Physics 31 Lecture 3 Main points of today s lecture: Gravitation potential energy GM1M PEgrav r 1 Tensile stress and strain Δ Y ΔL L 0 Bulk stress and strain: Δ V Δ P B Δ V Pressure in fluids: P ; Pbot Ptop + ρgh Buoyancy B W displaced ρ V f displaced g
Gravitational Potential Energy ΔPE mgh is valid only near the earth s surface or objects high h above the earth s surface, an alternate expression is needed (You would need calculus to derive it.) PE GG M r Zero potential energy is defined to be the value infinitely far from the earth If the total mechanical energy of an object exceeds zero, the object can escape the gravitation field of the earth. The escape velocity is when the E mech 0. t the Earth's surface: 1 GM Emech mv R mech Earth Earth If E 0 corresponding to v v, the object m E m escape can escape the Earth's gravitational field 1 GM m GM 0 mv v R R Earth Earth escape escape Earth Earth you will have loncapa problems on this. 9
Conceptual question rock, initially iti at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth. n observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Neglecting friction, the rock s speed when it hits the ground is + + a) twice GMEm b) three times 0 r c) four times also: KE0 0 d) six times e) eight times its speed at the top of the tower. ground ground KE 4 KE ground tower 1 mv 1 mv ground tower v ground tower v KE PE KE PE PE f f 0 0 0 PE PE 0 f f KE + PE 0 GME m KEf PEf GM rf Em 1 KE d r d r r 4r ground tower E KE GM tower Em 1 r r ground E r rtower tower 4
Stress and strain in solids ll materials deform when under stress, ie. when a force is applied to them. The atoms in a solid are held in place by forces that act like springs. If you pull on a bar of cross-sectional area with a force, the bar will stretch a distance ΔL.. In this case, the force is perpendicular to the surface, and the force and distance ΔL are related by Young s modulus Y: Y ΔL ; L 0 Stress; ΔL L 0 Strain The amount of force you need to apply to achieve the same strain increases with the cross sectional area of the bar. The force will be linearly proportional to the strain up to the elastic limit. Beyond this point, the bar will be permanently deformed. Stress units PaN/m. It is the force divided by the area, and has the units of pressure.
Example Bone has a Young s modulus of about 18 x 10 9 Pa. Under compression, it can withstand a stress of about 160 x 10 6 Pa before breaking. ssume that a femur (thigh bone) is 0.50 m long and calculate the length by which this bone can be compressed before breaking. Y 18 x 10 9 Pa a) 4.4 mm / 160 x 10 6 Pa b) 6.4 mm c) 1.6 cm d).0 cm L 0 ΔL 0.50 m? 6 L 160x10 N / m ΔL Y L 0 L 0 0 Δ L Y 6 0.5m 160x1010 N / m 4.4mm4 9 18x10 N / m If this force is uniformly applied, then /P, where P is the If this force is uniformly applied, then / P, where P is the pressure or force per unit area applied to the end of the femur. Units for pressure are N/m Pa. (Pa Pascals)
Bulk modulus If you drop a metal block in the ocean, it will sink to bottom where the pressure is much greater than that of the atmosphere. This will compress the block; the amount of compression is governed by the bulk modulus B. ΔP ΔV B V ΔV ; ΔP pressure change ; V Volume strain
Reading Quiz 1. The SI unit of pressure is. N B. kg/m C. Pa D. kg/m 3 Slide 13-6
Density Nuclear matter.66x10 17 Slide 13-1
Pressure in fluids, Pascal s principle Consider a vessel filled with liquid and a smaller rectangular volume V of fthe same liquid idcontained within the vessel. If the liquid is at rest, there can be no shear force on V, because it would cause the liquid to move, but there is a pressure force P perpendicular to all 6 surfaces. In equilibrium: 0 x:left right 0; left Pleft x;right Prightx Pleft Pright z: front back 0; front Pfront z; back Pbackz Pfront Pback W y: W 0; P ; top Ptop Pbot Ptop + bot top bot bot y y y More detailed considerations show that pressure always exerts a force perpendicular to a surface and that pressure is the same everywhere at the same height, i.e. P front P back, and P left P right and P front P right, etc. Defining density ρ of the liquid by ρm/v, and using V y h we can simplify the dependence of P on height as follows: W Mg ρvg ρhg y ρhg y Pbot Ptop + Pbot Ptop +ρgh y z y x
P bot P Pressure and Depth equation top + ρ gh If the top is open, P bot is normal atmospheric pressure 1.013 x 10 5 Pa 14.7 lb/in The pressure does not depend upon the shape of the container Gauge pressure is the difference between the pressure in the fluid and atmospheric pressure. If you connected a gauge which measures the pressure relative to atmospheric pressure to the fluid at depth h, then for this specific case: P P P ρgh guage bottom,h _ below _ surface atmosphere
Pascal s principal and hydraulic machines Because P, one can make hydraulic machines that can generate large forces. Units: One atmosphere (1 atm) 76.0 cm of mercury 1.013 x 10 5 Pa 1.013 x 10 5 N/m 14.7 lb/in Quiz: simple hydraulic lift is employed to lift a truck that weighs 30000N. If 1 1 cm and 1000 cm, what force 1 is needed to lift the truck? (Hint: re the pressures equal under each piston?) a) 3N P1 P ; this is pascal'ss principal b) 300N c) 30N 1 30000N 1 1 1cm d).0.33n 1 1000cm 30N