# Meenakshi Narain - Physics 5 - Lecture 22

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Meenakshi Narain - Physics 5 - Lecture 22
11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Announcements Homework due TODAY at the end of lecture Homework 12 based on ch 12 & 13, due on Dec 5th. I may post some practice problem suggestions for ch 14 and 15. Final Exam on Thursday, Dec 13, 9am-12pm, in BARHOL 168 Close book, close lecture notes Chapters 1-15, with emphasis on post midterm 2 material Emphasis of the exam: problem solving (6-8 problems) Simple calculator and one-page formula sheet allowed Questions? Suggestions? 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Simple Harmonic Motion Chapter 15
11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A block of mass 1 kg is attached to a spring with k=100 N/m and free to move on a frictionless horizontal surface. At t=0, the spring is extended 5 cm beyond its equilibrium position and the block is moving to the left with a speed of 1 m/s. What is the displacement of the block as a function of time? (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
What are the values of A, , and  ? A,  are determined by initial conditions (x,v at t=0) – they are not properties of the oscillating system. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three 10,000 kg ore cars are held at rest on a 30o incline using a cable. The cable stretches 15 cm just before a coupling breaks detaching one of the cars. Find (a) the frequency of the resulting oscillation of the remaining two cars and (b) the amplitude of the oscillation. (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
m = kg d = 0.15 m first compute spring constant of rope: then compute frequency of oscillation: the amplitude equals the original displacement from the equilibrium position of two cars on the rope: A = d  d0 = 0.05 m 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A 10 g bullet strikes a 1 kg pendulum bob that is suspended from a 10 m long string. After the collision the two objects stick together and the pendulum swings with an amplitude of 10o. What was the speed of the bullet when it hit the bob? 1.0 kg 1.0 kg 10o (3) 10 g v=? 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Energy of simple pendulum
using with Get velocity of bob at equilibrium point from total energy get velocity of bullet from momentum conservation 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
CO2 molecule Carbon dioxide is a linear molecule. The C-O bonds act like springs. This molecule can vibrate such that the oxygen atoms move symmetrically in and out, while the carbon atom is at rest. The frequency of this vibration is observed to be 2.83x1013 Hz. What is the spring constant of the C-O bond? In which other way can the molecule vibrate and at what frequency? If the amplitude is the same, in which mode is the energy of the molecule higher? m=12u m=16u (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
(3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
CO2 molecule a) b) c)  the energy is the same for both modes. Parallel springs See Lecture 10 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Which of the following spring arrangements will oscillate with the smallest angular frequency? Assume that all springs are identical. (1) (2) (3) (4) (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Driven Harmonic Oscillator
What happens if you have an oscillator, such as a mass on a spring, where an external force is acting on the system? Example: Motion of a building or bridge during an earthquake Essentially all objects have one or more “natural frequencies” that they will oscillate at if they are initially displaced from equilibrium Example: mass on a spring has a natural frequency given by If an oscillating external force is applied with angular frequency w close to the natural frequency w0, the results can be dramatic 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Driven Harmonic Oscillator
To get motion, use Newton’s 2nd law: Suppose the external force is sinusoidal: Eventually, the object’s motion will oscillate with frequency w since that’s the frequency of the applied force 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Driven Harmonic Oscillator
Solution for driven harmonic oscillator is somewhat more complicated than what we have done so far in Physics 5 With some effort, one can solve the equation of motion where the frequency and amplitude are given by: 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Driven Harmonic Oscillator
