LESSON 11

PHYS 1100 LESSON 10 FOR
MONDAY, AUGUST 23, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due and Return of Papers

IV Review of Lesson 9: Waves and Fluids
A. Waves
1. Length, l
2. Frequency, n
88 Mhz – 108 MHz FM
3.8 m
530 KHz – 1710 KHz
Light, v = c = 300,000,000 m/s
= pitch in sound
Sound, v = 340 m/s
3. Period, P = 1/n
4. Amplitude, A
5. Speed v = l n
6. Types
7. Other

B. Fluid Dynamics
1. Definition of “fluid”
2. And, defn of “dynamics”
3. Density; r = M/V; g/cm3.
4. Pressure; P = F/A; Pascal; 1.0 N/m2.
Pascal; 1013 mB = ATM; 14.7 lb/in2. Mm of Hg; inches of Hg; ATM
5. Bernoulli’s Law – pressure differences between surfaces
6. Archimede’s Principle – displacement of fluids by masses
And more….

V Lesson 10: Density, Archimedes, Bernoulli
Examples of Aviation: airplanes, gliders, helicopters, and similar.

VI Laboratory Exercise 10: Archimedes, Liquids, and Solids

VII HWK Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 530-534 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; Due 8/26

VIII Essay 4: Albert Einstein, due 8/26

LAB 10: ARCHIMEDES, SOLIDS, LIQUIDS (for Monday, August 23)

Physics Lab 10: Archimedes, Solids, and Liquids Name _____________
Monday, August 23, 2010 Dr Dave Menke, Instructor

I Title: Solids and Liquids

II Purpose: To observe the effects of solids in a liquid environment

III Equipment and Supplies:
500-mL beaker, Large clear glass, or similar
Water
A metallic cylindrical weight much denser than water
A pair of tongs
Paper Towels to clean up your mess
Metric Ruler
Paper Clip(s)
Straw
“Sharpie” style pen

IV Procedure
1. Determine the diameter, d, of the beaker or glass in cm. Record.
2. Determine the area, A, of the base of the beaker or glass (A = p r2, where r = ½ d). Record.
3. Determine the height/depth of the beaker or glass, h, in cm. Record.
4. Calculate the maximum volume of the beaker or glass in cm3. Record. [Do this by multiplying the height/depth that you found in #3 (h), with the area of the base of the glass or beaker, A, that you found in #2].
5. Fill beaker about ¾ full with water (measure ¾ h from the bottom of the beaker, record, and mark the outside of the beaker/glass with Sharpie Pen).
6. Be sure that the marked water line is level with the water.
7. Use a straw to remove some of the water, via capillary action (don’t use it to drink the water). This means to lower the straw into the water almost all the way to the bottom, place your finger over the top of the straw, and slowly remove it. (The water should stay in the straw, unless it is defective. DON'T THROW THE WATER AWAY THAT IS IN THE STRAW).
8. Note the new water line. Measure how far the water line dropped in millimeters. Convert to centimeters. Record.
9. Return the water in the straw to the beaker and put away the straw. Make sure the water line is the same as before; if not, then add a few more milliliters of water to make it so.
10. Carefully place ONE paper clip on the water’s surface – if you are good at it, the paper clip will float. How many times did you have to do this to make it work? Record.
11. How many paper clips can you put on the water before it breaks the surface tension? Record.
12. Explain what allows the paper clip(s) to float.
13. Remove the paper clip(s) and put them away. Make sure the water line is back to where you started.
14. Select a metal (lead) weight. Calculate its volume, in cm3. Record. (Do this by measuring the diameter of the base, then get the Area of the base in cm2, then measure the height, and multiply the area times the height.)
15. Slowly place (with tongs) the dense lead weight into the water; note the new water line. Measure how far the water line increased, in millimeters, above the water line. Convert to centimeters. Record.
16. Find the increased volume of the water from the addition of the weight in cm3. Do this by multiplying what you found in #15 with the Area that you found in #2. Record.
17. Compare what you just found in #16 with what you calculated in #14 (they should be the same.) Explain.

V Data & Observations
A. Diameter of the beaker or glass d = ______ centimeters
B. Area of the bottom of the beaker or glass = ___________cm2.
C. Height/depth of the beaker or glass, h = ______ centimeters
D. Volume of the beaker or glass, Vb = ____________ cm3.
E. The distance from the bottom of the beaker/glass to ¾ h = ________ cm.
F. Distance that the water dropped from capillary removal ______ cm
G. Number of attempts to make the paper clip float: _____ times
H. Number of paper clips before breaking surface _______ paper clips
I. Explanation of floating paper clips:
J. Volume of lead weight, Vc = ___________ cm3.
K. Distance that the water increased from submersion of dense weight: ______ cm.
L. Increased volume of the water due to the submersion of the weight: ___________cm3.
M. Comparison between #14 and #16.



VI Results:



VII Error Analysis
A. Qualitative Only
1. Personal
2. Systematic
3. Random
B. Quantitative: NA

VIII Questions
1. Why can the straw hold water, even if the bottom is open and gravity is pulling on it?
2. What can you conclude about denser materials in water?
3. In one paragraph of a few sentences, explain Archimedes' Principle.

