Physics 207: Lecture 25, Pg 1 Lecture 25 Today Review: Exam covers Chapters plus angular momentum, statics Exam covers Chapters plus angular momentum, statics Assignment Assignment  For Thursday, read through all of Chapter 18

Physics 207: Lecture 25, Pg 2 Angular Momentum Exercise A mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed  i = 5 rad / s in a circle of radius r i = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m. l What is the final angular velocity ? l Underlying concept: Conservation of Momentum riri ii

Physics 207: Lecture 25, Pg 3 Angular Momentum Exercise A mass m=0.10 kg is attached to a cord passing through a small hole in a frictionless, horizontal surface as in the Figure. The mass is initially orbiting with speed  i = 5 rad / s in a circle of radius r i = 0.20 m. The cord is then slowly pulled from below, and the radius decreases to r = 0.10 m. l What is the final angular velocity ? No external torque implies  L = 0 or L i = L c I i  i = I f  f I for a point mass is mr 2 where r is the distance to the axis of rotation m r i 2  i = m r f 2  f  f = r i 2  i / r f 2 = (0.20/0.10) 2 5 rad/s = 20 rad/s riri ii

Physics 207: Lecture 25, Pg 4 Example: Throwing ball from stool A student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector has a perpendicular distance d from the axis of rotation. l  What is the angular speed  F of the student-stool system after they throw the ball ? Top view: before after d Mv M I I FF r

Physics 207: Lecture 25, Pg 5 Example: Throwing ball from stool What is the angular speed  F of the student-stool system after they throw the ball ? l Process: (1) Define system (2) Identify Conditions (1) System: student, stool and ball (No Ext. torque, L is constant) (2) Momentum is conserved (check |L| = |r| |p| sin  for sign) L init = 0 = L final =  M v d + I  f Top view: before after d v M I I FF

Physics 207: Lecture 25, Pg 6 Walking the plank… l A uniform rectangular beam of length L= 5 m and mass M=40 kg is supported, but not attached, to two posts which are length D=3 m apart. A child of mass W=20 kg starts walking along the beam. l Assuming infinitely rigid posts, how close can the child get to the right end of the beam without it falling over? [Hint: The upward force exerted by the beam on the left cannot be negative – this is the limiting condition on how far the child can be to the right. Set up the static equilibrium condition with the pivot about the left end of the beam.] 3 m 5 m 40 kg 20 kg

Physics 207: Lecture 25, Pg 7 Walking the plank… l A uniform rectangular beam of length L= 5 m and mass M=40 kg is supported, but not attached, to two posts which are length D=3 m apart. A child of mass W=20 kg starts walking along the beam. l Sum of forces and torques are zero l As the child walks from left to right N L goes from positive to “negative” l If N L < 0 (a pull) then the board will tip. l Choose blue dot as pivot point. 400 N 200 N NRNR NLNL 0.5 m

Physics 207: Lecture 25, Pg 8 Walking the plank… l A uniform rectangular beam of length L= 5 m and mass M=40 kg is supported, but not attached, to two posts which are length D=3 m apart. A child of mass W=20 kg starts walking along the beam.   = 0 = -N L x x N R x 0 - d = d 200 d = 1 meter from pivot point which is 2.0 meter from right side distance to end (2 – 1) meters or 1 meter 400 N 200 N NRNR NLNL 0.5 m

Physics 207: Lecture 25, Pg 9 Velocity and Acceleration k x m 0 Position: x(t) = A cos(  t +  ) Velocity: v(t) = -  A sin(  t +  ) Acceleration: a(t) = -  2 A cos(  t +  ) by taking derivatives, since: x max = A v max =  A a max =  2 A

Physics 207: Lecture 25, Pg 10 Measuring g To measure the magnitude of the acceleration due to gravity, g, in an unorthodox manner, a student places a ball bearing on the concave side of a flexible speaker cone. The speaker cone acts as a simple harmonic oscillator whose amplitude is A and whose frequency  can be varied. The student can measure both A and . The equations of motion for the speaker are l If the ball bearing is in contact with the speaker, then  F y = ma y = - mg+ N so that N = mg + ma y N = mg - m  2 A cos(  t +  ) Maximum: N = mg + m  2 A Minimum: mg - m  2 A but must be > 0! Position: y(t) = A cos(  t +  ) Velocity: v y (t) = -  A sin(  t +  ) Acceleration: a y (t) = -  2 A cos(  t +  )

