Physics 220: Classical Mechanics

Transcription

1 Lecture /33 Phys 0 Physics 0: Classical Mechanics Lecture: MWF 8:40 am 9:40 am (Phys 4) Michael Meier mdmeier@purdue.edu Office: Phys Room 38 Help Room: Phys Room schedule on course webpage Office Hours: 9:40 am-0:40 am or by appointment Course Webpage: Textbook: College Physics, nd Ed, Volume, by N. Giordano Exams: Midterm July 9th, grades are on CHIP Final Aug 6th (0:30 am - :30 pm) Phys 4 Calendar: on course webpage

2 Lecture /33 Phys 0 Physics Topics Kinematics Forces Work Conservation of Energy Conservation of Momentum Angular Momentum M7 Density and Pressure Harmonic Motion

3 Lecture 3/33 Phys 0 Determine force of fluid on immersed cube Draw FBD» F B = F bot F top = P bot A P top A = (P bot P top )A = r liquid g d A = r liquid g V sub Archimedes Principle Buoyant force is the weight of the displaced fluid! F F B B m r displaced fluid liquid V g submerged g A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object

4 Lecture 4/33 Phys 0 Apparent Weight The minimum force required to lift a submerged object is called the apparent weight The apparent weight of the box is smaller than its true weight by an amount equal to the buoyant force acting on it Section 0.4

5 Lecture 5/33 Phys 0 Archimedes Example A cube of plastic 4.0 cm on a side with density = 0.8 g/cm 3 is floating in the water. When a 9 gram coin is placed on the block, how much does it sink below water surface? S F = m a F b Mg mg = 0 r g V disp = (M+m) g V disp = (M+m) / r h A = (M+m) / r h = (M + m)/ (r A) Mg = (5.+9)/( x 4 x 4) = 3.76 cm F b mg h M = ρ plastic V cube = 4x4x4x0.8 = 5. g

6 Lecture 6/33 Phys 0 Question 4 A crate is submerged in water. Which of the following free body diagrams is correct? A) C) N F gravity B) F B F gravity D) F gravity F B F gravity

7 Lecture 7/33 Phys 0 Question 5 A ship is using a chain to pull up its anchor in the ocean. The volume of the anchor is.4 m 3, and the anchor is made of steel and has a density of 7,800 kg/m 3. If the density of the ocean water is,000 kg/m 3, what is the buoyant force on the anchor? Which density is used to calculate the force of buoyancy on the anchor? A) density of steel B) density of water C) both the density of steel and the density of water F B = 3.5 kn

8 Lecture 8/33 Phys 0 Fluids in Motion The general theory of fluids in motion is very complex since the velocity of a fluid usually varies from place to place within the fluid Assumptions made to simplify: The fluid density is constant» The fluid is incompressible The flow is steady» The velocity is independent of time» It can still vary from point to point in space There is no friction» No viscosity There are no complex flow patterns» Such as turbulence

9 Lecture 9/33 Phys 0 Principle of Continuity Consider the velocity through the pipe The fluid enters the pipe on the left with speed v L It exits on the right with speed v R The amount of fluid that flows through the pipe must be conserved Volume flow rate: Q V t Av Mass flow rate: m t r Av Continuity: r A v r A v

10 Lecture 0/33 Phys 0 Question 6 Water is flowing out of a faucet as shown in the picture on the right. The water at has a velocity v and the diameter of is twice the diameter of. What is the velocity of the water at (v )? A) v = v /4 B) v = v / C) v = v D) v = *v E) v = 4*v

11 Lecture /33 Phys 0 Demo - Streamlines

12 Lecture /33 Phys 0 Flow Speed and Pressure Continuity Equation says fluid speeds up going through a smaller opening slows through al larger opening Acceleration due to change in pressure P > P Smaller tube has high velocity and low pressure Large tube has low velocity and high pressure

13 Lecture 3/33 Bernoulli s Equation Phys 0 Consider a tube where both area and height change W W W total total total W W F x P A x F x P A x W W * total * total KE PE (/ ) mv mgh (/ mgh ) mv W total W * total Bernoulli s Equation: PV PV P rv mgh r gh mv P mgh rv mv r gh

14 Lecture 4/33 Phys 0 Interpreting Bernoulli s Equation The pressure terms represent energy associated with pressure in the fluid A higher value of P corresponds to higher energy because the fluid can exert a higher force Bernoulli s equation says that the total mechanical energy of the fluid is conserved as it travels from place to place, but some of this energy can be converted from kinetic energy to potential energy or potential to kinetic Static Pressure (energy due to pressure) Total Pressure P rv r gh P rv r gh constant Dynamic Pressure (kinetic energy) Head Pressure (potential energy)

15 Lecture 5/33 Phys 0 Demo Venturi Tubes

16 Lecture 6/33 Phys 0 Question 7 Air moves over an airfoil as shown below. v top Top v bottom Bottom Where is the velocity of the air greater? A) The velocity on the top is greater B) The velocity on the bottom is greater C) They are the same velocity

