Notes on Archimedes' Lab and Principle

Derivations of Archimedes Principle

Archimedes Principle comes about due to the net force the pressure of the surrounding fluid exerts on an object in the fluid. Since the pressure is greater below the object is greater than the pressure above the object the upward force from pressure below the object is greater than the downward force form the pressure above the object. This creates a net upward force called the buoyant force, F_B. The magnitude of F_B can be derived as follows.

Pressure below, P_below is greater than Pressure above, P_above by:

DP =r_fluid x g x h of object submerged in fluid. (1)

where r_fluid is the mass per m³ of fluid

Since force = pressure times area, F_B ,the net upward force from the upward force of the pressure below - the downward force from the pressure above is given by the difference in pressure multiplied by the area of the horizontal face of the object:

F_B = DP x A (2) and we can use equation (1) to replace DP and get

F_B =r_fluid x g x h of object submerged in fluid x A (3)

The volume submerged is the:

length x width x height submerged

= h of object submerged in fluid x A

= Vol submerged, which we will call V_sub,

we can rewrite equation (3) as

F_B =r_fluid x g x V_sub (4)

Since r_fluid x g is the weight of a cubic meter of fluid

F_B is equal to the weight of V_sub m³ of fluid and can be thought of as the weight of the fluid ÒreplacedÓ (or as most books would have it, displaced) by the object being submerged.

Another way to look at this [courtesy of Doc Jr.] is that before the object is submerged, the fluid surrounding the space provides enough net force to hold up the fluid in this space. This net force is equal to the Mg on the fluid in this space or r_fluid x g x Volume of the space. When the fluid in that space is replaced (displaced) by the object, the surrounding fluid still supplies the same net force. This net force is F_B so

F_B = weight of fluid replaced ( displaced) by the object.

Floating objects

If an object is floating, then since acceleration in the vertical direction = 0,

F_net vertical = 0 and so F_B = - F_g = --Mg

F_B = Mg

Since F_B = r_fluid x g x V_sub and Mg = r_obj x g x V_obj

and V_sub can never be more than V_obj

then floating objects must have r_obj no greater than r_fluid

Example

If an object floats on a lake the pressure on the top is 1 atmosphere. The pressure on the bottom is:

1 atmosphere + r_air x g x h in air + r_water x g x h in water

so net force due to fluid pressure, F_B, is DP x A

or = (r_air x h_air + r_water x h_water) x g x A

Which is the same as the weight of air of the volume of the

object in air, plus the weight of water of the volume of the

object in water

Since r_air is so much less than r_water , we

usually ignore the r_air x h_air g x A

( i.e. weight of air of volume submerged in

air) term for things like boats and logs.

This would not be a good approximation for a helium filled balloon resting on a lake surface.

Apparent Weight

If you weigh an object in a vacuum the buoyant force is zero and your scale reads Mg. If you weigh it in air at the surface of the earth your scale will read Mg - F_Bair. The F_Bair term is usually negligible for objects other than balloons, etc. If you were to submerge an object in water the scale would read Mg - F_Bwater.

Archimedes' Principle Lab

In this lab you were to take several measurements:

The mass of an object, M_obj

The mass of the beaker, M_B

The mass of water and beaker, M_wB

The mass reading of the same water and beaker with a solid object suspended in it so it was fully submerged but did not hit the beaker bottom, M_wBobj

Height of water, h₀

Height of water with object h_wobj

Internal area of beaker bottom, A

Since we are studying the buoyant force and pressure we need to convert the balance readings to force (N). This requires converting grams to kg (i.e. dividing by 1000) and multiplying by g (= 9.8 N/kg), force = M in kg x 9.8

Note that M_wBobj is not the mass of something, it is the force supplied by the balance divided by g.

We can use these reading to try to confirm several things we learned about fluids. Your labs should address all 3 items below

1) Does P = r_fluid g h_fa ? where h_fa = height of fluid above.

if this is true and since F= PA then M_wB x g - M_B x g should = r_fluid g h₀ x A

2) Since the buoyant force is an upward force on the object, the object must exert a downward force on the fluid. This is in accordance with NewtonÕs third law:

F _{of 1 on 2} = -F _{of 2 on 1}. Since placing the object in the fluid places a downward force on the fluid in addition to the weight of the fluid and since there is still no vertical acceleration the balance must exert an additional force = F_B.

