#$&*
course phy 201
02/05 9
120123
The Introductory Problem Set problems for your course are at
http://vhmthphy.vhcc.edu/ph2introsets/default.htm
Problem Set 5 corresponds to work we are doing with gases and fluids. You should work through that set and make notes indexing topics, concepts, solution methods, etc.. By Feb. 2 you will be expected to be familiar with the problems in this set.
Problem Set 6 corresponds to work we are doing related to waves and optics. You will be expected to familiarize yourself with this set by Feb. 8.
General College Physics students should do a quick read-through of Chapters 10, 13, 14 and 15
University Physics students should do a quick read-through of Chapters 14, 17, 18, 19 and 20
You are expected to have made brief notes, at least recognizing things in the text that you think might be relevant to the systems we are working with in class.
`q001. Include a copy of your data for the domino measurements, including your estimated uncertainty:
****
m stick= meter stick, measured in centimeters
paper= paper ruler made by instructor
Domino A Domino B Domino C Domino D Domino E Domino F
Paper m stick paper m stick paper m stick paper m stick paper m stick paper m stick
Length 7.8 5.1 7.9 5.1 7.9 5.0 7.9 5.0 7.9 5.1 7.9 5.1
Width 3.9 2.5 3.9 2.5 3.9 2.5 3.9 2.5 3.9 2.5 3.9 2.5
Thick 1.3 0.8 1.4 0.9 1.4 1.0 1.4 1.0 1.4 0.9 1.1 0.7
OVERALL AVERAGE PERCENT UNCERTAINTY OVERALL AVERAGE PERCENT UNCERTAINTY
(1.9%+5.99%+11.98%)=19.87% (1.28%+2.56%+7.97%)=11.81%
#$&*
Based on your data, calculate the volume of each domino in cm^3, and in (cm_reduced)^3, where cm_reduced is the measurement made with your paper ruler.
****
v=abc=Ah
Domino A Domino B Domino c Domino D Domino E Domino F
m stick (in cm) 5.1*2.5*.8=10.2cm^3 5.1*2.5*.9=11.5cm^3 5.0*2.5*1.0=12.5cm^3 5.0*2.5*1.0=12.5cm^3 5.1*2.5*.9=11.5cm^3 5.1*2.5*.7=8.9cm^3
paper(cm_reduced)^3 7.8*3.9*1.3=39.5cm^3 7.9*3.9*1.4=43.1cm^3 7.9*3.9*1.4=43.1cm^3 7.9*3.9*1.4=43.1cm^3 7.9*3.9*1.4=43.1cm^3 7.9*3.9*1.1=33.9cm^3
#$&*
`q002. Describe your experience in working with the buoyant balance.
****
There is a greater pressure at the bottom of the object in the water than at the top of the object. Forces are not equal and opposite at the bottom of the object experiencing the buoyant force, therefore there is an upward force.
#$&*
`q003. Report all data obtained during class, and include a description of what was measured and how:
****
A 40.5 cm trim balanced on the bottom of a large plastic cup, on 2 dominos (lying flat together) with a paperclip between them. 2 large and 1 small paperclip placed atop 1 side of the trim. 2 large and 1 small paperclip dangling on the other side of the trim. A small plastic cup with water in it (so some of the paperclip would be submerged) placed beside the large placed cup balancing the trim. The trim would not balance on the system until the cup of water was placed so that the paperclip was in the water.
#$&*
`q004. Once the buoyant balance is constructed, a small mass is placed on the free end (i.e., the end on which the paperclips aren’t hanging into the water). The system is observed to rotate in such a way that this end moves lower. Taking the direction of this rotation as the positive rotational direction, we conclude that the added mass results in an additional positive torque on the system.
The system oscillates for a short time and comes to rest, with the end containing the added mass a little lower than before.
Once the system is at rest, the net torque on it is zero.
Explain how we know that the torque in the positive direction is greater than before the mass was added.
****
If the torque was not greater with the mass the system would not come to rest on the side with the added mass.
#$&*
Is the torque in the negative direction now greater or less than before we added the mass?
