1. Lecture 19: Introduction to Solids & Fluids

2. Questions of Yesterday 1) A solid sphere and a hoop of equal radius and mass are both rolled up an incline with the same initial velocity. Which object will travel farthest up the inclined plane? a) the sphere b) the hoop c) they’ll both travel the same distance up the plane d) it depends on the angle of the incline 2) If an acrobat rotates once each second while sailing through the air, and then contracts to reduce her moment of inertia to 1/3 of what is was, how many rotations per second will result? a) once each second b) 3 times each second c) 1/3 times each second d) 9 times each second

3. A 10.00-kg cylindrical reel with the radius of 0.500 m and a frictionless axle starts from rest and speeds up uniformly as a 5.00 kg bucket falls into a well, making a light rope unwind from the reel. The bucket starts from rest and falls for 5.00 s. Practice Problem What is the linear acceleration of the falling bucket? How far does it drop? What is the angular acceleration of the reel? Use energy conservation principles to determine the speed of the spool after the bucket has fallen 5.00 m 5.00 kg 10.0 kg 0.500 m

4. 4 States of Matter GasFluidSolid Plasma Definite Shape & Volume Molecules close together & slow Definite Volume Takes shape of container Molecules farther apart & faster Expands to fill any volume Takes shape of Container Molecules even farther apart & even faster Expands to fill any volume Takes shape of Container Made up of ions Fastest of all matter states

5. Density The amount of matter (mass) in a given volume Density Distance between molecules MVMV  =

6. Solids Definite Shape & Volume Applying force can change shape & size (deform) When force is removed -> original shape & size SOLIDS are ELASTIC Remind you of anything? x = 0 F S = -kx STRESS = ELASTIC MODULUS x STRAIN Force per Unit Area Elasticity of material Measure of Deformation

7. Length Elasticity Elastic Modulus and Induced strain depends on type of stress STRESS = ELASTIC MODULUS x STRAIN L0L0 LL F F L0L0 A F L0L0 L0L0 LL F = Y LL FAFA L0L0 LENGTH ELASTICITY Young’s Modulus Relative Length Change SI Units = N/m 2 = Pascal (Pa)

8. Volume Elasticity STRESS = ELASTIC MODULUS x STRAIN = -B VV FAFA V0V0 VOLUME ELASTICITY Bulk Modulus Relative Volume Change SI Units = N/m 2 = Pascal (Pa) F V0V0 F V0V0 VV

9. Pressure PRESSURE (P) Perpendicular Force per unit area When would this situation occur? Uniform force F is acting over entire surface area A in a direction perpendicular to surface area F P = FAFA A

10. Pressure PRESSURE (P) Perpendicular Force per unit area Fluids are NOT elastic -> do not return to initial state after being deformed But… Fluids do exerted force Uniform force F is acting over entire surface area A in a direction perpendicular to surface area P = FAFA F A

11. Pressure Fluids exert PRESSURE on submerged objects and the walls of their container Fluids are NOT elastic -> do not return to initial state after being deformed But… Fluids do exerted force Force exerted by fluid on a submerged object is always PERPENDICULAR to surface of object F A

12. Pressure The net FORCE acting on any portion of fluid is ZERO If a fluid is at rest in a container what do we know about it? F2F2 It is in EQUILIBRIUM, so… all points at the same DEPTH must be at the same PRESSURE F1F1 y 0 y1y1 y2y2 F 1 (y 1 ) = -F 2 (y 1 ) P 1 (y 1 ) = P 2 (y 1 )

13. Pressure The net FORCE acting on any portion of fluid is ZERO all points at the same DEPTH must be at the same PRESSURE y 0 y1y1 y2y2 P1AP1A P2AP2A Mg ∑F y = P 2 A - P 1 A - Mg = 0 MVMV  = P 2 = P 1 +  g(y 1 - y 2 )

14. Pressure Pressure P at depth h below the surface of a liquid open to the atmosphere is greater than atmosphere pressure (P 0 = 1.013*10 5 Pa) by the amount  gh Gas making up atmosphere exerts pressure on fluid y 0 h y1y1 P 2 = P 1 +  g(y 1 - y 2 ) What is the pressure at the surface of the fluid (open to the air)? P0AP0A PA P = P 0 +  gh

15. Pressure What if you change the pressure exerted at the surface? PASCAL’S PRINCIPLE A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the container F P = P 0 +  gh