END

SOLUTION SET 4 - COMPLETE

Chapter 15: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 530-534.

1. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.

So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.~ 8.1 x 102 N.

3. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.

9. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.

13. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.

19. A circular wine barrel is shown in figure 15-26 on page 530. The barrel has a diameter of 75 cm. If a net upward force exceeds 643 Newtons it will burst. A tube of 1.0 cm in diameter extends into the barrel through a hole in the top. Initially the barrel is filled to the top and the tube is empty above that level. What weight of water must be poured into the tube in order to burst the barrel? And why would anyone want to do that anyway?
Solution:
Pressure = force/area
Density = mass/volume
Force = m g = 643 N (in barrel)
W = weight of water
Force density = r g
PE = mgh
PEdensity = r g h
Pbarrel = Ptube
F/Abarrel = W/Atube
The weight of the water must be poured into the tube, such that the above holds true:

W = [(Atube)/(Abarrel)] F

Abarrel = p (0.375)2 = 0.4415625 m2 . (Barrel’s area)
Atube = p (0.005)2 = 0.0000785 m2 . (Tube’s area ). So,

W = [(Atube)/(Abarrel)] F = [(0.0000785)/(0.4415625)] (643 N) = [0.000178](643) = 0.114 N.

Or, W = 0.11 N.~ 1.1 x 10-1 N.


25. Referring to example 15-4 on page 507 in the book, suppose that some vegetable oil has been added to both sides of the U-tube. Not YouTube. On the right side of the tube the depth of oil is 5.0 cm as before. On the left side of the tube the depth is 3.0 cm. Find the difference in fluid level between the two sides of the tube.
Solution:
We know the density of the oil and the water, respectively:

roil = 9.2 x 102 kg/m3.
rwater = 103 kg/m3.

The pressure point at “C” is given by,

PC = Pat + rogh1 + rwgx

and the pressure point at “D” is given by,

PD = Pat + rogh2 + rwgy.

Now we must find the difference in fluid level between the two sides of the tube:

(y + h2) – (x + h1)

At equilibrium, PC = PD, or, Pat + rogh1 + rwgx = Pat + rogh2 + rwgy..This leads us to:

rogh1 + rwgx = rogh2 + rwgy. Because the atmospheric pressure, Pat, is on both sides.

An ordinary orangutan can also see that the acceleration of gravity, g, is in each term, so we can divide that out, and now end up with:

roh1 + rwx = roh2 + rwy. We can rewrite this as rw (x - y) = ro(h2 – h1).

(h2 – h1) = [(rw)/(ro)](x – y). Or we can invert it to be: (x – y) = [(ro)/(rw)](h2 – h1)].

Now, multiply both sides by negative one ( - 1), then add (h2 – h1) to both sides: Then we will have:

[(h2 – h1) - (x – y)] = (h2 – h1) - [(ro)/(rw)](h2 – h1)..And that can be factored to look like this:

(y + h2) - (x + h1) = (h2 – h1)[1 - (ro)/(rw)]. And this equals (5 cm – 3 cm)[1 - (9.2 x 102 kg/m3)/(103 kg/m3)] = (2 cm)[1 – 0.92] = (2 cm)(0.08) = 0.16 cm.~ 1.6 x 10-2 m.


29. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III

31. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.

39. A hydrometer is a device for measuring fluid density. It's constructed as shown in Figure 15-30 on page 532. If the hydrometer samples fluid 1, the small float inside the tube is submerged to level 1. When fluid 2 is sampled the float is submerged to level 2. Is the density of fluid1 greater, less, or the same as fluid 2's density? [This is how auto mechanics test the antifreeze level in your car engine. Since antifreeze (ethylene glycol) is denser than water, the higher the density of coolant in your radiator the more antifreeze protection the engine has.]
Solution:
How would I know? I'm not an auto mechanic. He he he.

A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.

Chapter 16: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 566-567

1. The official record for lowest temperature ever recorded on Earth was set at Vostok, Antarctica, on July 21, 1983. The temperature on that day fell to T = ( – 89.2° C), which is well below the temperature of dry ice (frozen CO2). Give this temperature in degrees Fahrenheit.
Solution: We use the “famous” equation that °F = (9/5) °C + 32°. So, let's “plug and chug” and that will be it: °F = (9/5) °C + 32° = (9/5)(-89.2°C) + 32° = (1.8)(-89.2°C) + 32° = (- 160.56°) + 32° =
- 128.56 °F.~ - 129 °F.

3.Normal body temperature for humans is T = 98.6° F. Convert this temperature to degrees:
a. Celsius (aka Centrigrade)
b. Kelvin
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°).
a. °C = (5/9)(°F – 32°) = (5/9)(98.6° – 32°). = (0.556)(98.6° – 32°). = (0.556)(66.6) = 37°C .
b. Using °K = °C + 273, then °C + 273 = 37 + 273 = 310°K.