Physics 207: Lecture 25, Pg 11 Bernoulli Equation  P 1 + ½  v  g y 1 = constant A 5 cm radius horizontal pipe carries water at 10 m/s into a 10 cm radius. (  water =10 3 kg/m 3 ) What is the pressure difference? P 1 + ½  v 1 2 = P 2 + ½  v 2 2  P = ½  v ½  v 1 2  P = ½  v v 1 2  and A 1 v 1 = A 2 v 2  P = ½  v 2 2  – (A 2 /A 1  2  = 0.5 x 1000 kg/m x 100 m 2 /s 2 (1- (25/100) 2 ) = Pa VV Ideal Fluid

Physics 207: Lecture 25, Pg 12 A water fountain l A fountain, at sea level, consists of a 10 cm radius pipe with a 5 cm radius nozzle. The water sprays up to a height of 20 m.  What is the velocity of the water as it leaves the nozzle?  What volume of the water per second as it leaves the nozzle?  What is the velocity of the water in the pipe?  What is the pressure in the pipe?  How many watts must the water pump supply?

Physics 207: Lecture 25, Pg 13 A water fountain l A fountain, at sea level, consists of a 10 cm radius pipe with a 5 cm radius nozzle. The water sprays up to a height of 20 m. l What is the velocity of the water as it leaves the nozzle? Simple Picture: ½mv 2 =mgh  v=(2gh) ½ = (2x10x20) ½ = 20 m/s l What volume of the water per second as it leaves the nozzle? Q = A v n = x 20 x 3.14 = m 3 /s l What is the velocity of the water in the pipe? A n v n = A p v p  v p = Q /4 = 5 m/s l What is the pressure in the pipe? 1atm + ½  v n 2 = 1 atm +  P + ½  v p 2  1.9 x 10 5 N/m 2 l How many watts must the water pump supply? Power = Q  g h = m 3 /s x 10 3 kg/m 3 x 9.8 m/s 2 x 20 m = 3x10 4 W

Physics 207: Lecture 25, Pg 14 Fluids Buoyancy l A metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a massles rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm 3 and the water is 1.0 gm/cm 3 l What is the average density of the cylinder? l What was the tension in the rope when the cylinder was submerged in the oil?

Physics 207: Lecture 25, Pg 15 Fluids Buoyancy l r = 0.5 m, h= 4.0 m  oil = 0.9 gm/cm 3  water = 1.0 gm/cm 3 l What is the average density of the cylinder? When T = 0 F buoyancy = W cylinder F buoyancy =  oil g ½ V cyl. +  water g ½ V cyl. W cylinder =  cyl g V cyl.  cyl g V cyl. =  oil g ½ V cyl. +  water g ½ V cyl.  cyl = ½  oill. + ½  water What was the tension in the rope when the cylinder was submerged in the oil? Use a Free Body Diagram !

Physics 207: Lecture 25, Pg 16 Fluids Buoyancy r = 0.5 m, h= 4.0 m V cyl. =  r 2 h  oil = 0.9 gm/cm 3  water = 1.0 gm/cm 3 l What is the average density of the cylinder? When T = 0 F buoyancy = W cylinder F buoyancy =  oil g ½ V cyl. +  water g ½ V cyl. W cylinder =  cyl g V cyl.  cyl g V cyl. =  oil g ½ V cyl. +  water g ½ V cyl.  cyl = ½  oill. + ½  water = 0.95 gm/cm 3 What was the tension in the rope when the cylinder was submerged in the oil? Use a Free Body Diagram!  F z = 0 = T - W cylinder + F buoyancy T = W cyl - F buoy = g (  cyl. -  oil ) V cyl T = 9.8 x 0.05 x 10 3 x  x x 4.0 = 1500 N

Physics 207: Lecture 25, Pg 17 Fluids Buoyancy r = 0.5 m, h= 4.0 m V cyl. =  r 2 h  oil = 0.9 gm/cm 3  water = 1.0 gm/cm 3 l What is the average density of the cylinder? When T = 0 F buoyancy = W cylinder.  cyl = ½  oill. + ½  water = 0.95 gm/cm 3 What was the tension in the rope when the cylinder was submerged in the oil? Use a Free Body Diagram!  F z = 0 = T - W cylinder + F buoyancy T = W cyl - F buoy = g (  cyl. -  oil ) V cyl T = 9.8 x 0.05 x 10 3 x  x x 4.0 = 1500 N