Example: Mass on spring Notice how amplitude and phase change as the frequency of the external force crosses the natural frequency Example: Vibrations of a solid object Solid objects typically have one or more natural frequencies that they oscillate at If the damping is small, large oscillations occur when driven at these natural frequencies These vibrations can be a considerable source of stress on the object! 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Oscillations Summary We get periodic motion when force acts to push object back towards equilibrium position Many problems exhibit simple harmonic motion Energy exchanged between kinetic and potential energy, total mechanical energy unchanged for undamped oscillations Correspondence between simple harmonic motion and uniform circular motion Amplitude of oscillations decays with a damping force Driven oscillations exhibit a “resonance” at the natural frequency 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Fluids Chapter 14 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Fluids What is a fluid? A substance that flows Examples include liquid, gas, plasma, etc A simple fluid can withstand pressure but not shear Density Unit: kg/m3 Examples: density of water 1000 kg/m3 ; air 1.21 kg/m3 A material’s specific gravity is the ratio of the density of the material to the density of water at 4°C. What is special about water at 4°C? Water is most dense at that temperature. Aluminum has a specific gravity of 2.7 – it is 2.7 times more denser than water at 4°C. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A table of densities Material Density (kg/m3) Interstellar space 10-20 Air (20°C, 1 atm.) 1.21 Water (4°C, 1 atm.) 1000 Sun (average) 1400 Earth (the planet) 5500 Iron 8700 Mercury (the metal) 13600 Black hole 1019 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Pressure Pressure Unit: pascal (Pa)=N/m2 Examples: 1 atm=1.01x105 Pa=760 torr=14.7 lb/in2 Torricelli (torr) is defined as the pressure of mm Hg. Blood pressure: 70/120 torr At any point in a fluid at rest, the pressure is the same in all directions. If this were not true there would be a net force on the fluid and it could not be at rest. The force due to fluid pressure acts perpendicular to any surface. Else there would be a force component along the surface which would accelerate the fluid. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Atmospheric pressure At atmospheric pressure, every square meter has a force of 100,000 N exerted on it, coming from air molecules bouncing off it! Why don’t we, and other things, collapse because of this pressure? We have an internal pressure of 1 atmosphere. Objects like tables do not collapse because forces on top surfaces are balanced by forces on bottom surfaces, etc. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Fluid-Statics Static equilibrium A simpler expression Where p0 is the pressure at the surface, and h is depth of the liquid The pressure at any point in a fluid is determined by the density of the fluid and the depth. It does not depend on any horizontal dimension of the fluid or its container. It also does not depend on the shape of the container. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Measuring pressure The relationship between pressure and depth is exploited in manometers (or barometers) that measure pressure. A standard barometer is a tube with one end sealed. The sealed end is close to zero pressure, while the other end is open to the atmosphere. The pressure difference between the two ends of the tube can maintain a column of fluid in the tube, with the height of the column being proportional to the pressure difference. pressure at bottom of column = atmospheric pressure 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Mercury/Water barometer
atmospheric pressure pushes Hg column up  unit mm-Hg (=torr) Thus Atmospheric pressure pushes the Hg column up by 101.3 kPa/133 Pa/mm = 760 mm Water: Thus atmospheric pressure pushes the water column up by 101.3 kPa/9.8 Pa/mm = 10.3 m another unit: 1 bar = 105 N/m2 in calculations only use N/m2 = Pa (SI unit) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Gauge Pressure Gauges measure pressure relative to atmospheric pressure absolute pressure = gauge pressure + atmospheric pressure manometer (height of column of liquid measures gauge pressure) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Blood pressure A typical reading for blood pressure is 120 over 80. What do the two numbers represent? What units are they in? 120 mm Hg (millimeters of mercury) is a typical systolic pressure, the pressure when the heart contracts. 80 mm Hg is a typical diastolic pressure, the blood pressure when the heart relaxes after a contraction. 760 mm Hg is typical atmospheric pressure. The blood pressure readings represent gauge pressure, not absolute pressure – they tell us how much above atmospheric pressure the blood pressure is. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A delicious drink sits on the patio. From your balcony several stories up you manage to lower a straw into the glass, which is m below you. Can you syphon up the drink? 15 m yes, but I will have to suck really hard probably not, but a vacuum pump could no, this is not possible I don’t know (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Water pressure At the surface of a body of water, the pressure you experience is atmospheric pressure. Estimate how deep you have to dive to experience a pressure of 2 atmospheres. h works out to 10 m. Every 10 m down in water increases the pressure by 1 atmosphere. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Rank by pressure A container, closed on the right side but open to the atmosphere on the left, is almost completely filled with water, as shown. Three points are marked in the container. Rank these according to the pressure at the points, from highest pressure to lowest. A = B > C B > A > C B > A = C C > B > A C > A = B some other order 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Blaise Pascal ( ) A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished everywhere in the fluid and to the walls of the container 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A container is filled with oil and fitted on both ends with pistons. The area of the left piston is 10 mm2; that of the right piston is 10,000 mm2. What force must be exerted on the left piston to keep the 10,000 N car on the right at the same height? =10000 N 10 N 100 N 10,000 N 106 N 108 N =? (3) =10 mm2 =10000 mm2 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Pascal’s Principle Hydraulic lever (see diagram on right) Something has to give… Since the liquid is incompressible, the volume drop on the left is equal to that rises on the right, ie. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Pascal placed a long thin tube vertically into a wine barrel. When the barrel and tube were filled with water to a height of 12 m, the barrel burst. (a) what is the mass of the water in the tube? (b) what is the net force exerted onto the lid of the barrel? (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Imagine holding two identical bricks under water. Brick A is just beneath the surface. Brick B is at a greater depth. The force needed to hold brick B in place is larger the same as smaller than the force required to hold brick A in place (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
The Buoyant Force With fluids, we bring in a new force. The buoyant force is generally an upward force exerted by a fluid on an object that is either fully or partly immersed in that fluid. Let’s survey your initial ideas about the buoyant force. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
The Buoyant Force A wooden block with a weight of 100 N floats exactly 50% submerged in a particular fluid. The upward buoyant force exerted on the block by the fluid … has a magnitude of 100 N has a magnitude of 50 N depends on the density of the fluid depends on the density of the block depends on both the density of the fluid and the density of the block 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Learning by Analogy Our 100 N block is at rest on a flat table. What is the normal force exerted on the block by the table? To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Apply this to Buoyant force
Apply the same method when the block floats in the fluid. What is the magnitude of the buoyant force acting on the block? To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Reviewing the normal force
We stack a 50-newton weight on top of the 100 N block. What is the normal force exerted on the block by the table? To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. The block presses down farther into the table (this is hard to see). 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Buoyant force We stack a 50-newton weight on top of the 100 N block. What is the buoyant force exerted on the block by the fluid? To answer this, we apply Newton’s Second Law. There is no acceleration, so the forces balance. The block presses down farther into the fluid (this is easy to see). 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Apply Newton’s Second Law
Even though we are dealing with a new topic, fluids, we can still apply Newton’s second law to find the buoyant force. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Beakers The wooden block, with a weight of 100 N, floats in all three of the following cases, but a different percentage of the block is submerged in each case. In which case does the block experience the largest buoyant force? 4. The buoyant force is equal in all three cases. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Beakers What does the free-body diagram of the block look like? What is the difference between these fluids? The density 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A block of weight mg = 45.0 N has part of its volume submerged in a beaker of water. The block is partially supported by a string of fixed length. When 80.0% of the block’s volume is submerged, the tension in the string is 5.00 N. What is the magnitude of the buoyant force acting on the block? 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Apply Newton’s Second Law
The block is in equilibrium – all the forces balance. Taking up to be positive: 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Water is steadily removed from the beaker, causing the block to become less submerged. The string breaks when its tension exceeds 35.0 N. What percent of the block’s volume is submerged at the moment the string breaks? 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Apply Newton’s Second Law
The block is in equilibrium – all the forces balance. Taking up to be positive: The buoyant force is proportional to the volume of fluid displaced by the block. If the buoyant force is 40 N when 80% of the block is submerged, when the buoyant force is 10 N we must have 20% of the block submerged. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
After the string breaks and the block comes to a new equilibrium position in the beaker, what percent of the block’s volume is submerged? what does the free-body diagram look like now? 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Apply Newton’s Second Law