9. A world record for the greatest change in temperature was set in Spearfish, North Dakota on January 22, 1943. At 7:30 AM, the temperature was T = ( - 4° F); two minutes later, at 7:32 AM, the temperature was T =+ 45° F. Calculate the mean rate of temperature change, per second, using Celsius or Kelvin scales (they use the same units); Hint: find DT.
Solution: We use the “famous” equation that °C = (5/9)(°F – 32°). At the lowest temperature, °C = (5/9)(- 4° – 32°). = (0.5555)( - 36) = - 20°C. At the highest temperature, °C = (5/9)(45° – 32°). = (0.5555)(13) = 7.2°C. Then, DT. = [7.2 - ( -20)] = 27.2 °C. in two minutes. Or, 27.2 / (120 sec) =
+ 0.227 °C/s.

13. Referring to Table 16 - 1 on page 545, which would be more accurate for an all-season, outdoor use: a tape measure made of steel or one made of aluminum?
Solution: Steel (iron) because it will shrink and expand less.

19. It is desired to slip an Aluminum ring over a steel bar as shown in Figure 16-14 on page 567. At T = 10.00° C, the inside diameter of the Aluminum ring is da = 4.000 cm, and the diameter of the steel rod is ds = 4.040 cm. The steel bar maintains a constant temperature of Ts = 10.0° C.
a. In order for the ring to slip over the bar, should the ring be heated, or cooled? Explain.
b. Find the temperature, Ta, of the ring at which it fits over the over the bar.
c. Who is the Lord of the Rings?
Solution: In this case, we must consult the coefficients of linear expansion, or, a. If the aluminum expands at a faster rate than the steel as we heat them, then we should heat the aluminum. Otherwise, cool it.
a. a(Al) = 24 x 10- 6 per degree K; and a(Steel) = 12 x 10- 6 per degree K. This means that Aluminum expands faster, so we heat the aluminum.
b. The Aluminum ring will NOT fit over the Steel bar at this temperature (10° C), so we need it to expand to exceed the diameter of the steel, i.e. Da > ds. So, Da > 4.040 cm, such that Da = (4.000 cm )+ Dd. According to the linear expansion formula, DL = aL0DT. So, for this problem, Dd = a(da DT). And, of course, Dd = ds – da = 0.040 cm; DT = Tf – Ti = Tf - 10.00° C. Where We are looking for Tf. So, let's re-write this as:

(Dd)/ (ada ) = DT = (Tf – Ti). And now, we can re-arrange that to get Tf.

Tf = Ti + (Dd)/ (ada ) = 10.00° C + (0.040 cm) / [(24 x 10- 6)(4.000 cm)] =

10.00° C + (0.040 cm) / [(96 x 10- 6)] = 10.00° C + (4.17 x 102 °C) = 427°C.

c. Exactly.

25. An aluminum saucepan with a diameter of d = 25.0 cm and a height of h = 6.0 cm, is filled “to the brim” with H2O. The initial temperature of the pan and water is T = 19° C. But then the pan with the water in it is placed on a range where it is heated to T = 88° C.
a. Will the water overflow from the pan, or will it decrease? Explain.
b. Calculate the volume of water that either overflows or decreases.
c. Go make some popcorn. Turn off the stove.
Solution: In this case, we must consult the coefficients of linear expansion for Aluminum, or, a. and volumetric expansion for water, or b. If the aluminum expands at a faster rate than the water as we heat them, then the water level will go down. Otherwise, it will overflow. We will use both DL = aL0DT for Aluminum, and DV = bV0DT for water.
a. a(Al) = 24 x 10- 6 per degree K; and b(water) = 0.21 x 10- 3 per degree K. This means that the water will expand faster in volume than the Aluminum will in linearity. The pot will overflow. By 24 cm3 .
b. The volume of the pan (and thus, the water) is the area of the pot, which is the area of the circle of the pan, multiplied by its height, or, (h)(p r2) where r = ½ d. Thus, the volume is:

V = (h)(p r2) = (6.0 cm)(p)(12.5)2 = (6.0 cm)(3.14)(156.25) = (6.0 cm)(3.14)(156.25) = 2945 cm3. So, this is the “original” volume, or V0. DT = 88° C – 19° C = 69° C.

Thus, DV(water) = bV0DT = (0.21 x 10- 3 )(2945)(69) = 42.7 cm3.

But the Aluminum pot will expand, too, by:

DV(Al) = 3aV0DT = (3)(2.4 x 10- 5)(2.945 x 103)(6.9 x 101) =

(3)(2.4)(2.945)(6.9)(10- 5)(103)(101) = 146.31 x 10- 1. = 14.63 cm3.

Finally, overall change of volume is [DV(water) - DV(Al)] = (42.7) - (14.63) = 28 cm3.

c. Okay.



