So sorry this is so late, more is coming.
I hope to be caught back up by Sunday" "Physics II Class 120111
(This document contains class notes of 1/09/12 and 1/11/12. It is duplicated under the date 120111).
Your best attempt on the questions posed here will be due by the end of the day on Friday, 1/20/12.
To raise a vertical water column 10 cm requires about an additional 1.0% of atmospheric pressure.
To achieve a 1% increase in the pressure of a confined gas at constant volume requires that its absolute temperature be raised 1%.
If the absolute temperature of a confined gas increases by 1% without changing its pressure, its volume increases by 1%.
For our purposes the freezing and boiling points of water at atmospheric pressure are defined, respectively, to be 0 Celsius and 100 Celsius. This is not the standard
SI definition. You are not yet expected to be prepared to understand the standard SI unit, so for the moment we are using the ‘old’ definition. The two definitions
agree to about 5 significant figures.
A mole of ideal gas at atmospheric pressure at 0 Celsius occupies 22.4 liters; for ballpark calculations you could use either 20 or 25 liters as a basis for
estimation.
To raise the temperature of a mole of a monatomic ideal gas by one degree Celsius requires about 21 Joules, provided the gas is allowed to expand at constant
pressure. If the gas is confined to a constant volume then it doesn’t have to do the work of expansion, and requires only 3/5 as much energy. If the gas is diatomic
then every degree of temperature increase requires about 8 Joules more than if it was monatomic (the extra energy is needed because diatomic molecules can and do spin
as a result of their collisions with other molecules, which takes additional energy).
The change in the potential energy of an object or system (designated `dPE) between two events is defined to be equal and opposite to the work done by conservative
forces on the system between those events. Examples of conservative forces are gravitational, electrostatic and magnetic forces, as well as ideal elastic forces.
The weight of an object is the force exerted on it by gravity. The force exerted on an object by gravity in the vicinity of the Earth’s surface will, in the absence
of other forces, accelerate that object toward the center of the Earth at 9.8 meters / second^2, which is also equal to 980 centimeters / second^2 and close to 32 feet
/ second^2. We use g to stand for this acceleration. Since the unopposed gravitational force gives the object this acceleration, the gravitational force on the
object must be F_grav = m g, where m is its mass. Thus the weight of an object of mass m, near the surface of the Earth, is weight = m g.
If we raise the object its displacement is in the direction opposite its displacement, so that the gravitational force does negative work on it.
The equation of motion of an object undergoing simple harmonic motion is x(t) = A cos(omega * t + theta_0), where omega is the angular frequency of the motion, A the
amplitude and theta_0 the initial angular position of the circular-model reference point. In the absence of other information theta_0 may be taken to be zero. In SI
units the angular frequency is equal to the frequency of the motion in cycles per second, multiplied by the 2 pi radians in a cycle.
In an elastic collision, kinetic energy is conserved. An object which collides elastically with a much more massive object loses negligible kinetic energy.
`q000. Report the data you obtained in lab on 1/09 and 1/11. Include a brief but clear description of what you did, and report the data in a concise, organized,
self-explanatory table.
****
JAN 9
System: Holding a wood-like strip at two points and ""shaking"" to measure the frequency (oscilation/sec)
Data: Horizonatal (w/ gravity)
Distance between end1 and hand1 (H1) --> 24.5cm
H1 and H2 --> 73cm
H2 and end2 --> 25.5cm
>>>22 osc in 7 sec
: Vertical (perpendicular to gravity)
Distance between end1 and hand1 (H1) --> 36.5cm
H1 and H2 --> 47.5cm
H2 and end2 --> 39cm
>>>12 osc in 2.3 sec
System: Balancing a wood-like strip on [a domino on a cd on a cup] at two seperate points and ""releasing weight"" to see how long it oscilates
Data: Distance between end1 and structure1 (S1) --> 24.5cm
S1 and S2 --> 73cm
S2 and end2 --> 25.5cm
>>>took 18.3 sec to stop
: Distance between end1 and structure1 (S1) --> 36.5cm
S1 and S2 --> 47.5cm
S2 and end2 --> 39cm
>>>took 20.4 sec to stop
JAN 11
System: Balancing a wood-like strip on 3 structures to try to create the longest lasting oscilation
Data: Distance between end1 and structure1 (S1) --> 52.5cm
S1 and S2 --> 92cm
S2 and S3 --> 71cm
S3 end3 -->28cm
>>> 43 sec
: Distance between end1 and structure1 (S1) --> 74cm
S1 and S2 --> 81.5cm
S2 and S3 -->79.5 cm
S3 and end3 --> 12cm
>>> 61 sec
: Distance between end1 and structure1 (S1) --> 17cm
S1 and S2 --> 69cm
S2 and S3 --> 135cm
S3 and end3 --> 24.5cm
>>> 16 sec
: Distance between end1 and structure1 (S1) --> 56.5cm
S1 and S2 --> 58.5cm
S2 and S3 --> 71.5cm
S3 and end3 --> 59cm
>>> 56 sec
: Distance between end1 and structure1 (S1) --> 17cm
S1 and S2 --> 115cm
S2 and S3 --> 102.5cm
S3 and end3 --> 11cm
>>> 36 sec
#$&* Please note the following: Your response to the question should be inserted in the ‘middle line’ which is the line following the **** and before the #$&*.