Physics 207: Lecture 25, Pg 18 A new trick l Two trapeze artists, of mass 100 kg and 50 kg respectively are testing a new trick and want to get the timing right. They both start at the same time using ropes of 10 m in length and, at the turnaround point the smaller grabs hold of the larger artist and together they swing back to the starting platform. A model of the stunt is shown at right. l How long will this stunt require if the angle is small ?

Physics 207: Lecture 25, Pg 19 A new trick l How long will this stunt require? Period of a pendulum is just  = (g/L) ½ T = 2  (L/g) ½ Time before ½ period Time after ½ period So, t = T = 2  (L/g) ½ = 2  sec Key points: Period is one full swing and independent of mass (this is SHM but very different than a spring. SHM requires only a linear restoring force.)

Physics 207: Lecture 25, Pg 20 Example l A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass. The clay sticks. What is the new amplitude? k m k m (≡X eq ) M M

Physics 207: Lecture 25, Pg 21 Example l A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass. What is the new amplitude? Sequence: SHM, collision, SHM ½ k A 0 2 = const. ½ k A 0 2 = ½ mv 2 + ½ k (A 0 /2) 2 ¾ k A 0 2 = m v 2  v = ( ¾ k A 0 2 / m ) ½ v = (0.75*200*4 / 1 ) ½ = 24.5 m/s Conservation of x-momentum: mv= (m+M) V  V = mv/(m+M) V = 24.5/3 m/s = 8.2 m/s k m k m (≡X eq ) M M

Physics 207: Lecture 25, Pg 22 Example l A Hooke’s Law spring, k=200 N/m, is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. A 1.0 kg mass is initially attached to the spring however, at a displacement of 1.0 m a 2.0 kg lump of clay is dropped onto the mass. The clay sticks. What is the new amplitude? Sequence: SHM, collision, SHM V = 24.5/3 m/s = 8.2 m/s ½ k A f 2 = const. ½ k A f 2 = ½ (m+M)V 2 + ½ k (A i ) 2 A f 2 = [(m+M)V 2 /k + (A i ) 2 ] ½ A f 2 = [3 x /200 + (1) 2 ] ½ A f 2 = [1 + 1] ½  A f 2 = 1.4 m Key point: K+U is constant in SHM k m k m (≡X eq ) M M

Physics 207: Lecture 25, Pg 23 Fluids Buoyancy & SHM l A metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm 3 and the water is 1.0 gm/cm 3 l Refer to earlier example l Now the metal cylinder is lifted slightly from its equilibrium position. What is the relationship between the displacement and the rope’s tension? l If the rope is cut and the drum undergoes SHM, what is the period of the oscillation if undamped?

Physics 207: Lecture 25, Pg 24 Fluids Buoyancy & SHM l Refer to earlier example Now the metal cylinder is lifted  y from its equilibrium position. What is the relationship between the displacement and the rope’s tension? 0 = T + F buoyancy – W cylinder T = - F buoyancy + W cylinder T =-[  o g(h/2+  y) A c +  w gA c (h/2-  y)] + W cyl T =-[ghA c (  o +  w )/2 +  y gA c (  o -  w )]+ W cyl T =-[W cyl +  y gA c (  o -  w )]+ W cyl T = [ g A c (  w -  o ) ]  y If the is rope cut, net force is towards equilibrium position with a proportionality constant g A c (  w -  o ) [& with g=10 m/s 2] If F = - k  y then k = g A c (  o -  w ) =  x10 3 N/m

Physics 207: Lecture 25, Pg 25 Fluids Buoyancy & SHM l A metal cylinder, 0.5 m in radius and 4.0 m high is lowered, as shown, from a rope into a vat of oil and water. The tension, T, in the rope goes to zero when the cylinder is half in the oil and half in the water. The densities of the oil is 0.9 gm/cm 3 and the water is 1.0 gm/cm 3 l If the rope is cut and the drum undergoes SHM, what is the period of the oscillation if undamped? F = ma = - k  y and with SHM ….  = (k/m) ½ where k is a “spring” constant and m is the inertial mass (resistance to motion), the cylinder So  = (1000  /4 m cyl ) ½ = (1000  / 4  cyl V cyl ) ½ = ( 0.25/0.95 ) ½ = 0.51 rad/sec T = 3.2 sec