29. Consider the apparatus the Joule used in his experiments on the mechanical equivalent of heat, shown in Figure 16-8 on page 550. Suppose both blocks have a mass of m1 = m2 = 0.95 kg and that they fall through a distance of h = 0.48 meter.
a. Find the expected rise in temperature, DT, of the water, given that 6200 Joules are needed for every increase of DT = 1.0° C. Give your answer in degrees Celsius (°C).
b. Would the temperature rise in degrees Fahrenheit (°F) be greater than, or less than, the result in part (a)? Explain.
c. Find the rise in temperature, +DT, in Fahrenheit degrees (°F).
Solution: In each case, the stored gravitational energy is PE = m g h = (0.95 kg)(9.8 m/s2)(0.48 m) = 4.4688 J. Since there are two of them, then the PE = 2(4.4688) = 8.9376 J.
a. If all the PE were converted to heat (which it is not), then 8.9376 J of heat went into the water. And if 6,200 J are needed for +1.0° C, then the rise is +[(8.9376) / (6200)](1.0° C) = + 0.00144° C which is negligible.
b. Greater.
c. Using the “famous” relationship: we find that each degree rise in Celsius is the same as each 1.8° rise in Fahrenheit. So, +DT(Fahrenheit) = (1.8)(0.00144) = + 0.002595° F.

31. Two objects are made of the same material, but have different temperatures, T1 and T2. Object 1 has a mass of “m” while object two has a mass of “2m.” If the objects are brought into thermal contact,
a. Is the temperature change of object 1, DT1, greater, lesser, or equal to the temperature change of object 2, DT2?
b. Choose the best explanation from among the following three choices:
I The larger object gives up more heat and therefore its temperature change is greatest
II The heat given up by one object is taken up by the other. Since the objects have the same heat capacity the temperature changes are the same
III One object loses heat of magnitude Q the other gains heat of magnitude Q. With the same magnitude of heat involved the smaller object has the greater temperature change.
Solution: DQ = m C DT, this is the change of heat energy relative to Temperature. In this case, Q is heat energy; m is the mass; C is the specific heat capacity, and you know T.
a. The smaller one.
b. III

39. If 2200 Joules of heat are added to a 190-gram object, its temperature increases by +DT = 12° C.
a. Find the heat capacity of the object.
b. Find the specific heat of the object.
Solution: We already know that DQ = m C DT, so...
a. DQ = m C DT, and we have DQ, m, and DT. We need only to find C: Let's re-write to isolate C:

C = (DQ) / (m)(DT) = (2200 J) / [(0.190 kg)(12)] = (2200 J) / (2.28) ~ 965 Joules / kg-deg = 9.65 x 102 kg-deg.

b. The same, 9.65 x 102 kg-deg..

END

HOMEWORK SET 4

Chapter 15: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 530-534.

1. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.

3. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.

9. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.

13. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?

19. A circular wine barrel is shown in figure 15-26 on page 530. The barrel has a diameter of 75 cm. If a net upward force exceeds 643 Newtons it will burst. A tube of 1.0 cm in diameter extends into the barrel through a hole in the top. Initially the barrel is filled to the top and the tube is empty above that level. What weight of water must be poured into the tube in order to burst the barrel? And why would anyone want to do that anyway?

25. Referring to example 15-4 on page 507 in the book, suppose that some vegetable oil has been added to both sides of the U-tube. Not YouTube. On the right side of the tube the depth of oil is 5.0 cm as before. On the left side of the tube the depth is 3.0 cm. Find the difference in fluid level between the two sides of the tube.

29. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.

31. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
39. A hydrometer is a device for measuring fluid density. It's constructed as shown in Figure 15-30 on page 532. If the hydrometer samples fluid 1, the small float inside the tube is submerged to level 1. When fluid 2 is sampled the float is submerged to level 2. Is the density of fluid1 greater, less, or the same as fluid 2's density? [This is how auto mechanics test the antifreeze level in your car engine. Since antifreeze (ethylene glycol) is denser than water, the higher the density of coolant in your radiator the more antifreeze protection the engine has.]

Chapter 16: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, on pp 566-567.

1. The official record for lowest temperature ever recorded on Earth was set at Vostok, Antarctica, on July 21, 1983. The temperature on that day fell to T = ( – 89.2° C), which is well below the temperature of dry ice (frozen CO2). Give this temperature in degrees Fahrenheit.

3.Normal body temperature for humans is T = 98.6° F. Convert this temperature to degrees:
a. Celsius (aka Centrigrade)
b. Kelvin

9. A world record for the greatest change in temperature was set in Spearfish, North Dakota on January 22, 1943. At 7:30 AM, the temperature was T = ( - 4° F); two minutes later, at 7:32 AM, the temperature was T =+ 45° F. Calculate the mean rate of temperature change, per second, using Celsius or Kelvin scales (they use the same units); Hint: find DT.

13. Referring to Table 16 - 1 on page 545, which would be more accurate for an all-season, outdoor use: a tape measure made of steel or one made of aluminum?

19. It is desired to slip an aluminum ring over a steel bar as shown in Figure 16-14 on page 567. At T = 10.00° C, the inside diameter of the ring is d = 4.000 cm. The bar maintains a constant temperature of T = 10.0° C.
a. In order for the ring to slip over the bar, should the ring be heated, or cooled? Explain.
b. Find the temperature, T, of the ring at which it fits over the over the bar.
c. Who is the Lord of the Rings?