This F_B must be equal to (M_wBobj x g- M_wB x g).

Again, since F = PA we should also find that:

(M_wBobj x g- M_wB x g) = r_fluid g (h_wobj - h₀) x A.

3) F_B = volume of object (its all submerged) x r_water x g and

r_water x g = 9800 N/m³. Thus we can find volume of object by:

V_obj = F_B/ 9800 = (M_wBobj x g- M_wB x g)/9800.

We can use this to find the density, r_obj by

r_obj = M_obj/ V_obj

You can compare this to density tables to see if your density makes sense. Your objects were iron or brass or perhaps lead or zinc.

Projects

Honors students (solo efforts or groups of no more than 3 are allowed)have the opportunity to improve their grades by submitting a project this quarter. The project can be either something you construct related to physics or a lab an honors class could perform. The criteria are the same as last quarter.
How much did you learn?
How much will some one else learn from performing your lab or seeing your project and hearing you explain it?
The quality and polish of your project.

The work must be your own [no cutting and pasting], and the principles of physics for your project or lab must be clearly explained. I will try to estimate the effort put into it in terms of hours. It can replace up to 12 points of your weakest category. If all you categories are strong it can amount to 3 extra credits. Note that this is subject to change depending on how the grades work out. A+ is not a reasonable average grade for a class. Again, I will be the sole judge; whining is forbidden.

I'll try to get this and last quarter's project folks to present right after our midyear exams.

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Rotational Motion

Honors students must read this. It might help AP students as well.

When things travel in a circle using our normal linear description of motion is very difficult. The direction of travel changes continually. For example, when you travel on a merry ground you may start out going west but then your velocity changes from west bound to south bound, then to east bound and then north bound, etc. So even with this simple motion your velocity is changing in a complex way, and so is your position.

However, it would be very simple to describe your motion using two numbers. One is your distance from the center of the merry ground, i.e. the radius of the circle your motion describes, and the second could be the angle theta, that your radius makes with the east . Then we could describe your position easily, saying for example, you are 7 meters from the center at an angle of 1.5 radians from east. We could describe your rate of motion as the rate of change of this angle. This would be called omega, your angular speed (radians/s) . If you measure the angle in radians, the distance you travel as you go around the circle is just radius x change in angle, i.e. distance = r delta Theta. Then your (tangential) speed (the old fashioned m/s) is just r x omega.

Try this for Monday:

Your are 3 m from the center of a carousel. The carousel turns half a turn, i.e. pi radians. How far did you move?

The carousel makes one complete rotation in 10 seconds, how far did you move in those ten seconds?

How fast were you going in m/s? This is your tangential velocity.

How many radians did it turn in those ten seconds? [ answer is 2 pi] How many radians did it turn in one second?

What was your angular speed (its called omega) in radians / second?

Was your tangential speed (m/s) equal to omega times 3 m?

AP Labs

Cleanup for AP labs has been somewhere between poor and nonexistent. I just removed carbon paper taped to the floor for four lab groups. These groups are now starting out with a grade of -5 for the Ballistic pendulum.

Failure to return materials and equipment to proper locations will be treated similarly in the future. Please ask before returning equipment in case the next class needs the setup, BUT DO NOT ASSUME YOU DO NOT HAVE TO CLEAN UP AFTER YOURSELF. YOUR MOTHER DOES NOT WORK HERE!

Centripetal Force Lab

For this lab, you were supposed to compare Fc i.e. mstopper x omega squared x r with mwasher x g since Fc is supposed to be provided by Ft, which should equal Mwasher g.

If you had large discrepancies tape binding on the plastic tube, touching the string, and letting the washers swing so that they also had an ac, are all likely culprits. You might try redoing it for an improved grade.

Greek letters in reports

You can use the symbol font to get the correct greek letters in your report. For example D in the symbol font is Delta; w in symbol font will give you omega; Q will give you Theta.

