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Every good lab report has as its primary function the
goal of convincing the teacher that the student writing the lab report
understands what is going on. If you can focus your report to this end you will
produce a much better lab report. In any case, every lab report should include
each of the following items:
- A clear statement of the goal of the lab!
- A background theory section where the writer
explains the theory behind the lab including particularly the definitions of
any new terminology.
- A well organized depiction of the data collected in
the lab. [Including estimated error on ALL measurements!]
- A clear demonstration of any calculations done in
the lab.
- A well written conclusion that should include:
- What was the goal of this part of the lab?
- What was done in order to reach the goal?
- What were the results of this lab? [Including
specific references to supporting evidence such as bar graphs.]
- Did you reach the goal/goals of the lab as
stated above and how does your evidence lead you to this
conclusion?
- Finally, write your lab report in a narrative form
as if you are speaking to the reader!
All lab reports are due two class days after the completion
of the lab activity. All lab reports turned in early [the day before the lab is
due] will receive an extra 10% while all late labs [labs turned in after the
completion of the class period on the day that the lab is due] will be penalized 30%!
The calculation of error in a lab consists of finding out how
much of the error resulting from measurements in the laboratory shows up
in the resulting calculations of the lab. For example, suppose that you were
measuring the velocity of a cart rolling across the floor. Since the velocity of
a moving object is given by v = Dd/Dt
and since each measurement made in the lab is going to contain inevitable
measurement error, the resulting velocity will also have error associated with
it. In general we make the assumption that the errors in the lab are always
maximized [meaning that we always assume the worst]. For example, suppose the
distance along the floor that the cart moves is measured to be Dd
= 2.50 +/- .05 meters while the time interval is measured to be Dt
= 0.72 +/- 0.03 seconds. To calculate the velocity of the cart you would, of
course, divide the distance moved by the cart by the time interval as below.
v = Dd/Dt
= 2.50 m/0.72 sec = 3.47 m/sec
To determine the error on this velocity you will make the
assumption that everything that could have gone wrong, did go wrong in the worst
possible way. In the case of division that means that the numerator increases
while the denominator decreases. Both of which will make the velocity bigger
[maximized!]
vmaximized = (2.50 + 0.05) m/(0.72-0.03) sec =
2.60 m/0.69 sec = 3.77 m/sec
If you then take the difference between the maximized value
and the value that you got without accounting for error, the difference between
these two numbers will be equal to the error on the calculated velocity!
Dv = vmaximized
- v = 3.77 - 3.47 = 0.30 m/sec
And therefore the resulting velocity will
be v = 3.47 +/- 0.30 = 3.5 +/- .3 m/sec.
Note that the final calculated velocity has been rounded off
to the first digit of error. After all, who cares what the value is in the
hundredth's place when you barely know the number in the tenths place! You
should always round off your final answer to the appropriate number of
significant digits.
Finally, what do you do when you are not dividing? Below are
listed the calculations you will probably encounter in this course.
-
Adding - Just add the errors together.
(3.8 +/- 0.3) m + (5.3 +/- 0.2) m = 9.1 +/- 0.5 m
-
Subtracting - Just add the errors together.
(7.8 +/- 0.3) m - (5.3 +/- 0.2) m = 2.5 +/- 0.5 m
-
Multiplying - Calculate, calculate with error & take
the difference as in dividing.
Area = (7.8 +/- 0.3) m x (5.3 +/- 0.2) m
Areamaximized = (7.8 + 0.3) m x (5.3 + 0.2) = 44.6 m2
Areawithouterror = 7.8 m x 5.3 m = 41.3 m2
DArea =
Areamaximized - Areawithouterror = 44.6 m2
- 41.3 m2 = 3.3 m2
And so finally Area = (41.3 +/- 3.3) m2 = 41 +/- 4 m2
[Note how again the number has been rounded off so as to be consistent with
the error!]
I will normally be available until the 4:00 o'clock bus on
Tuesday's and Thursday's. I am usually available every other day for at least a
few minutes after school. In addition, I am also usually available before school
by about 7:10 AM in room C208. If cannot meet at these times I can also be
reached via email at JimTHX@home.com or by
AOL instant messenger under the name JimTHX.
Normally worked missed by absence must be made up within two
weeks of the return school. There are, however, two exceptions:
-
For long term absences work is to be made up within 4
weeks of the return to school.
-
For absences of a single day work is to be submitted
within one day of the return to school. [With the exception of absences on a
lab day where two weeks again applies.]
To blow up basketballs, of course!