25. An aluminum saucepan with a diameter of d = 25.0 cm and a height of h = 6.0 cm, is filled “to the brim” with H2O. The initial temperature of the pan and water is T = 19° C. But then the pan with the water in it is placed on a range where it is heated to T = 88° C.
a. Will the water overflow from the pan, or will it decrease? Explain.
b. Calculate the volume of water that either overflows or decreases.
c. Go make some popcorn. Turn off the stove.

29. Consider the apparatus the Joule used in his experiments on the mechanical equivalent of heat, shown in Figure 16-8 on page 550. Suppose both blocks have a mass of m1 = m2 = 0.95 kg and that they fall through a distance of h = 0.48 meter.
a. Find the expected rise in temperature, DT, of the water, given that 6200 Joules are needed for every increase of DT = 1.0° C. Give your answer in degrees Celsius (°C).
b. Would the temperature rise in degrees Fahrenheit (°F) be greater than, or less than, the result in part (a)? Explain.
c. Find the rise in temperature, +DT, in Fahrenheit degrees (°F).

31. Two objects are made of the same material, but have different temperatures, T1 and T2. Object 1 has a mass of “m” while object two has a mass of “2m.” If the objects are brought into thermal contact,
a. Is the temperature change of object 1, DT1, greater, lesser, or equal to the temperature change of object 2, DT2?
b. Choose the best explanation from among the following three choices:
I The larger object gives up more heat and therefore its temperature change is greatest
II The heat given up by one object is taken up by the other. Since the objects have the same heat capacity the temperature changes are the same
III One object loses heat of magnitude Q the other gains heat of magnitude Q. With the same magnitude of heat involved the smaller object has the greater temperature change.

39. If 2200 Joules of heat are added to a 190-gram object, its temperature increases by +DT = 12° C.
a. Find the heat capacity of the object.
b. Find the specific heat of the object.

END

LESSON 9

PHYS 1100 LESSON 9 FOR
THURSDAY, AUGUST 19, 2010

I Introduction

II Logistics – Running Grades

III Turn in Assignments Due: Homework Sets 1, 2, 3. Essays 1, 2, 3, etc. and Return of Papers

IV Test 3

V Mind Game 3

VI Review of Lesson 8: Inertia, Torque, Gravity
A. Inertia is kg m2.
B. Torque is N-m
C. Gravity:

1. Kepler – P2 (m + M) = [(4 p2)/G] r3.

This is Kepler’s 3rd law dealing with the period of a planet, P, with a mass, m, traveling around the Sun, with a mass, M, at a distance between the Sun and the planet equaling “r”. However, it applies to moons, asteroids, stars, satellites, blah blah.

Period is in seconds, mass in kilograms, distance in meters (mks). The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

2. Newton – F = GmM/r2.

This part of Newton’s 2nd Law, which deals with the force, F, of gravity.

Little ‘m’ is the mass of a smaller object, such as human; big “M” is the mass of the larger object, such as planet Earth; r is the distance between the center of the first object (like a human’s navel) and the center of the second object (like the distance from the human to Earth’s core, i.e., the radius of Earth.

As a force, instead of a period, this law applies both to moving objects, like planets, moons, stars; and to object that are not moving, like people standing on Earth, or between two bowling balls in a bowling alley.

The Universal Gravitational Constant is G = 6.67 x 10-11 N m2 / kg2.

VII Lesson 9: Waves and Fluids
A. Waves
1. Length
2. Frequency
3. Period
4. Amplitude
5. Speed
6. Types
7. Other

B. Fluid Dynamics
1. Definition of “fluid”
2. And, defn of “dynamics”
3. Density
4. Pressure
5. Bernoulli’s Law
6. Archimede’s Principle
7. And more….

VIII Laboratory Exercise 9: Kepler’s Laws


IX HWK Assignment 4: Read Chapters 15-16 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 530-534 and do Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65, and 67 on pp 566-569; Due 8/26

X Essay 4: Albert Einstein, due 8/26

LAB 9 for Thursday, August 19

PhysicsLab9 Thursday, August 19, 2010 Name __________________
Dr Dave Menke, Instructor

I Title: Applying Kepler's Laws to Planetary Orbits

II Purpose: To determine the distance, orbital period, and velocity of the planets given, using Kepler's laws of Motion. To determine radio communication wait time in space.

Background information:
The German genius physicist, Herr Johannes Kepler (1571-1630), determined his three laws of motion in separate research, and they were published separately. However, one of his laws is P2 = k a3, where P is the period of revolution of the object; "k" is a constant, and the letter "a" represents the semi-major axis of the planet's elliptical orbit (kind of an average). If the orbit were a circle, then "a" would be the radius of that circle.

The constant, k, is equal to the number 1.00, but ONLY if the period, P, is in Earth Years; AND if the semi-major axis, "a," is given in astronomical units (AU). Otherwise, the constant
k = [4p2 /G]/(M + m), where "G" is the constant of universal gravitation, [G = 6.673 x 10-11 N-m2 /kg2 .] The letter "M" refers to the mass of the heavier object (Sun) and the letter "m" refers to the mass of the lighter object (planet) in kilograms. The units of period, P, would be in seconds. The letter "N" is a unit of forced called a "Newton."