Help after school

Please be advised that I have faculty meetings on the first and second Wednesdays of the month. I will not be available for help until 3:30 on these days.

Notes and Corrections for Fluids

Correction to Fluid Solution for problem 62 b. The velocity squared term for the smaller diameter is 16 squared or 256 not just 16. I believe this gives a pressure in the narrow line of 173,000 Pa

Specific gravity is simply the ratio of a material's ( solid or fluid) density to that of water. It may also be taken to mean the density in units of grams per cubic cm. Note that water has a specific gravity of 1 and a density of 1 g/cubic cm

Problem 3 on the problem sheet is worded poorly. It should ask: "What is the power output required from a motor to drive this pump?". The work done by the pump per second (i.e. the pump output power) , divided by the efficiency, gives the power that must be supplied by the motor.

Problem 7 on Unit 7 Problem sheet: The top of the ice would be level with the surface of the water. In other words it is just barely submerged. The guy doesn't get his feet wet, but he comes awfully close.

Solution Set for Fluid Text problems

Here are the solutions for the problems from the text. There are ten GIF images in all. Let me know if this worked.

Torque Lab

For the Torque and CoM Lab: just fill out the package I gave you including answering all questions. If you did not complete Part 6, that will have to be okay. If you did part 5a you have all the data you need to do parts 5b and 5 c. Note that 5 b will give a large difference between cw and ccw torques because you are leaving out part of the torque the meter stick generates. That is the point of 5b, to show what you get if you do not include everything you should.

Food Drive

All students should bring canned goods as often as possible, especially my first period class, since we want the bagels!

Rotational and translational Equilibrium

Some of you are making the DREADED MISTAKE. Do not think that no net torque means no rotation can occur. If the net torque is zero the angular speed, omega, is constant; it can be zero, but it does not necessarily have to be zero. It's just like linear (translational) motion: if Fnet is zero, v is constant, not necessarily zero. Equilibrium means a and alpha are zero and v and omega are constant. Static equilibrium does mean alpha, a, omega and v are all zero.

Honors Lab Reports

Lab reports continue to lag pretty badly. Many are now 3 weeks late. You will lose 3 OR MORE points for each week a report is late however 10 points off is still a lot better than a zero!

Labs on parallel and perpendicular components should show predicted and measured values for Fgperp and Fg parallel and a for each of the three runs, not just the first run. The predicted and measured numbers should be compared and good agreement cited or poor agreement explained. This is true for all labs reports.

Fluid Machines see prob 5 on prob sheet

Fluid Machines see prob 5 on prob sheet:

Fluid machines can be analyzed using conservation of energy which in this case can be taken as work on the machine equals work done by the machine since the energy of the machine itself doesn't change. Fluids do work by exerting a pressure times an area over a distance i.e.

W= P x A xd

so if work is done on (energy put into) the machine at point 1 and the machine does work at point 2, we have:

P1 x A1 x d1 = P2 x A2 x d2

Honors Homework

For Thursday night: do the power and efficiency worksheet.

For the weekend : do the Unit 6 problem sheet 1 through 9

Nuclear reaction problems (for honors and other interested parties)

Honors Section 1 try these; Honors Section 2 got these in class

Nuclear Collisions Problems

Data
1 eV = 1.6x10-19 J
1 Mev = 1.6 x10-13 J
q for a proton = 1.6x10-19 Coulombs
the mass of a proton is 1.67x10-27 kg
the mass of a neutron is 1.67x10-27 kg
D has one proton and one neutron
T has one proton and two neutrons,
He 3 has one neutron and two protons
He 4 has two neutrons and two protons,
Li has 3 protons and 3 neutrons
c = 3x108 m/s
see Appendix for energies produced by reactions

In collision between charged nuclei, the electrostatic potential energy due to the charges of the protons is given by

PEES = 9x109 q1q2/r where r is the distance between the two nuclei and q are the charges for each nucleus.