Follow this practice throughout the course. Never add anything to or delete anything from any line that begins with #$&* or ****.
`q001. What would be the potential energy change of a 10 gram mass of water whose vertical position changes by 40 centimeters?
****
PE - equal and opposit to the work done by the conservative force
looking for d'PE of 10g of water that lowers 40cm
if it's lowering then it's PE is decreasing as well
d'PE = - (row->desity) * (d'Volume) * (g->gravity) * (y->height of water)
= - (1 g *cm^3) * (mass/density) * (9.8 m/sec^2) * (40cm)
= - (1 g *cm^3) * (10g / 1g*cm^3) * (9.8 m/sec^2) * (40cm)
= - 392000g*cm^2/sec^2
#$&*
@&
Good but watch your signs:
A change of +40 cm would be in the direction opposite the conservative gravitational force, so the work done by the conservative force would be negative.
The PE change would therefore be positive.
*@
`q002. If the potential energy of 10 grams of water, which we will consider to be initially at rest, changes by 100 000 g cm^2 / s^2 (which is equal to .01 kg m^2 /
s^2 or .01 Joules),
and if this PE loss is converted to KE, then how fast is the water moving?
****
PE of 10g of water changes by 100,000gcm^2/sec^2
PE is converted to KE
velocity of water?
d'Ke = d'PE = - (-p*d'V*g*y) = p*d'V*g*y
= (1/2 * m * V_f^2) - (1/2 * m * V_o^2)
= (1/2 * 10g * V_f^2) - (1/2 * 10g * 0cm/sec) = 100,000gcm^2/sec^2
V_f = 141.4323562 cm/sec
#$&*
`q003. A stream of water spurts out of the side of a vertical cylinder, falling 60 centimeters while traveling 40 centimeters in the horizontal direction. Assuming
that the horizontal velocity of the water remains constant, what was the speed of the water as it exited the cylinder?
****
falls 60cm
""goes"" 40cm
exiting speed?
V_exit^2 = 2gy
= 2 * (980cm/sec^2) * (water depth? ... 60cm ...)
= 342.928564 cm/sec
#$&*
`q004. If the frequency of the oscillation of a strip of plastic between two dominoes separated by 40 cm is 5 cycles per second, with amplitude 2 cm, then assuming
that each point on the strip undergoes SHM:
What is the equation of motion of the point halfway between the dominoes?
****
separated by 40cm
5 cycles per sec = 10 pi rad per sec
2cm amplitude
SHM
x(t) = A cos(omega * t + theta_0)
= 2cm cos [(10pi rad / sec) * t + 0]
#$&*
Calculate or estimate the amplitude of motion for a point on the strip which is ¼ of the way between the dominoes.
****
1/4 of 2cm = 1/2 cm
#$&*
Based on your estimate what is the equation of motion of this point?
****
x(t) = A cos(omega * t + theta_0)
= 1/2cm cos [(10pi rad / sec) * t + 0]
#$&*
What is the equation of motion of a point on the strip which is 1/6 of the way between the dominoes?
****
x(t) = A cos(omega * t + theta_0)
= 1/3cm cos [(10pi rad / sec) * t + 0]
#$&*
@&
The shape of the strip is not a straight line, but is rather concave down. So your 1/2 and 1/3, while reasonable, are a little low. As long as you understand this it's not a problem.
*@
If x is the coordinate of a point on the strip, as measured from the domino on the left, what is the equation of motion at this point?
****
(x,y)
x is the distance from the left
y is the amlitude at that point
the location of the point with determine both of these points (as shown in previous problems)
x(t) = A cos(omega * t + theta_0)
#$&*
`q005. If a ball of mass 10 grams, moving at 500 cm / s in a direction perpendicular to a wall, strikes the wall and bounces off elastically, then what is its
momentum change from the moment it contacts the wall until it comes to rest?