One astronomical unit, AU, equals 150,000,000 kilometers (approximately). The speed of light, "c," is equal to 300,000 km/sec (approximately).

III Equipment -- Each Student Will Have:
1. pen, pencil, or quill;
2. calculator, abacus, or mainframe computer;
3. personal hard drive (brain);
4. personal printer (hand at end of arm);
5. graph paper
6. etc.

IV Procedure
1. Given the value of an astronomical unit, and both the perihelion and aphelion distances for several planets (in the DATA TABLE below), find for each planet given:
a. the semi-major axis (a) of its elliptical orbit, in AU’s;
b. the period of revolution about the Sun, P, in years;
c. the average velocity of all the planets listed, around the Sun (in this case,
assume all have circular orbits).

2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that
instant, if you are given the distance from Earth to Ganymede, find out how long it will take for a radio signal to reach Earth. (The distance to Ganymede will be similar, but not exact, to the distance to Jupiter. Consult the Data section below).

MORE…

Physics Lab 9, page 2

3. With the same conditions, calculate the typical response time from an
astronaut on Ganymede to Houston's Johnson Space Flight Center. (Round trip wait time).

V Data & Calculations
Below is a table of the distances that the indicated planets are from the Sun, at their closest point, perihelion, and at their most distant point, aphelion. The units are in Astronomical Units, also known as A.U.'s, where 1.0 A.U. = 149,500,000 kilometers = 1.49 x 1011 meters, which is the average distance that Earth is from the Sun.


Planet Perihelion (AU) Aphelion (AU) SemiMajor, a (AU) Period, P (years)
Mercury 0.31 0.47
Venus 0.72 0.73
Earth 0.98 1.02
Mars 1.39 1.66
Jupiter 4.95 5.46


[Perihelion + Aphelion]/2 = a = semi-major axis

2pr = circumference

velocity = 2pr/P = circumference/Period

VI Results
The purpose was / was not achieved because

VII Error Analysis
A. Quantitative Error
Find the Average Percent Error. Do this by first finding the percent error for each of the five planets, i.e., find the percent error for Mercury, then for Venus, etc., and then add all five percent errors and divide by 5 to get Average Percent Error.

Percent error = [|True – Yours| / True ] x 100% =

What is the “truth” as far as periods go? Look it up on Wikipedia.org or some other source, like an astronomy book. Or, ask your instructor

B. Qualitative Error
Sources of error are enumerated as:
Personal –
Systematic –
Random --

VIII Questions
1. Find the average orbital velocity of each of the above planets, in km/sec. To do this, you must assume the orbits are circles in the first order approximation.

2. Pretend that we have traveled to Ganymede (a large moon of Jupiter). At that instant, if the distance from Earth to Ganymede is 5.38 AU, find out how long it will take for a radio signal to reach Earth.

3. With the same conditions, calculate the typical response time from an astronaut on Ganymede to Houston’s Johnson Space Flight Center. (Round trip wait time).

4. Use the Copernican equation (below) to determine the Sidereal (true) Period of revolution, P, from the, Synodic (observed) period, S, for Jupiter and for Saturn. The Synodic Period of Jupiter and Saturn are 13 months and 12.5 months respectively.

1/P = 1- (1/S)

SOLUTION SET 3 - COMPLETE

PHYSICS 1100
SOLUTION SET 3
Problems and Conceptual Exercises


Chapter 10: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, and 65 on pp 323-327

1.Convert the following degree angles to radians:
a. 30°
b. 45°
c. 90°
d. 180°
Solution:
Since, by definition, 2 p radians = 360° , then half 360° = 180° would be p, and half that, 90°, would be p/2, and 45° is half of 90° so, 45° = p/4; and 30° is 1/3 of 90°, so 30° would be 1/3 of p/2 = p/6.
a. p/6
b. p/4
c. p/2
d. p

3. Find the angular speed (aka angular velocity, or, w) of
a. the minute hand of Big Ben in London
b. the hour hand of same.
Solution:
a. the minute hand of Big Ben, or any other clock, makes 360° or 2 p radians every hour, or, every 3,600 seconds. So, its angular velocity is w = 2 p radians / 3600 seconds = (2)(3.14)/3600 =
1.74 x 10-3 radians/sec.

9. The Crab Nebula has a pulsar (pulsating neutron star) embedded in it, which rotates every 33 ms. Find the angular velocity, w, of this pulsar in radians/sec.
Solution:
The period of rotation, P = 33 ms = 3.3 x 10-5 sec; frequency is n = 1/P = 1/(3.3 x 10-5 sec) = 3.03 x 104 Hz.; and angular velocity is w = 2 p n = (2)(3.14)(3.03 x 104) = 1.9 x 105 rad/s.

13. An object at rest begins to rotate with a constant angular acceleration, a. After a time, t, the object has an angular velocity of w. What was the angular velocity at ½ t?
Solution:
w = a t, so at half the time, w = ½ a t.