Physics 207: Lecture 25, Pg 26 Underdamped SHM If the period is 2.0 sec and, after four cycles, the amplitude drops by 75%, what is the time constant  if  ≡ 2m / b ? if Four cycles implies 8 sec So 0.25 A 0 = A 0 exp(-8b / 2m) ln(1/4)= -8   = -8/ ln(1/4) = 5.8 sec

Physics 207: Lecture 25, Pg 27 Anti-global warming or the nuclear winter scenario Assume P/A = P = 1340 W/m 2 from the sun is incident on a thick dust cloud above the Earth and this energy is absorbed, equilibrated and then reradiated towards space where the Earth’s surface is in thermal equilibrium with cloud. Let e (the emissivity) be unity for all wavelengths of light. l What is the Earth’s temperature?   P =  A T 4 =  (4  r 2 ) T 4 = P  r 2  T = [ P / (4 x  )] ¼    = 5.7  W/m 2 K 4  T = 277 K (A little on the chilly side.)

Physics 207: Lecture 25, Pg 28 Time for a swim l At solar equinox, how much could you reasonably expect for the top 1 meter of water in a lake at 45 deg. latitude to warm on a calm but bright sunny day assuming that the sun’s flux (P/A) is 1340 W/m 2 if all the sun’s energy were absorbed in that layer ? l At 45 deg., sin 45 deg or x 1340 or 950 W/m 2 l But only 6 hours of full sun l So Q available is 6 x 60 x 60 seconds with 950 J/s/m 2 l Q = 2 x 10 7 J in one square meter Q = m c  T with m  10 3 kg/m 3 and c = 4.2 x 10 3 J/kg-K l about 5 K per day.

Physics 207: Lecture 25, Pg 29 Q, W and  E TH l 0.20 moles of an ideal monoatomic gas are cycled clockwise through the pV sequence shown at right (A  D  C  B  A). (R = 8.3 J/K mol) l What are the temperatures of the two isotherms? l Fill out the table below with the appropriate values. (a) pV = nRT T = pV/nR T A = x 0.10 / (0.20 x 8.3) T A = 1000 K T B = 500 K (b)  E TH =W + Q Ideal Gas at constant V Q = n C V  T

Physics 207: Lecture 25, Pg 30 Ch. 12 |L|=mvr perpendicular

Physics 207: Lecture 25, Pg 31 Ch. 12

Physics 207: Lecture 25, Pg 32 Hooke’s Law Springs and a Restoring Force Key fact:  = (k / m) ½ is general result where k reflects a constant of the linear restoring force and m is the inertial response (e.g., the “physical pendulum” where  = (  / I) ½

Physics 207: Lecture 25, Pg 33 Simple Harmonic Motion Maximum kinetic energy Maximum potential energy

Physics 207: Lecture 25, Pg 34 Resonance and damping l Energy transfer is optimal when the driving force varies at the resonant frequency. l Types of motion  Undamped  Underdamped  Critically damped  Overdamped

Physics 207: Lecture 25, Pg 35 Fluid Flow

Physics 207: Lecture 25, Pg 36 Density and pressure

Physics 207: Lecture 25, Pg 37 Response to forces

Physics 207: Lecture 25, Pg 38 States of Matter and Phase Diagrams

Physics 207: Lecture 25, Pg 39 Ideal gas equation of state

Physics 207: Lecture 25, Pg 40 pV diagrams

Physics 207: Lecture 25, Pg 41 Thermodynamics

Physics 207: Lecture 25, Pg 42 Work, Pressure, Volume, Heat T can change! In steady-state T=constant and so heat in equals heat out

Physics 207: Lecture 25, Pg 43 Gas Processes

Physics 207: Lecture 25, Pg 44 Lecture 25 Exam covers Chapters plus angular momentum, statics Exam covers Chapters plus angular momentum, statics Assignment Assignment  For Thursday, read through all of Chapter 18