In order for the two charges to approach to where the strong forces take over they must get within 10-14 meters.
1) Calculate the kinetic energy the following reactions require.
2) Assuming the kinetic energy is shared equally by the two fuel nuclei, what are their velocities?
3) Using the energies for reactions from the Appendix find the decrease in mass between the products and the fuel
4) Find the velocities of the products (see Appendix for energies produced by reactions)


D +D -> T (1.01 MeV) + p ( 3.02 MeV) (50%)
D +D -> He3 (0.82 MeV) + n ( 2.45 MeV) (50%)
D +He3 -> He4 (3.6 MeV) + p (14.7 MeV)
He3+T -> He4 (4.8 MeV) + D ( 9.5 MeV) (43%)
He3+T -> He4 (0.5 MeV) + n ( 1.9 MeV) + p (11.9 MeV) (6%)

Appendix:
If you want to use fusion as a source of energy, the reaction chosen must satisfy several criteria. It must:
be exothermic. This one is obvious, but it limits the reactants to the low Z side of the curve of binding energy. It also makes helium He-4 the most common product because of its extraordinarily tight binding, although He3 and T also show up.
involve low Z nuclei. This is because the electrostatic repulsion must be overcome before the nuclei are close enough to fuse.
have two reactants. At anything less than stellar densities, three body collisions are too improbable.
have two or more products. This allows simultaneous conservation of energy and momentum without relying on the (weak!) electromagnetic force.
and conserve both protons and neutrons. The cross sections for the weak interaction are too small.
This makes the list of candidates pretty short. The most interesting reactions are the following.
1) D +T -> He4 (3.5 MeV) + n (14.1 MeV)
2) D +D -> T (1.01 MeV) + p ( 3.02 MeV) (50%)
3) D +D -> He3 (0.82 MeV) + n ( 2.45 MeV) (50%)
4) D +He3 -> He4 (3.6 MeV) + p (14.7 MeV)
5) T +T -> He4 + 2 n + 11.3 MeV
6) He3+He3 -> He4 + 2 p
7) He3+T -> He4 + p + n + 12.1 MeV (51%)
8) He3+T -> He4 (4.8 MeV) + D ( 9.5 MeV) (43%)
9) He3+T -> He4 (0.5 MeV) + n ( 1.9 MeV) + p (11.9 MeV) (6%)
10)D +Li6 -> 2 He4 + 22.4 MeV (11) p +Li6 -> He4 (1.7 MeV) + He3 ( 2.3 MeV)
11)He3+Li6 -> 2 He4 + p + 16.9 MeV
12)p +B11 -> 3 He4 + 8.7 MeV

Work = area under Force vs distance curve

Just as we found the displacement was the area under the velocity vs time curve, we can use the same reasoning to show work is the area under the force vs distance curve.

In other words:
Since W = F x d (provided d and F are in the same direction) if we look at a strip of area under the force curve. That strip of area is F high and delta distance wide; so it has an area of F x delta distance. This is the work done over the delta distance. If we add up all these strips we get the work done for all the delta distances which is the work done for the total distance.

New Quiz Dates

Because of illnesses and absences for the Spanish field trip, the Honors and AP quizzes will be given on Thursday. The honors quiz will include some things about e = mc squared and collisions of particles as well as simple energy and work problems. The AP classes will begin fluids on Wednesday after a little rotational motion review.

The AP quiz for Section 4 (6 th period) will be a partner quiz. There will be 3 or 4 short questions and three or four problems to be answered by the group. The there will be a fourth or fifth short question and a fourth or fifth problem to be answered by individuals. The group answers and the lowest score for the individual questions will be used to calculate the group score.

If you would like a similar partner test for your class, let me know.

Error in AP Solution Sheet

Kevin Lee pointed out an error in the solution I provided to Problem 3 on R-2. I took the torque due to tne mass of the beam as counterclockwise, but it is a clockwise torque and it should appear on the same side of the torque balance equation as M2. Thus
M2g x 1.75 + 200 x .5 = 250 x 1 or M2= (250-100)/1.75

Solution to AP Chapt 7 prob 55

Solution to Chapt 7 Problem 55 b

Remember that sin theta = r/L and cos theta is h/L [r is radius or Dx and h is height or Dy) and that theta is angle with y axis here.