19. A ceiling fan is rotating at 0.96 revolutions per second. When turned off, it slows uniformly to a stop in 2.4 minutes.
a. how many revolutions does it do in this 2.4 minutes?
b. Find the number of revolutions the fan must make for its speed to decrease from 0.96 rev/s to 0.48 rev/sec.
Solution:
Initially it has a radial velocity of w i = (0.96)(2p ) radians in 1.0 sec = (0.96)(6.28)/(1.0 sec) = 6.0288 rad/sec. Then, 2.4 minutes (144 sec) later, the fan has stopped, and its radial velocity is w f = 0.0 rad/s. The angular acceleration here (actually, a deceleration) is a = (w f - w i)/Dt = (0.0 - 6.0288)/(144 sec) = - 4.19 x 10-2 rad/s2. It's negative, or a deceleration.

We can now use the angle equation, for any angle, Ɵ, this is true:

Θ = Θ0 + w0 t + ½ a t2 = (in radians)
a. How many revolutions (each revolution is 2 p radians) in 2.4 min (144 sec)? Assuming the initial angle is zero radians, i.e., Θ0 = 0.0 radians, then:

Θ = 0 + (6.0288)(144s) + (0.5)(- 4.19 x 10-2)(144)2 = 0 + 868.1472 - 434.4192 = 433.728 radians. Dividing this by 2 p radians will give us the number of revolutions:

433.728 radians / 2 p radians per rev = 69.06 revs.

b. We can now use the angular velocity equation, this is true:

(w f 2 - w I2 ) = 2 a Θ , or if we re-write this, we will have:

Θ = (w f 2 - w I2 )/(2 a) = [(0.48)2 – (0.96)2] /(2)(- 4.19 x 10-2) = [(0.2304) – (0.9216)] / (- 8.38 x 10-2) = (- 0.6912)/(-8.38 x 10-2) = + 0.0825 x 102
= 8.25 radians, or 1.31 revolutions.


25. When a carpenter shuts off his circular saw, the 10.0-inch (25.4 cm) diameter blade slows from 4440 rpm to 0.00 rpm in 2.50 seconds.
a. Find the angular acceleration, a, of the blade.
b. What's the distance traveled by a point on the rim of the blade during the deceleration?
c. What is the magnitude of the net displacement of a point on the rim of the blade during the deceleration?
Solution:
a. The angular acceleration is equal to the final angular velocity minus the initial angular velocity, all divided by the amount of time it takes to stop, or, a = (w f - w i)/Dt . We know that the final angular velocity, w f = 0.00 rad/s because it is stopped. But what is the initial radial velocity, w i ? Well, 1.0 revolution per minute, or, 1 rpm = 2 p radians per minute, or, 6.28 radians per minute. And that is the same as 6.28/60 seconds = 1.05 x 10-1 radian per second. Therefore, 4,440 rpm would equal (4440)(1.05 x 10-1) = 464.72 rad/s. And that is the initial angular velocity, w i.
We go back to the relationship, a = (w f - w i)/Dt, to find: a = (0.0 rad/s – 464.72 rad/sec)/(2.50 sec) = - 185.888 rad/s2, or ( - 1.86 x 102 rad/s2). It is negative, because it is slowing down, or, decelerating.

29. When standing at the top of a tall building,
a. Is your angular velocity, w, greater, less, or the same as if you were at sea level?
b. What is the best explanation?
I. The angular velocity is the same no matter one's height
II. On the top of the building, one is further from the axis of rotation
III. You spin faster closer to Earth's center.
Solution:
a. Since the top of the tall building has to make one complete circle of 360° in 24 hours, it has to have the same radial velocity as the ground.
b. #I

31. Two kids ride on the merry-go-round shown in Conceptual Check point 10-1 on page 306 in the book. Sally is 2.0 meters from the axis of rotation while Sue is 1.5 meters from same. If the merry-go-round completes one rev per 4.5 s,
a. Find the angular velocity, w, of each kid; v = w r
b. Find the linear velocity, v, of each kid
Solution:
a. Sally and Sue have the same angular velocity, where w = 2 p n, where n = 1/P, and P = 4.5 s, or, = (2)(3.14) /(4.5) = 1.4 rad/s.

39. A Ferris wheel with a radius of 9.5 m rotates at a constant rate, completing one revolution every 36 seconds. Find the direction and magnitude of a passenger's acceleration...
a. at the top
b. at the bottom
Solution:
First of all, the acceleration is radially, towards the center of the Ferris wheel at all times, but it would be greater at the top since you add the acceleration of gravity to it, and less at the bottom as you subtract the acceleration of gravity. And, of course, a = v2/r. We know r = 9.5 meters. But to find v, we invoke the relationship v = w r. If a = v2/r, then we replace v with w r and we have: a = (w r)2/r = w2 r. But w = 2 p n = 2 p / P = (2)(3.14)/(36s) = 1.74 rad/s. And thus, w2 r = (1.74)2(9.5) = 28.9 m/s2.
a. So, at the top, the acceleration is 28.9 + 9.8 = 38.7 m/s2 radially.
b. At the bottom, the acceleration is 28.9 – 9.8 = 19.1 m/s2 radially.