The block is in equilibrium – all the forces balance. Taking up to be positive: The buoyant force is proportional to the volume of fluid displaced by the block. If the buoyant force is 40 N when 80% of the block is submerged, when the buoyant force is 45 N we must have 90% of the block submerged. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Archimedes’ Principle
While it is true that the buoyant force acting on an object is proportional to the volume of fluid displaced by that object. But, we can say more than that. The buoyant force acting on an object is equal to the weight of fluid displaced by that object. This is Archimedes’ Principle. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A Floating Object When an object floats in a fluid, the downward force of gravity acting on the object is balanced by the upward buoyant force. Looking at the fraction of the object submerged in the fluid tells us how the density of the object compares to that of the fluid. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Beaker on a Balance A beaker of water sits on a scale. If you dip your little finger into the water, what happens to the scale reading? Assume that no water spills from the beaker in this process. 1. The scale reading goes up 2. The scale reading goes down 3. The scale reading stays the same 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Blocks We have three cubes of identical volume but different density. We also have a container of fluid. The density of Cube A is less than the density of the fluid; the density of Cube B is exactly equal to the density of the fluid; and the density of Cube C is greater than the density of the fluid. When these objects are all completely submerged in the fluid, as shown, which cube displaces the largest volume of fluid? 1. Cube A 2. Cube B 3. Cube C 4. The cubes all displace equal volumes of fluid 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Blocks Each cube displaces a volume of fluid equal to its own volume, and the cube volumes are equal so the volumes of fluid displaced are all equal. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Blocks Which object has the largest buoyant force acting on it? 1. Cube A 2. Cube B 3. Cube C 4. The cubes have equal buoyant forces 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Three Blocks Each cube displaces an equal volume of the same fluid, so the buoyant force is the same on each. 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Two identical glasses are filled to the brim with water. One of the two glasses has a ball floating in it. Which glass weighs more? The glass without the ball The glass with the ball The two weigh the same (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
A boat carrying a large boulder is floating in a lake. The boulder is thrown overboard and sinks. What happens to the water level in the lake (relative to the shore)? it sinks it rises it remains the same (3) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Cartesian diver The diver is an object in a sealed container of water. Air in the diver makes it buoyant enough to barely float at the water's surface. When the container is squeezed, the pressure compresses the air and reduces its volume. This permits more water to enter the diver, resulting in it being less buoyant and sinking. regular coke and diet coke 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

The origin of the buoyant force
The net upward buoyant force is the vector sum of the various forces from the fluid pressure. Because the fluid pressure increases with depth, the upward force on the bottom surface is larger than the downward force on the upper surface of the immersed object. This is for a fully immersed object. For a floating object, h is the height below the water level, so we get: 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

When the object goes deeper
If we displace the object deeper into the fluid, what happens to the buoyant force acting on it? Assume the fluid density is the same at all depths. The buoyant force: increases decreases stays the same 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

When the object goes deeper
If the fluid density does not change with depth, all the forces increase by the same amount, leaving the buoyant force unchanged! 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Archimedes’ Principle
Buoyant force Objects that float Dry wood, ice, some plastics, oil, wax (candles)… Boats made of woods, ceramic, steel, or any other materials, as long as they are hollow enough Objects that sink Rocks, sands, clay, metal, etc. Any material with density larger than water Apparent weight (example: submerged object weighs less Apparent weight=actual weight - buoyant force What is your apparent weight in water? (no more than a few pounds!) 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Unbalancing the forces
If we remove the balance between forces, we can produce some interesting effects. Demonstrations of this include: The Magdeburg hemispheres (see below) Crushing a can 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Crush a can Remember that this is just the collective effect of a bunch of air molecules! 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22

Meenakshi Narain - Physics 5 - Lecture 22
Summary Density and pressure of fluids Air pressure, blood pressure and underwater pressure Pascal’s Principle Archimedes’ Principle 11/28/2007 Meenakshi Narain - Physics 5 - Lecture 22