Lecture 10 1/34 Phys 220 Physics 220: Classical Mechanics Lecture: MWF 8:40 am 9:40 am (Phys 114) Michael Meier mdmeier@purdue.edu Office: Phys Room 381 Help Room: Phys Room 11 schedule on course webpage Office Hours: 9:40 am-10:40 am or by appointment Course Webpage: www.physics.purdue.edu/phys220/ Textbook: College Physics, 2 nd Ed, Volume 1, by N. Giordano Exams: Midterm July 9th, grades are on CHIP Final Aug 6th (10:30 am - 12:30 pm) Phys 114 Calendar: on course webpage
Lecture 10 2/34 Phys 220 Physics Topics Kinematics Forces Work Conservation of Energy Conservation of Momentum HW 8 & 9 Torque (rotational) HW 8 & 9 Energy (rotational) HW 8 & 9 Angular Momentum (rotational) HW 8 & 9 Density and Pressure
Lecture 10 3/34 Momentum Phys 220 Impulse change in momentum p p f p i Conservation of Momentum Elastic objects bounce off momentum conserved v m v m v m v m1 1i 2 2i 1 1 f 2 2 f p related to Force F t p F t KE conserved Inelastic objects bounce off and KE lost to heat momentum conserved KE not conserved v m v m v m v KE m1 1i 2 2i 1 1 f 2 2 f i KE f ma Totally Inelastic objects stick together and KE lost to heat momentum conserved KE not conserved m v m v m v ( m m ) 2 v KE m v 1 1i 2 2i 1 f 2 f 1 f 1 2 m v 1 2 1i 1 m2v 2 2 2i 1 m1v 2 2 1 f i KE f 1 2 m 2 v 2 2 f
Lecture 10 4/34 Phys 220 Linear and Rotational Motion Linear Motion Rotational Motion x θ v ω a α m I F = m a τ = I α p = m v L = I ω KE translational = (1/2) m v 2 KE rotational = (1/2) I ω 2
Lecture 10 5/34 Phys 220 Kinetic Energy An object has Translational KE KE translational = (1/2) m v 2 if the pivot point (or center of mass) is moving Rotational KE KE rotational = (1/2) I ω 2 if the object is rotating about a pivot point (or center of mass) KE rot KE trans
Lecture 10 6/34 Phys 220 How to Solve Torque Problems Draw a picture with forces Pick a pivot point Pick a direction for positive torque If a force will make the object spin in the positive direction, then the torque is positive Write down the equation : N I 1 N2 Fg crate Fg beam r F r F sin 0 R θ F N 1 N 2 Fg oncrate F g on beam
Lecture 10 7/34 Phys 220 How to Solve Torque Problems N 1 N 2 Fg oncrate F g on beam N 1 R 4 R 1 =0 R 2 θ 2 N 2 F g on crate θ 4 R 3 θ 3 F g on beam : N 1 0 R 2 N N 2 2 Fg crate 2 sin R 3 Fg m beam crate I 0 g sin 3 R 4 m beam g sin 4 0
Lecture 10 8/34 Phys 220 Physics Topics Kinematics Forces Work Conservation of Energy Conservation of Momentum Torque (rotational) Energy (rotational) Angular Momentum (rotational) M7 Density and Pressure
Lecture 10 9/34 Phys 220 Example 1 - A A student of mass 100 kg is sitting on a freely rotating stool and is holding a rotating bicycle wheel. Initially (Picture A), the angular velocity of the bicycle wheel is 10 rad/s and its mass is 1 kg and radius is 0.5 m. The student then moves the bicycle wheel as shown (Picture B) and begins to rotate at 1 rad/s. What is the total angular momentum of the system at B? A B L total at B = 0 kg*m 2 /s
Lecture 10 10/34 Phys 220 Example 1 - B A student of mass 100 kg is sitting on a freely rotating stool and is holding a rotating bicycle wheel. Initially (Picture A), the angular velocity of the bicycle wheel is 10 rad/s and its mass is 1 kg and radius is 0.5 m. The student then moves the bicycle wheel as shown (Picture B) and begins to rotate at 1.5 rad/s. What is the magnitude of person s angular momentum at B? A B L person at B = 2.5 kg*m 2 /s
Lecture 10 11/34 Phys 220 Example 1 - C : Question 1 A student of mass 100 kg is sitting on a freely rotating stool and is holding a rotating bicycle wheel. Initially (Picture A), the angular velocity of the bicycle wheel is 10 rad/s and its mass is 1 kg and radius is 0.5 m. The student then moves the bicycle wheel as shown (Picture B) and begins to rotate at 1.5 rad/s. What is the change in the person s angular momentum and the change in the wheel s angular momentum? A) L P = 0 kg*m 2 /s B) L P = 2.5 kg*m 2 /s C) L P = 5.0 kg*m 2 /s L w = -2.5 kg*m 2 /s L w = -2.5 kg*m 2 /s L w = -5.0 kg*m 2 /s A B