Velocity is defined to be the change in the position of an
object divided by the time interval over which this change occurs. For
example, if you rode your bicycle 25 feet in 5 seconds your average
velocity over the 5 second period would be v = 25 feet/5 seconds = 5 ft/sec.
To find your instantaneous velocity [in other words your
velocity at a particular moment in time] you would need to make the time
interval very small. For example suppose you moved 0.25 feet in 0.05 seconds
your instantaneous velocity would be v = 0.25 feet/0.05 sec = 5 ft /sec. Now
looking at these two answers you may say, "but aren't they the
same?", but they are not! In the first case the speed of the bicycle
could have varied considerably over the 5.0 second time interval and so you
don't really know how fast the bicycle was going at any particular moment
within the 5 second interval. On the other hand in the second case, where the
time interval was 0.05 seconds, the speed of the bicycle could not have
realistically changed significantly during the given 0.05 second time
interval. Therefore, the calculated velocity of the bicycle during the second
time interval is the actual velocity throughout the interval! [The
"speedometer" in an automobile gives you the instantaneous value!]
Instantaneous acceleration is almost the same.
Average acceleration equals the change in the velocity divided by the time
interval over which the change occurs. If this interval is relatively large
the resulting acceleration will be average while if the interval is very small
the acceleration will be instantaneous. For example if the speed of your
bicycle goes from 5 feet/sec to 25 feet/sec over a 4.0 second time interval
the average acceleration will be a = (25 ft/s - 5 ft/s)/4.0 seconds = 5.0
ft/sec^2. To make this instantaneous you need to measure the change in
velocity over a MUCH smaller time interval. For example, if the bicycle's
velocity goes from 5 ft/sec to 6 ft/sec over a time interval of t = 0.2
seconds the acceleration would be a = (6 ft/sec - 5 ft/sec)/0.2 sec = 5.0
ft/sec^2. Again, the same mathematical answer, but a very different idea! In
the first case the acceleration was the average over a fairly large
interval of time during which the acceleration of the bicycle could have
realistically changed considerably. In the second case, however, the
acceleration during the 0.2 second interval would not realistically have
changed very much during such a small interval of time and therefore the
calculated acceleration is the "instantaneous" value.
In order to solve a problem involving you should go through
the following steps:
-
First, determine if energy conservation can be used to
solve the problem. In order for this to be so the type of energy in the
problem must be changing from one form to another and the system needs to be
closed, meaning that either there are no outside forces acting on the system
or that all outside forces have been accounted for.
-
Second, determine what types of energy are present at the
beginning of the problem and what types of energy are present at the end of
the problem.
-
If there are no unaccounted outside forces acting on the
system make the total of all the energy types at the beginning of the
problem equal to the sum of all the energy types at the end.
-
If there is an outside force, such as friction, make the
difference between the total energy at the beginning and the total energy at
the end equal to the work done by the outside force.
-
Finally, solve for whatever information is missing!
Example problem without any unaccounted for
outside forces!
Suppose that a 2.0 kg ball is thrown from the top of a
building, which is 125 meters high, with a velocity of 42.0 m/sec at an angle of
35o above the horizontal. What will be the velocity of the ball just
as it reaches the ground?
-
There are no unaccounted forces here so the total energy at
the beginning must be equal to the total energy at the end.
-
At the beginning there are two types of energy present.
Gravitational potential energy [GPE] because the ball is at the top of a
building and kinetic energy [KE] because the ball is moving since it was
thrown.
-
At the end the only energy remaining is kinetic because the
height at the end is zero since the ball strikes the ground.
-
If you make these two energies equal you get that KE + GPE at
the beginning is equal to just KE at the end. Just make them equal and solve for
velocity final!
-
1/2mvo2 + mgDh
= 1/2mvf2 solve for velocity final vf.= 64.9
m/sec
Example with a significant outside force,
friction!
Suppose that a mass of 4.50 kg is compressed a distance of
25.0 cm. against a horizontally mounted spring, which has a spring constant of k
= 440 N/m, and is then released. This mass slides a along a horizontal surface
which has a coefficient of sliding friction of m =
0.38. How far along this horizontal surface will the mass slide until it comes
to a halt?
-
At the beginning the only type of energy present is
elastic potential energy [EPE].
-
At the end there is no energy because the mass has come
to a halt.
-
If you calculate the energy at the beginning and subtract
from it the energy the difference must be equal to the work done by the
frictional force.
-
In this case 1/2kDx2
= Ffd = mgmd
-
1/2(440)(.25)2 = 4.5(9.8)(0.38)d where d =
0.82 m is the distance that the mass slides.