16. Pressure What happens if you apply a force F 1 to one side of this apparatus? PASCAL’S PRINCIPLE A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the container F2F2 F1F1 A2A2 A1A1

17. Pressure PASCAL’S PRINCIPLE A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the container F1F1 x2x2 x1x1 F 1 /A 1 = F 2 /A 2 F2F2

18. Pressure How does  x 1 compare to  x 2 ? F1F1 x2x2 x1x1 Fluids have a definite volume (incompressible) ->  V = 0 F 1  x 1 = F 2  x 2 What does this tell you about the work done on the fluid? F2F2 F 1 /A 1 = F 2 /A 2

19. Buoyancy What allows an object to float in a fluid? V M Is the object in equilibrium?

20. Buoyancy V ∑F y = B - m obj g = 0 m What allows an object to float in a fluid? Is the object in equilibrium? mg B

21. Buoyancy V ∑F y = P 2 A - P 1 A - M fluid g = 0 M What allows an object to float in a fluid? Is the object in equilibrium? Mg P1AP1A P2AP2A ∑F y = B - m obj g = 0 B = P 2 A - P 1 A =  fluid V fluid g BUOYANT FORCE

22. Buoyancy V ∑F y = P 2 A - P 1 A - M fluid g = 0 m What allows an object to float in a fluid? Is the object in equilibrium? mg B B = P 2 A - P 1 A =  fluid V fluid g ∑F y = B - m obj g = 0 If the object is in equilibrium with the fluid… BUOYANT FORCE =  fluid  obj V fluid V obj

23. Buoyancy What if the object is rising or sinking? B =  fluid V fluid g B - m obj g > 0 V m mg B =  fluid  obj V fluid V obj V m mg B V m B a = 0 aa B = m obj g B - m obj g < 0 (  fluid -  obj )V obj g > 0(  fluid -  obj )V obj g < 0

24. Properties of an Ideal Fluid NONVISCOUS no internal friction between adjacent layers An Ideal Fluid is… INCOMPRESSIBLE constant density STEADY velocity, density, pressure at each point is constant in time Ideal Fluid Motion is… WITHOUT TURBULENCE Angular velocity about center of each element is zero All points can translate but not rotate

25. Ideal Fluid Motion How does v 1 compare to v 2 ? Equation of Continuity v2v2 A2A2 A1A1 x2x2 x1x1 v1v1 Mass is conserved  M 1 =  M 2  1 A 1 v 1 =  2 A 2 v 2 = v 1  t Fluid is incompressible A 1 v 1 = A 2 v 2 Volume of fluid leaving 1 = Volume of fluid entering 2 in the same time interval

26. Ideal Fluid Motion Is energy conserved in an ideal fluid? v2v2 P1A1P1A1 x2x2 x1x1 v1v1  M 1 =  M 2 W 1 = P 1 A 1  x 1 What is the work done on the fluid? W = P 1 V - P 2 V y2y2 y1y1 P2A2P2A2 W 2 = -P 2 A 2  x 2

27. Ideal Fluid Motion v2v2 A2A2 P1A1P1A1 x2x2 x1x1 v1v1 Is the energy of the fluid changing? What types of energy are present? W = P 1 V - P 2 V y2y2 y1y1 P2A2P2A2 W fluid =  KE +  PE P 1 + (1/2)  v 1 2 +  gy 1 = P 2 + (1/2)  v 2 2 +  gy 2

28. Ideal Fluid Motion v2v2 A2A2 P1A1P1A1 x2x2 x1x1 v1v1 BERNOULLI’S EQUATION The sum of pressure, kinetic energy per unit volume, and potential energy per unit volume is equal at all points along the streamline W = P 1 V - P 2 V y2y2 y1y1 P2A2P2A2 P 1 + (1/2)  v 1 2 +  gy 1 = P 2 + (1/2)  v 2 2 +  gy 2

29. Questions of the Day 1) Two women of equal mass are standing on the same hard wood floor. One is wearing high heels and the other is wearing tennis shoes. Which statement is NOT true? a) both women exert the same force on the floor b) both women exert the same pressure on the floor c) the normal force that the floor exerts is the same for both women 2) A boulder is thrown into a deep lake. As the rock sinks deeper and deeper into the water what happens to the buoyant force? a) it increases b) it decreases c) it stays the same