Lecture 10 12/34 Phys 220 Example 1 - D : Question 2 A student of mass 100 kg is sitting on a freely rotating stool and is holding a rotating bicycle wheel. Initially (Picture A), the angular velocity of the bicycle wheel is 10 rad/s and its mass is 1 kg and radius is 0.5 m. The student then moves the bicycle wheel as shown (Picture B) and begins to rotate at 1.5 rad/s. Finally, the student flips the bicycle wheel over (Picture C). What is the total angular momentum of the system at C? A) 2.5 kg*m 2 /s B) 5.0 kg*m 2 /s C) 0 kg*m 2 /s D) unknown A B C
Lecture 10 13/34 Phys 220 Example 1 - E : Question 3 A student of mass 100 kg is sitting on a freely rotating stool and is holding a rotating bicycle wheel. Initially (Picture A), the angular velocity of the bicycle wheel is 10 rad/s and its mass is 1 kg and radius is 0.5 m. The student then moves the bicycle wheel as shown (Picture B) and begins to rotate at 1.5 rad/s. Finally, the student flips the bicycle wheel over (Picture C). What is the magnitude of angular velocity of the person at C? A) 10 rad/s B) 1.5 rad/s C) 0 rad/s D) 5 rad/s E) unknown A B C
Lecture 10 14/34 Phys 220 Physics Topics Kinematics Forces Work Conservation of Energy Conservation of Momentum Torque (rotational) Energy (rotational) Angular Momentum (rotational) Density and Pressure
Lecture 10 15/34 Phys 220 States of Matter Fluids Solid Hold Volume Hold Shape Liquid Gas Hold Volume Adapt Shape Adapt Volume Adapt Shape Compressible and uncompressible fluids
Lecture 10 16/34 Phys 220 Pressure Pressure = F normal / A Scalar Units:» Pascal (Pa) = N/m 2» Atmosphere: 1 atm = 101.3 kpa» 1 atm ~ 10 N/cm 2 ~ 15 lb/in 2 Force due to molecules of fluid colliding with container. Change in Impulse = p
Lecture 10 17/34 Phys 220 Pascal s Principle A change in pressure at any point in a confined fluid is transmitted everywhere in the fluid Hydraulic Lift F1 F2 P = F 1 / A 1 on right P1 P 2 A P = F 2 / A 2 on left 1 A2 Since P is same, set equal F 1 / A 1 = F 2 / A 2 F 2 = F 1 (A 2 / A 1 ) Can get LARGE force! Volume is conserved A 1 d 1 = A 2 d 2 d 2 = d 1 A 1 / A 2 Work is the SAME: W = Fd F 2 d 2 = F 1 (A 2 /A 1 )*d 1 (A 1 /A 2 ) = F 1 d 1
Lecture 10 18/34 Phys 220 Density Mass density of an object of mass m and volume V Density = Mass/Volume = m/v Units = kg/m 3 Densities of some common things (kg/m 3 ) Water 1000 kg/m 3 1 g/cm 3 (water) m V V Ice 917 kg/m 3 0.917 g/cm 3 (floats on water) Blood 1060 kg/m 3 1.060 g/cm 3 (sinks in water) Lead 11,300 kg/m 3 11.3 g/cm 3 Copper 8890kg/m 3 8.9 g/cm 3 Mercury 13,600 kg/m 3 13.6 g/cm 3 Aluminum 2700 kg/m 3 2.7 g/cm 3 Wood 550 kg/m 3 0.55 g/cm 3 (floats on water) Air 1.29 kg/m 3 0.00129 g/cm 3 Helium 0.18 kg/m 3 0.00018 g/cm 3 Uranium 19,000 kg/m 3 19.0 g/cm 3 m
Lecture 10 19/34 Phys 220 Point A F A A A Question 4 F B Point B A B Pressure is constant in this pipe from point A to point B What happens to the force at B compared to the force A, given that the area at B is smaller than the area at A? A) The force at B is less than the force at A B) The force at B is greater than the force at A C) Both forces at A and B are the same force
Lecture 10 20/34 Phys 220 Gravity and Pressure Consider a small piece of the fluid Draw Forces on the fluid element y-direction: y P 2 A mg P 1 A = 0 P 2 A (Ad)g P 1 A = 0 x P 2 (d)g P 1 = 0 P 2 = P 1 + gd m V m Ad Pressure under fluid P = P atmosphere + gd Basically weight of air + weight of fluid P 1 P2 g h2
Lecture 10 21/34 Phys 220 Pressure versus Depth For a fluid in an open container: The pressure is the same at a given depth, independent of shape of the container The fluid level is the same in a connected container
Lecture 10 22/34 Phys 220 Question 5 Two identical light containers are filled with water. The first is completely full of water, the second container is filled only ½ way. Compare the pressure each container exerts on the table. A) P 1 > P 2 B) P 1 = P 2 C) P 1 < P 2 P = F/A = mg / A Cup 1 has greater mass, but same area 1 2 Under water P = P atmosphere + g h
Lecture 10 23/34 Phys 220 Demo Pressure Forces
Lecture 10 24/34 Phys 220 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1.01310 5 N/m 2 = 14.7 lbs/in 2 Example 1: Disc r = 2.4 in A = p r 2 = 18.1 in 2 (0.01167 m 2 ) F = P A = (14.7 lbs/in 2 )(18.1 in 2 ) = 266 lbs Example 2: Sphere r = 0.1 m A = 4 p r 2 = 0.125 m 2 F = 12,000 N (over 2,500 lbs)!