Science Department (732)792-7200 ext 8619
All one dimensional momentum problems can be solved using the equation:
m1v1 + m2v2 = m1v3
+ m2v
where m1 and m2 are the two objects in the collision,
v1 and v2 are the corresponding velocities of these two
masses BEFORE the collision while v3 and v4 are the
corresponding velocities of these two objects AFTER they have collided.
Inelastic Collisions
The first step here will be to make diagrams showing both objects before as
well as after the collision.
For example, suppose that a cart, which has a mass of 6.0 kg. [m1],
is moving toward the right with a velocity of 12.0 m/sec [v1] when it
collides with a second cart, which has a mass of 4.0 kg [m2], and
which is moving toward the left with a velocity of 8.0 m/sec [v2].
The diagram before will show the two masses moving in opposite direction as
shown below.
from which you get m1v1 + m2v2=
6.0kg(12m/sec) + 4.0kg(-8.0m/sec) . Note that v2 is negative in the
equation because the second cart is moving towards the LEFT before the
collision!
Now suppose that this collision is totally inelastic. This means that the
two object stick together after the collision and that significant amounts of
kinetic energy are lost in the collision. If so, the diagram after the collision
will appear as shown below.
from which you get m1v3 + m2v4 =
(m1 + m2)v3 = (6.0kg + 4.0kg)v3
where v3 is the unknown velocity after
the collision. Since the total vector momentum BEFORE the collision must be
equal to the total vector momentum AFTER the collision you
merely make the two quantities above equal to one another!
m1v1 + m2v2 = (m1 +
m2)v3 which then becomes
6.0kg(12m/sec) + 4.0kg(-8.0m/sec) = (6.0kg + 4.0kg)v3
72 kg.m/sec - 32 kg.m/sec = 40 kg.m/sec = (10kg)v3
Which if you then solve for v3 becomes v3 = 40 kg.m/sec/10kg.
= 4.0 m/sec
Elastic Collisions
Elastic collisions are somewhat more complicated. The diagram before and the
calculations before are the same as for inelastic collisions. However in an
elastic collision you need to be concerned about the kinetic energy after the
collision which in an elastic collision is the same both before and after the
collision.
For example, suppose that a cart, which has a mass of 6.0 kg. [m1],
is moving toward the right with a velocity of 12.0 m/sec [v1] when it
collides with a second cart, which has a mass of 4.0 kg [m2], and
which is moving toward the left with a velocity of 8.0 m/sec [v2].
The diagram before will show the two masses moving in opposite direction as
shown
below.
from which you get m1v1 + m2v2=
6.0kg(12m/sec) + 4.0kg(-8.0m/sec) . Note that v2 is negative in the
equation because the second cart is moving towards the LEFT before the
collision!
After the collision, however, the carts will NOT stick together but will
instead bounce off of one another. How they bounce off of one another will
depend on the conditions before the collision. Below is just one example of what
might happen.
In this case the momentum after the collision will be given by m1v3
+ m2v4 = 6.0kg(v3) + 4.0(v4). Again
you will make the momentum before equal to the momentum after giving you
m1v1 + m2v2 = m1v3
+ m2v4
but this time there will be two unknown variables v3
and v4! So what are you to do? You NEED
a second equation and there are two possibilities available. One relatively
difficult, the other fairly easy.
The difficult approach is to make use of the fact that this is an elastic
collision. This means that the total kinetic energy BEFORE the collision should
be equal to the total kinetic energy AFTER the collision as given by the
equation below.
1/2m1v12
+ 1/2m2v22 = 1/2m1v32
+ 1/2m2v42
and then solve this simultaneously with
m1v1 + m2v2 = m1v3
+ m2v4
And although this is doable, since it involves the squares in the kinetic
energy equation it can become an algebraic pain in the neck! So, what is the
alternative? The alternative is to make use of what is called the coefficient of
elasticity [or coefficient of restitution - same thing, different name]. The
coefficient of elasticity, represented by the letter e,
relates the relative velocity between the two objects AFTER the collision to the
relative velocity BEFORE the collision and is given by the equation:
-e = (v4-v3)/(v2-v1)
where (v4-v3) is the relative velocity [remember
"relative" means to "take the difference"] between the two
objects after the collision and where (v2-v1) is the
relative velocity between the two objects before the collision. The negative
sign is there because the objects reverse direction when they bounce off of one
another. The coefficient of elasticity will be exactly e = 1 if the collision is
elastic, e < 1 if the collision is inelastic [in fact e = exactly 0 if the
collision is perfectly inelastic!] and will be e > 1 if the interaction is an
explosion [more about explosions later!].