****
Torque in the negative direction would be greater as well.
#$&*
What caused the torque in the negative direction to change?
****
When the added mass added torque to the positive direction, it caused the other side to lose some of the boyuant force which changed the torque.
#$&*
@&
As you correctly said the negative torque had to increase to balance the increased positive torque.
The loss of buoyant force did indeed increase the negative torque. How was it that the loss of buoyant force had this effect?
*@
Suppose we add a square of paper having mass 0.6 gram to the ‘free’ side of the system, at a point 20 cm from the balancing point. How much additional torque will then be acting on the system?
****
1g=0.001kg F=ma T=moment arm*force
0.6g=0.0006kg 20cm=0.2m
F=0.6g*980(cm/s^2)=588g(cm/s^2) or F=0.0006kg*9.8(m/s^2)=0.00588kg(m/s^2)
T=20cm*588g(cm/s^2)=11760g(cm^2/s^2) or T=0.2m*0.0059kg(m^2/s^2)=0.00118kg(m^2/s^2)
=11760 erg or =0.00118 Newton
#$&*
@&
Units are mostly, but not completely right.
F = .00588 kg m/s^2, as you say.
This is .00588 Newtons.
This is also equal to 588 g cm / s^2 = 588 dynes.
Multiplying this force by the 20 cm moment arm you get
torque = 11760 g cm^2 / s^2, which you would express as 11760 cm * dynes.
When calculating energy the unit g cm^2 / s^2 is also called the erg, but we're calculating torque here so we use cm * dynes.
Using .00588 Newtons and 0.20 meters we also get
.00118 kg m^2 / s^2
in agreement with your result.
We would call the unit kg m^2 / s^2 the meter * Newton.
In an energy calculation we would call the kg m^2 / s^2 a Newton * meter or Joule.
*@
Assuming that the paper clips are suspended from a point 20 cm from the balancing point, on the end opposite the ‘free end’, what must be the change in the torque on this side once the system has come to rest in its new position?
****
If the torques are equal and opposite to have a net torque of 0, the torque would be -11760 erg=-0.00118 Newton
#$&*
@&
Good. With correct units, -11760 cm * dynes or .00118 m * N.
*@
How is it that the torque on this end changes?
****
Net force changes as the boyuant force changes. Boyuant force change because some the the paperclips that were in the water are not now so the paperclips have more weight.
#$&*
@&
Gravity exerts the same force on the paperclips regardless. So their weight doesn't change.
However the total force on that side does change by the amount you have calculated, due to the lesser buoyant force
*@
Does the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
****
Weight increased because there is not as much of a boyuant force because less of the paperclips are submerged in the water.
#$&*
Does the torque due to the weight of the paper clips on this end change? If so does it increase or decrease, and by how much?
****
Yes, the torque changes, torque increased negatively because the other side was set as the positive direction.
#$&*
@&
Negative torque increased because the buoyant force, which tends to rotate the system in the positive direction, decreased.
*@
Does the buoyant force on the paper clips on this end change? If so does it increase or decrease, and by how much?
****
The boyuant force decreases on the paperclips because there isn't as much of the paperclips in the water.
#$&*
Does the buoyant force on the paper clips on this end change? If so does it increase or decrease, and by how much?
****
#$&*
Suppose we add a square of paper having mass 0.6 gram 3 cm on each edge, to the ‘free’ side of the system, at a point 20 cm from the balancing point. A sheet of this paper, 8.5 inches x 11 inches, has a mass of 5 grams. By how much does the torque in the positive direction change as a result?
****
#$&*
If the hanging paper clips are 15 cm from the balancing point, then after the system has come to rest, by how much has the weight of the paperclips changed, and by how much has the buoyant force on the paper clips changed?
****
#$&*
`q005. BB’s are bouncing back and forth, without loss of energy, between two parallel walls, one on the left and one on the right. Assume that the walls are 36 cm apart, and that gravitational influences have negligible effect. A BB shot from one wall, straight toward the other, will therefore keep bouncing back and forth between the walls forever. The speed of the BB, when it’s not in contact with the wall, will always be the same as the speed from which it was shot. (This is of course an idealization; the coefficient of restitution for an actual BB colliding with a real wall is less than 1 so energy will in the real world be lost with each collision).