Lecture 10 25/34 Phys 220 Atmospheric Pressure You buy a bag of potato chips in Lafayette and forget them (unopened) under the seat of your car. You drive out to Denver Colorado to visit a friend for Thanksgiving, and when you get there you discover the lost bag of chips. The odd thing you notice right away is that the bag seems to have inflated like a balloon (i.e. it seems much more round and bouncy than when you bought it). How can you explain this? P U P L In Lafayette Due to the change in height, the air is much thinner in Denver. Thus, there is less pressure on the outside of the bag of chips in Denver, so the bag seems to inflate because the air pressure on the inside is greater than on the outside. P U P D In Denver
Lecture 10 26/34 Phys 220 Pressure and Depth Barometer: a device to measure atmospheric pressure Pressure in the liquid at points A and B is the same P A = Atmospheric Pressure P B = 0 P P A atm g h g (0) P A atm P B g h B (0) g h g h p 1 =0 Measure h, determine p atm Example: Mercury = 13,600 kg/m 3 p atm = 1.01 x 10 5 Pa h = 0.757 m = 757 mm = 29.80 (for 1 atm) p 2 =p atm A B h
Lecture 10 27/34 Phys 220 Question 6 Suppose you have a barometer with mercury and a barometer with water. How does the height h water compare with the height h mercury? A) h water is much larger than h mercury B) h water is a little larger than h mercury C) h water is a little smaller than h mercury ρ water = 1,000 kg/m 3 ρ mercury = 13,600 kg/m 3 D) h water is much smaller than h mercury p 1 =0 mercury 13. 6 water h P atm g h mercury h water P 13.6 atm 13.6 h water g mercury p 2 =p atm h
Lecture 10 28/34 Phys 220 Pumping Water Vacuum Pump at surface High Pressure Pump underground P bottom g h bottom P atm P top g h top P bottom g h bottom g h 0 g h P P g h 5 Patm 1.01310 P h 10.33 m h g (1000)(9.8) bottom bottom P g P top atm atm top no limit!
Lecture 10 29/34 Phys 220 Manometer A device to measure gas pressure P B = P B = P C + g d P = P B - P C = g d The difference in mercury level d is a measure of the pressure difference.
Lecture 10 30/34 Phys 220 Demo Fluid Density and Pressure
Lecture 10 31/34 Phys 220 Question 7 A B Two dams of equal height prevent water from entering the basin. Compare the net force due to the water on the two dams. A) F A > F B B) F A =F B C) F A < F B F = P A, and pressure is gh. Same pressure, same area same force even though more water in B!
Lecture 10 32/34 Phys 220 Archimedes Principle Determine force of fluid on immersed cube Draw FBD» F B = F 2 F 1 = P 2 A P 1 A = (P 2 P 1 )A = g d A = g V Buoyant force is the weight of the displaced fluid! F F B B m 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
Lecture 10 33/34 Phys 220 Sink or Float The buoyant force is equal to the weight of the liquid that is displaced If the buoyant force is larger than the weight of the object, it will float; otherwise it will sink We can calculate how much of a floating object will be submerge If the object is in equilibrium V V F m V g liquid displaced object displaced B object liquid object object g V object g y Does an object float or sink? F = ( fluid - solid ) g V fluid > solid F > 0 object rises fluid = solid F = 0 neutral buoyancy fluid < solid F < 0 object sinks
Lecture 10 34/34 Phys 220 Exam Question A non-uniform beam has a mass of 5 kg and a length of 10 m. The beam is supported on the left end by a 32 N force and on the right end by a 17 N force. How far is the center of mass of the beam from the left end? A) 1.68 m B) 4.65 m C) 6.53 m D) 8.32 m E) 3.47 m