17 Lecture 7/33 Phys 0 Question 8 Air moves over an airfoil as shown P top r v top r g h top Assume that the height of the airfoil is small Where is the pressure greater? bottom P top P bottom Top Bottom top A) The pressure on the top is greater B) The pressure on the bottom is greater C) They are the same pressure P P top r v r v bottom r g P bottom h bottom r v bottom

18 Lecture 8/33 Phys 0 Airplane Wing The speed of the air over the top of the wing is higher than over the bottom of the wing An increase in speed causes a decrease in pressure F total = (P bot P top ) A This produces the lift on the airplane wing Ptop r vtop r g htop Pbottom r vbottom r g hbottom There is a difference in height, but» The high air speeds and small difference in height indicates the pressure difference is due almost entirely to the speed terms in Bernoulli s equation Bernoulli s equation applied to the airplane wing assumed an incompressible fluid Air is compressible There is about a 5% difference due to differences in density» This means this result for lift is accurate to within about 5%

19 Lecture 9/33 Phys 0 Pitot-Static Tube Used on airplanes to calculate the velocity of the airplane height difference is extremely small Velocity of air entering the tube is zero g h v P g h v P r r r r 0 since v v P P r r ) ( P P v

20 Lecture 0/33 Phys 0 Turbulence When fluid velocities are large, the flow patterns can be very complex and fluctuate with time The picture on the left shows vortices produced in the baseball s wake Dimples on a golf ball cause turbulent flow that reduces the size of the wake and results in less drag

21 Lecture /33 Phys 0 Demo Ball in Tube

22 Lecture /33 Phys 0 Viscosity Real fluids have frictional forces internal to the fluid and between the fluid and the walls of the container These are called viscous forces These forces depend on the speed of the fluid They are often negligible for small speeds In many cases, however, viscosity is important Consider the fluid flowing through a tube The fluid molecules near the wall of the tube experience strong forces from the molecules in the wall The molecules close to the wall move relatively slowly Thick fluids experience large forces and have a large viscosity honey Thin fluids have much smaller viscosities water

23 Lecture 3/33 Phys 0 Simple Harmonic Motion Harmonic motion is a type of motion in which an object moves along a repeating path over and over again position velocity

24 Lecture 4/33 Phys 0 Simple Harmonic Oscillation Force exerted by a spring is directly proportional to the amount by which it is stretched or compressed. F spring = - k x x is the displacement from the relaxed position and k is the constant of proportionality always trying to restore its original length

25 Lecture 5/33 Phys 0 Analogy to Circular Motion What does moving in a circle have to do with moving back and forth in a straight line? x = A cos = A cos (wt) since = w t y x A A 7 -A x 7 3 8

26 Lecture 6/33 Phys 0 Analogy to Circular Motion Period = T (seconds per cycle) Frequency = f = /T (cycles per second) Angular frequency = w = f = /T Demo Scotch Yoke Model

27 Lecture 7/33 Phys 0 Analogy to Circular Motion

28 Lecture 8/33 Phys 0 Analogy to Circular Motion x(t) = Acos(wt) x(t) = Asin(wt) V(t) = -Awsin(wt) a(t) = -Aw cos(wt) =-w x(t) V(t) = Awcos(wt) a(t) = -Aw sin(wt) =-w x(t) For spring: F = ma = -kx, so a(t) = -(k/m)x(t). By analogy, we have w = k/m x max = A v max = Aw a max = Aw

29 Lecture 9/33 Phys 0 Potential Energy in Spring Force of spring is conservative F = -k x Force PE = -W W = -½ k x work PE= ½ k x x Work done only depends on initial and final position

30 Lecture 30/33 Phys 0 Energy Conservation A mass is attached to a spring and set to motion. The maximum displacement is x=a SW nc = KE + PE 0 = KE + PE or Energy PE+KE is constant! Energy = ½ k x + ½ m v At maximum displacement x=a, v = 0 Energy = ½ k A + 0 At zero displacement x = 0 Energy = 0 + ½ mv m Since Total Energy is same ½ k A = ½ m v m v m = sqrt(k/m) A PE S 0 x=0 m x x

31 Lecture 3/33 Phys 0 Example Question 9 A 0. kg mass is on a spring. The mass is displaced a certain distance from equilibrium and let go. The graph on the right shows the position vs time of the mass. Which point on the graph matches the position of the mass on the spring? x=0 E B A D C +x

32 Example Lecture 3/33 Phys 0 A 0. kg mass is on a spring. The mass is displaced a certain distance from equilibrium and let go. The graph on the right shows the position vs time of the mass. How does the graph match with the position of the mass on the spring? x=0 +x

33 Lecture 33/33 Phys 0 Vertical Mass and Spring If we include gravity, there are two forces acting on mass. With mass, new equilibrium position has spring stretched d At equilibrium: F=kd mg=0 d = mg/k Let this point be y=0 At a displacement, F=k(d-y) mg=-k y Total energy: E=/k(d-y) +mgy+/mv =/kd +/ky +/mv Same as the horizontal case, except for a new equilibrium position