A BB is shot toward the right wall from the position of the left wall. A second BB is shot at the same velocity from the same position, just as the first BB hits the right wall, from which it rebounds toward the left wall without any loss of speed.
How far from the left wall will the two BB’s be when they meet up?
****
36cm/2=18cm from the left wall. They meet in the middle.
#$&*
Assuming that they narrowly miss one another and continue bouncing back and forth between the walls, how far from the left wall will they be the next time they pass one another?
****
18cm form the left wall, they will always meet in the middle.
#$&*
At what distances from the left wall will they pass for the third, fourth and fifth time?
****
18 cm, always 18 cm which is the halfway point.
#$&*
As they continue bouncing back and forth, assuming their paths are just far enough apart to avoid collision, how many ‘passing points’ will there be between the two walls?
****
18 cm from the left wall and right wall
#$&*
If the second BB was just a little slower than the first, how would the ‘passing point’ change as time goes on?
****
The passing point would continuously change.
#$&*
Suppose now that the second BB is fired at the instant the first BB is halfway to the right wall. How far from the left wall will they be when they meet?
****
27 cm, because 18 cm is the halfway point and half of that is 9 cm. 18cm+9cm=27cm
#$&*
If we continue firing a new BB every time the last BB reaches the halfway point, assuming that the BB’s always manage always to narrowly miss one another, at what distances from the left wall will BB’s pass one another?
At the instant of firing, at what possible distances from the left wall will all the other BB’s be located?
****
At 9cm,18cm,and 27cm from the left wall
#$&*
At the instant the first BB passes by the second, at what possible distances from the left wall will all the other BB’s be located?
****
9cm, 18cm, and 27cm
#$&*
In what other ways might the firing be timed so that when two BB’s pass, every other BB is at the same distance from the left wall as at least one other BB?
****
By firing when the BB's are at the left wall, right wall, 9cm point, 18cm point, and 27cm point.
#$&*
`q006. If an ideal gas is kept at constant volume, then the percent change in its pressure is the same as the percent change in its absolute temperature. Every additional 10 cm of water supported in a thin vertical tube requires an increase in pressure equal to about a 1% of standard atmospheric pressure.
Based on this, how much additional pressure do you estimate corresponds to one unit on your ‘squeeze scale’?
****
(1/2)%
#$&*
@&
1/2 % of an atmosphere would be correct.
*@
`q007. If the initial absolute temperature of the gas is 300 Kelvin, then if it is heated enough to raise a column of water 40 cm high, then:
By how much will its temperature have changed?
****
1 K change=1% atmospheric pressure change so,
4 K change = 4% atmospheric pressure change
1 K change raises water 10cm so,
40cm/10=4
300+4=304 K
#$&*
@&
A 1% increase in the pressure of a confined gas at constant volume implies a 1% increase in absolute temperature.
So a 4% increase in pressure implies a 4% increase in absolute temperature.
What was the absolute temperature?
What's 4% of this?
*@
If the bottle contains 500 milliliters of air, how much energy would be required to achieve this change? Note that the volume of water in a thin tube can be regarded as negligible, so that the volume of the gas will remain unchanged. (You should know how how to find approximately how many moles are contained in 500 milliliters of air at atmospheric pressure and typical ambient temperatures).
****
#$&*
@&
How many moles in a 500 ml container?
How much energy per mole, per degree?
How much energy is therefore required?
*@
How much would the temperature of the air in the bottle have to change to have the same effect as one unit on your ‘squeeze scale’?
****
#$&*
@&
Your estimate was 1/2 % of 1 atmosphere.
The original pressure was 1 atmosphere, so the change was 1/2 % of the original pressure.
What was the original temperature?
How much change do you conclude?
*@
*#&!*#&!
@&
Very good on most questions.
See my notes, and give yourself maybe 20-30 minutes to see what you can revise.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
*@