Solving the equation for coefficient of elasticity and the equation for
momentum conservation simultaneously is MUCH easier task than solving the
equation for momentum conservation simultaneously with the equation for energy
conservation as you can see below!
-e = (v4-v3)/(v2-v1)
becomes -1 = (v4-v3)/(-8m/sec-12m/sec)
= (v4-v3)/(-20m/sec)
If you solve this for v4 you
will get v4 = 20 + v3.
This can then be substituted into the momentum conservation equation
m1v1 + m2v2 = m1v3
+ m2v4
6.0kg(12m/sec) + 4.0kg(-8.0m/sec) = 6.0kg(v3)
+ 4.0(v4) = 6.0kg(v3)
+ 4.0(20 + v3)
40 = 6v3 + 80 + 4v3
leading to -40 = 10v3
solving for v3
v3 = -40/10 = -
4.0 m/sec
Finally, just substitute the value you got for v3
back into the equation from above and solve for v4!
v4 = 20 + v3
= 20 m/sec + -4.0 m/sec = 16
m/sec! Done!
One very strong hint! After solving for v3
and v4 always substitute back into the
original momentum conservation equation to make sure your answers really are
correct!
m1v1 + m2v2 = m1v3
+ m2v4
6.0 kg(12 m/sec) + 4.0 kg(-8.0 m/sec) = 6.0 kg(-4.0
m/sec) + 4.0 kg(16 m/sec)
72 kg m/sec - 32 kg m/sec = -24 kg m/sec + 64 kg m/sec
40 kg m/sec = 40 kg m/sec they are equal!
The position of any simple harmonic oscillator can be given by A
= Ao*cos(wt) where Ao is the maximum
displacement of the oscillator from the equilibrium position and where w
is the angular velocity of the oscillator. The angular velocity can be
determined from the period where w
= 2p/T where T is
the period of the oscillation. [Remember that the period of a simple pendulum is
given by T = 2p*(l/g)1/2
and the period of a mass vibrating on a spring is given by T
= 2p*(m/k)1/2!]
The velocity of this same oscillator can be determined from v = vo(-sin(wt))
where vo is the maximum velocity of this oscillator and can be
determined from vo = Ao*w.
The acceleration of this oscillator can be determined from a = ao(-cos(wt))
where ao is the maximum acceleration of the oscillator and can
determined from ao = Ao*w2.
What all of this means is that ALL you have to know, if you wish to
determine the position, velocity and acceleration of any simple harmonic
oscillator, is the maximum amplitude Ao
and the angular velocity w!
For example, suppose that you have a 3.0 kg mass attached to the end of a
vertical spring which has a spring constant of k = 450 N/m. This mass is then
lifted a distance of 15.0 cm above the equilibrium point and is then released
after which the mass vibrates up and down in simple harmonic motion.
The amplitude Ao of this oscillation is just Ao
= 0.15 m. while the period of the oscillation is given by T = 2p(m/k)^1/2
= 2*p*(3.0/450)^1/2 = 0.51 sec. From the period you
can easily determine the angular velocity from w
= 2p/T = 2p/0.51 = 12.3
rad/sec.
Now that you know these two pieces of information the equations for the
position, velocity and acceleration can be determined!
A = Ao*cos(wt) =
A = 0.15 m*cos(12.3t)
v = vo*(-sin(wt)) = Ao*w*(-sin(wt))
= 0.15 m*12.3 rad/sec*(-sin(12.3t) =
v = 1.85 m/sec*-sin(12.3t) where 1.85 m/sec is the maximum
velocity!
a = ao*(-cos(wt) = Ao*w^2*(-cos(wt))
= 0.15 m*(12.3 rad/sec)^2*(-cos(12.3t))
a = 22.7 m/sec2*-cos(12.3t) where 22.7 m/sec2
is the maximum acceleration!
Finally, don't forget to put you calculator into radian
mode. Otherwise, your answers will make no sense!
Determine the position, velocity and acceleration of this oscillator 25
seconds after the mass was released.
A = 0.15 m*cos(12.3t) = 0.15 m*cos(12.3 rad/sec*25
sec) = 0.15 m*0.93 = 0.14 m
v = 1.85 m/sec*(-sin(12.3 rad/sec*25 sec) = 1.85
m/sec*0.37 = 0.68 m/sec
a = 22.7 m/sec2*(-cos(12.5 rad/sec*25 sec)
= 22.7 m/sec2*(-.93) = -21 m/sec2
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SE·dA = 4·p·k·qenclosed
According to Gauss's Law the amount of electric flux passing through any
closed surface is dependent only upon the total charge contained within that
surface. [This is much like saying that the amount of light passing through a
closed surface is dependent only on the number of light bulbs contained within
the surface.] If there is not net charge within the closed surface than the
total flux passing outward through the surface will be exactly zero. [Or
again, in the case of light, if there are no light bulbs contained with the
closed surface there can be no net outflow of light through the surface.]