****
p = mv
= 10g * +500m/sec
= +5,000g*m/sec
it hits the wall with a momentum of +5,000g*m/sec
stops for a split second when it hits the wall to change direction
and bounces of with the same amount of force
at the wall and at rest its momentum is zero then the ultimate change in momentum is 0
#$&*
What is its momentum change between the time it comes to rest and the time it loses contact with the wall?
****
at rest: 0
loses contact w/ wall: -5,000g*m/sec
-5,000g*m/sec
#$&*
What is its momentum change between its first contact and its last contact with the wall?
****
+5,000g*m/sec hit
-5,000g*m/sec leave
0
#$&*
@&
The ball loses 5000 g cm / s as it comes to rest, so that change is -5000 cm/s.
When it rebounds the change is also -5000 g cm/s.
So the net change is -10 000 g cm / s.
*@
If the ball bounces elastically off of another wall and returns to the original wall every .1 second, what average force does it exert on that wall?
****
F = ma
= 10g *
V = 500m/sec
d't = .1 sec
#$&*
In the above, be sure you have treated velocity as a vector quantity. Mainly this means that you need to declare a positive direction and abide by your declaration.
`q006. If the temperature of the air in a 2-liter bottle increases by 10 degrees Celsius, how much energy is required? Air consists of a mix of gases, most of which
are diatomic.
****
diatomic gas requires 29 J = (21J + 8J) on energy to increase every degree of temperature
if it's in a concealed container then it takes about 17.4 J = [(3/5)(21J + 8J)]
to raise the temp by 10 degrees Celsius in a closed 2-liter bottle, it would take 10 * 17.4 J = 174 J
#$&*
@&
You've got the right energy for a mole of air. However the 2-liter bottle contains only ballpark .1 mole, and would therefore require only about 17 Joules of energy.
*@
`q007. A plastic tube has inner diameter about 3 cm.
What is the volume inside a 10-cm length of this tube?
****
V = pi * r^2 * h
= pi * 1.5cm^2 * 10cm
= 71 cm^3
#$&*
What percent is this of the volume of a 500-milliliter bottle?
****
about 14%
#$&*
How many 10-cm lengths of this tube would be required to match 1% of the volume of a 500-millileter bottle?
****
about 7
#$&*
If the air in a 500-milliliter bottle is heated until the water in the tube rises 10 cm, by what percent did the volume of the bottle increase, and by what percent do
you think the temperature of the air had to increase?
****
the volume of the actual bottle did not increase a significant amount
the air increased in volume which pushed the water out through the tube
it requires and additional 1.0% atmospheric pressure to raise a vertical water column 10cm
#$&*
@&
The water that goes into the tube is no longer in the bottle, so the volume of the air in the bottle does increase.
That 1% figure is for a thin tube. This tube is not thin. The volume of a 10 cm segment of the tube is very significant for this system.
*@
`q008. A plastic strip resting on a series of dominoes, including dominoes at its ends, is expected to resonate if the dominoes are equally spaced. What are the
first five domino spacings expected to produce harmonic resonance in a strip of plastic 4 meters long?
****
4 meters = 400 cm
400 cm / 4 = 100 cm
place a domino on each end ( 2 dominos total so far)
in between these two dominos place the remaining three dominos 100cm apart from one another
#$&*
@&
So 100 cm is a possible spacing.
The question asks for five spacings, so you need four more.
*@
`q009. A certain aluminum rod is supported at the middle. When the end of the rod is struck sharply in a direction perpendicular to the rod, it produced a bell-like
sound with a frequency of 300 cycles / second and amplitude .3 millimeters.
What is the maximum speed of the simple harmonic motion of a point on the end? (Note that you can determine this if you think about the SHM: what is the amplitude
and frequency of the SHM of this point, and how do you use the amplitude and frequency of the SHM to find the maximum speed?)
****
amplitude = .3 mm
frecquency = 300 cycles / sec = 600pi rad / sec
x(t) = A cos(omega * t + theta_0)
= .3 mm cos [(600pi rad / sec) * t + 0]
V = change in position wrt clock time
= 2(amplitude) / t
= 2(.3mm) / t
= .6mm /t
#$&*"
@&
At maximum speed the oscillating point has a speed that matches that of the reference-circle point.
What is the speed of the reference-circle point?
*@
end document
@&
Very good. but check my notes. I'll leave it to you whether to submit a revision. If you do be sure to mark your insertions before and after with &&&&&&&.
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