (about an axis through the sun) (F and r are anti-parallel). • L=Iω=(mr2)(v/r)=mrv is constant • Area swept out: ΔA=½rvΔt => ΔA/Δt=½rv=½mrv/m=½L/m => constant!

  5. Chapter 9 concepts • No net force means no translational acceleration • No net torque means no rotational acceleration • Object is said to be in equilibrium if no net force and no net torque • Study of forces on objects at rest is called statics • Most solids are elastic (for a range of forces) and obey Hooke's law (F= -kx) (more on this in Chapter 11)

  6. Fluid dynamics • When apply a force to a solid it accelerates, doesn't deform (much) • When apply a force to a liquid or gas, it flows, does not keep a fixed shape • In physics-speak liquids and gasses are known as fluids • Fluid dynamics is the subject of what happens to a fluid when you apply a force

  7. Density • An object's density is its mass per unit volume: ρ=m/V • Specific gravity (SG) is the density divided by the density of water • ρwater= 1000 kg/m3 = 1 g/cm3 • Example

  8. Your turn What is the mass of a small gold brick 3 cm x 4 cm x 5 cm? (ρgold= 19,300 kg/m3) Answer: m=ρV= (19,300 kg/m3)(0.03 m)(0.04 m)(0.05 m) = 1.16 kg

  9. Pressure • Pressure is force per unit area: P=F/A • Units: 1 N/m2= 1 Pa (Pascal) • Named after Blaise Pascal (1623-1662) • English unit: pounds per square inch (psi): 1 lb/in2 = 6900 Pa • Average atmospheric pressure at sea level provides another unit: 1 atm = 1.013 x 105 Pa = 14.7 lb/in2 • Example

  10. Gauge Pressure • Most pressure gauges (like tire gauges) measure the pressure relative to that of the atmosphere • Absolute pressure = atmospheric pressure + gauge pressure: P=PA+ PG • Example: If your tire gauge reads 32.1 psi, then the absolute pressure inside your tire is P=14.7 psi + 32.1 psi = 46.8 psi

  11. Your turn Suppose your tire gauge reads 30 psi and the inside surface area of your tire is 900 square inches. What is the total force on the inside walls of your tire? Answer: P=14.7 psi + 30 psi = 44.7 psi; F=PA=(44.7 lb/in2)(900 in2)= 40200 lb Note: the force on the outside of the tire will be P=14.7 psi; F=PA=(14.7 lb/in2)(900 in2)= 13200 lb

  12. Fluid statics • Study of fluids at rest (not flowing) • Consider a small volume of water inside of a beaker • Forces must balance in all directions or else the fluid would flow

  13. Two key ideas of fluid statics • Fluids exert pressure in all directions • The force a fluid exerts on a surface is directed perpendicular to the surface • (If the force had a component along the surface, from Newton's 3rd law, the fluid would flow along the surface and not be at rest.)

  14. Variation of pressure with depth • Assume fluid is incompressible (density of fluid does not change with pressure) (Note: not a good assumption for gases.) • In order to be at rest, the pressure must balance the weight of the fluid

  15. P0 Variation of pressure with depth h mg P1 • Top pressure force plus weight equals bottom (upward) pressure force: P0A + mg = P1A • Relate mass to density: m=ρV= ρAh • P0A + (ρAh)g = P1A => P0 + ρgh = P1 • P1 - P0 = ρgh => ΔP= ρgΔy • Pressures at equal depths h are equal: P= ρgh • Example

  16. Your turn What is the gauge pressure 5 cm below the surface? Answer: PG =P-PA=P-P0= ρgh= (1000 kg/m3)(9.8 m/s2)(0.05 m)= 490 Pa

  17. Pascal's principle • Pressure applied to a confined fluid increases the pressure throughout by the same amount • External pressure acting on a fluid is transmitted to all points of the fluid • Principle behind the hydraulic lift

  18. Hydraulic lift example • Suppose the car in the figure weighs 10,000 N and the area of the lift where the car sits is 10 m2. If the area where the force Fin is applied is 0.5 m2, find the force Fin needed to lift the car. • Pout= Fout/Aout = (10,000 N)/(10 m2) = 1000 N/m2; Pin= Pout => Fin/Ain =1000 N/m2 => Fin=(1000 N/m2)Ain =(1000 N/m2)(0.5 m2)= 500 N • A 500 N force can lift a 10,000 N car using pressure in an incompressible fluid.

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