Then we have for y direction: Ft cos theta =Ft h/L =Mg and for x direction: Ft sin theta = Ft r/L=Fc = M v^2/r We are given v and L ( and M but it cancels out here) so h r and theta are unknowns. Given h or r we can find theta from inverse cos or sin.

We have from y direction Ft = Mg/(h/L) and from x direction Ft = v^2/(r^2/L) setting the two equations for Ft equal to each other we have: Mg/(h/L) =M v^2/(r^2/L) and the Ls and Ms cancel so g/h = v^2/r^2

By Pythagorean Theorem r^2=L^2-h^2 so now we have g/h = v^2/(L^2-h^2).
Since we are after h, lets invert the equation and get h/g= (L^2-h^2) / v^2
multiply both sides by v^2 and collecting terms on one side as for a quadratic equation: h^2+(v^2/g)h –L^2= 0
the solution to this quadratic is [- v^2/g+/- squareroot(v^4/g^2 + 4 L^2)}/2

Putting numbers in this awful mess ( and I’ll use 9.8 here even though it hurts me to do so) we get h [-16 /9.8 +/- sqrt(256/96+9)]/2= [-1.63 + sqrt (11.67)]/2 = 1.79/2 = 0.893; h/L =cos theta = .595 and theta = 53.5 degrees.
and r = 1.5 sin 53.5 = 1.21m

Lets check

does Fc /(sin 53.5)= Ft = Mg/cos 53.5 ?

v^2 is 16 (M v^2/r )/sin 53.5= (16/1.21)/.804= 16.5 M
Mg/cos 53.5 = 1.68 Mg = 16.5 M so we are right but there must be a better way to do this problem

Comments on AP lab reportsfor last 2 weeks

Momentum /collision lab. Since there was no lab sheet provided, you must write a complete report with all sections as described in the course guide.

Stair climb lab. Most of the work was due to changing PE, which depended on the change in height x m x g. The change in height was 4.27 m. The work associated with changing PE did not depend on the time but only the mass x g. The power then depended on work divided by time and thus increased with decreasing time to do the climb. However, you were asked to find the additional work and power associated with the kinetic energy needed. The KE did depend on the time. Also it depended on the total, not just vertical, distance. We approximated this as 10.04 m. While the KE was negligible for the slow cases, it was marginally significant for the fast cases and this should have been calculated and discussed.

Ballistic pendulum. ( Sect 4 hasn't done it yet) Lab reports should discuss why the equations used are valid and when and why momentum was conserved and energy wasn't ( inelastic collision between ball and catcher) and when energy was conserved and momentum wasn't (pendulum travel after collision).

New requirements for minilab reports

As of this week, I am revising the requirements for minilab reports. From now on they must have a purpose statement and some error analysis, just as in full reports. Please remind me to state this in your class and write it on the board. However, consider this an official notice.

Matter as a form of energy

Just over 100 years ago, Albert Einstein developed his theory of special relativity. The theory was based on the idea that the laws of physics applied in the same way to all reference frames moving at constant velocities RELATIVE to one another. Since the speed of light (and all other electromagnetic radiation) in a vacuum is determined by the the physical laws of electromagnetics, then it is constant in every frame of reference, no matter how fast that frame is traveling with respect to the source of the light. In other words, the speed with which a light ray from a light bulb passes you is 300,000,000 meters per second regardless of whether you are running towards the light bulb or away from it, or standing still. This somewhat bizarre fact is just the tip of the iceberg.

Einstein discovered another result of the equivalence of reference frames traveling at constant relative velocities. The mass of an object depends on the frame from which it is viewed. Thus, its mass depends on its velocity or, more to the point, its kinetic energy effects it mass. With some brilliant insight he developed the fact that mass is transformable to energy at the rate that change in energy ( potential or kinetic or both ) is equal to change in mass times the square of the speed of light. Since the symbol "c" is used to mean the speed of light, this fact can be written as the energy content, E , of matter with mass, M, is given by :

E=Mc^2

We will discuss this a bit more as we explore energy in honors. AP classes will have to wait until April or May to get more on this subject in class.