The total net flux flowing out [or into] a closed surface is given by F
= 4*p*k*q where k is the electric force constant [k
= 9.0 x 109 Nm2/C2]and q is the charge
contained within the closed surface. If the intensity of the electric field E
is constant everywhere on the closed surface and is always perpendicular to
the closed surface then the total flux through the surface will also be equal
to the product of the electric field strength E and the surface area A of
the closed surface or F = E*A. If you put these two
together you get F = E*A = 4*p*k*qenclosed.
The real trick to using Gauss's Law is the proper selection of the closed
surface so that the above conditions are met [E is constant AND perpendicular
to the surface at ALL points!] and therefore you can completely avoid any
integration.
Point Source
For example, if the source of the electric field is a point source,
such as a proton, electron, ion, charged particle of dust, charged pith ball,
etc. the logical closed surface that fully encloses the point charge, is equal
distance from the particle at all points [so that the electric field E is
constant at all points] and surrounds the point source so that all electric
field lines reaching the spherical surface are perpendicular to the surface of
the spherical surface [since the charged particle is at the center of the
sphere all field lines E leaving the charged particle follow radii of the
sphere which are, by definition, always perpendicular to the surface of the
sphere]. In this case the surface area of a sphere is A = 4*p*r2
and so Gauss's Law becomes F = E*[4*pi*r^2] = 4*p*k*qenclosed.
If you solve this for E you get Coulomb's Law for point sources E = k*q/r2.
Linear Charge
If the charge is a linear source rather than a point source the geometry is
quite different. The shape of the surface which is equidistant from a line is
not a sphere, but is rather a cylindrical shell. The surface area of a
cylindrical shell [excluding the ends through which no flux passes in any
case] is given by A = 2*p*r*h where h is the length
of the cylinder. If the line charge has a charge density l
then the total charge contained in a length h would be q = l*h.
Putting these two quantities into Gauss's Law you get
F = E*[2*p*r*h] = 4*p*k*[l*h]
If you now solve this for the electric field E through the surface of the
cylinder you get E = 2*k*l/r. Note that here the
electric field strength E near a linear source drops off with the inverse of
the distance from the linear source while the field strength E near a point
source, as above, drops off with the inverse square of the distance r.
Planar Charge
Finally if the source is planar with a charge density sigma spread uniformly
across the surface of a plane which has a surface area A the Gaussian
surface can readily be pictured at a tin can through the planar surface with
the charged plane passing through the center of the tin can and with the
surface of the plane parallel to the flat surfaces of the can. In this case
flux passes out only through the flat ends of the can because the field lines
leaving a planar surface are always perpendicular to the surface [if the field
lines were not exactly perpendicular to the surface of the plane the field
lines would then have to have components parallel to the surface of the plane
which would result in currents flowing in the surface of the plane which is
inconsistent with the premise here of electrostatics!] . In this case the
total area through which the flux passes is 2*a where a is the surface area of
one flat surface of the tin can. The total charge contained within the tin can
is equal to the charge density on the surface of the plane times the area of
the plane intersected by the tin can q = s*a. Now
putting both of these in Gauss's Law again you get
F = E*[2*a] = 4*p*k*s*a
If you solve this for the electric field you get E = 2*p*k*s.
Notice that the expression above for the electric field in the case of a
charged, large planar surface does NOT have the variable r in it, indicating
that there is NO dependence on distance! The electric field strength near a charged planar
surface is dependent ONLY on the surface charge density s
and is therefore constant with distance from the planar surface. This applies
ONLY when the dimensions of the planar surface are significantly larger than
the distance from the plane! Once you get far enough away from the plane to
"see the edges" it no longer behaves like a plane and Gauss's Law no
longer applies.
 |
One way to determine the electric field near an infinite
sheet of charge is to imagine that the infinite sheet
is made up of concentric rings of charge. If you then
integrate [add up] all of the rings from 0 m to infinity
you can then determine the electric field near that
infinite sheet.
The general expression for the electric file of a ring
at some point along the central axis of the ring can
be derived from the initial definition of electric
field dE=kdq/r^2 for a ring this becomes:
|
In order to determine the electric field near a disk
you need to next integrate this function between zero
and
R [where R is the radius of the disk].
To determine the electric field near an infinite plane
just change the limit of the integration to infinity
instead of R.
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This site last revised:
February 13, 2005.