53. Tons of interplanetary dust and debris fall into Earth's atmosphere every day. What is the effect on Earth's Moment of Inertia?
Solution:
The moment of inertia of a homogenous sphere is I = 2/5 M R2. A ton of debris is about 900 Kg, so, let's say 3000 kg fall each day. Earth is 6 x 1024 kg, and a few tons is 3 x 103 kg. It's like 3 x 10-18%, so the effect is minimal.

63. Find the rate at which the rotational K.E. of Earth is decreasing. The “rate” of energy means energy over time, or, Joules/sec = watts. Earth's Moment of Inertia is 0.331 MÅ RÅ2, where MÅ = 6 x 1024 kg and R¿ = 6.4 x 106 m. Its period of rotation increases 2.3 x 10-3 seconds every century. (The iconic symbol for “Earth” is Å).
Solution:
Rotational K.E. is half the moment of inertia times radial velocity squared, or, K.E. = ½ I w2. So, the Earth's K.E. is the following: K.E. = ½ (0.331 MÅ RÅ2)w2 = ½ (0.331 MÅ RÅ2)(2p/P)2. Unless the Earth's mass or radius changes a great deal (see problem 53), the only thing different in this equation in 100 years from now will be the period, P.

Now the period is Pi. In 100 years, the period will be Pf = Pi + 0.0023 sec. Thus, K.E.f – K.E.i
= [(½) (0.331 MÅ RÅ2)(2p)2][(1/Pf)2 – (1/Pi)2].
Currently, the Earth rotates with a period of Pi = 86164.09053s and in 100 years from now, it will be Pf = Pi + 0.0023 sec = 86164.09053s + 0.0023 = 86164.09283s.

So, (Pf)2 = (86164.09283s)2= 7,424,250,893, and 1/(Pf)2 = 1.346937239 x 10-10.

And, (Pi)2 = (86164.09053s)2= 7,424,250,497, and 1/(Pf)2 = 1.346937119 x 10-10.

Thus, [(1/Pf)2 – (1/Pi)2] = [(1.346937239) - (1.346937119)] x 10-10 = 0.00000012 x 10-10 = 1.2 x 10-17.

And therefore, K.E.f – K.E.i = [(½) (0.331 MÅ RÅ2)(2p)2](1.2 x 10-17).

But, what is [(½) (0.331 M¿ R¿2)(2p)2]? Well, [(½) (0.331 M¿ R¿2)(2p)2] = [(0.5)(0.331)(6 x 1024)(6.4 x 106)2(2p)2] = [(0.5)(0.331)(6 x 1024)(40.96 x 1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(0.5)(0.331)(6)(1024)(40.96)(1012)(39.4384)] = [(1604)(1036)] = 1.604 x 1033

And, thus, my dear friends, K.E.f – K.E.i = (1.604 x 1033)(1.2 x 10-17) = 1.925 x 1016Joules.

But we are looking for rate of loss, so we need to find Power, P = DK.E. / 1 century = (1.925 x 1016Joules)/(100 years). And how many seconds are in 100 years? There are about 86164.09053 s in a day, and 365.2422 days in a year, so, in 1 year = (86164.09053s/d)(365.2422 days d/y) = 31,470,761.99 seconds in one year, and 3,147,076,199 in 100 years.

Finally, the rate, or power is P = (1.925 x 1016Joules)/(3.147076199 x 109 sec) = 0.612908589 x 107 watts. = 6.129 x 106 watts, or 6,129 KiloWatts.

65. Consider the physical situation shown in “Conceptual Checkpoint 10-5” on page 318 in the book. Suppose this time a “sphere” is released from rest on the frictionless surface. When the ball comes to rest on the no slip surface, is its height greater, less or equal to the height from which it was released?
Solution:
Less, because it loses energy as it rolls up the friction surface.


Chapter 11: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 65 and 77 on pp 366-371

1. To tighten a spark plug, it's recommended that a torque of 15 N m be applied. If a mechanic tightens the spark plug with a wrench that is 25 cm long, what is the force needed?
Solution:
We know that torque, t, is equal to the force applied, F, multiplied by the length of the lever, or, length of the moment arm, or, length of the handle, or whatever you want to call it, r. the relationship looks like this:
t = r F

Here, we are asked to find the force, F. If we re-write the relationship above, we have:

F = t/r. Do we know the torque, t? Yes, it is given as 15 N m. Do we know the length of the moment arm, r? Yes, it is given as 25 cm = 0.25 m. So all we have to do is “plug and chug” to get the answer, and we don't have to search for clues to find anything else:

F = t/r = (15)/(0.25) = 60 N.
3. A bowling trophy of mass 1.61 kg is held at arm's length, a distance of 0.65 m from the shoulder joint. What torque does the trophy exert about the shoulder if the arm is
a. horizontal
b. 22.5° below the horizontal?
Solution:
In reality, torque is the length of the moment arm, r, multiplied by the force, F, then multiplied by the sine of the angle between them, Θ. Or, in other ways of putting it,

t = r F sin Θ. If it is a right angle, Θ = 90°, then the sin 90° = 1.0, and we don't worry about it.
a. t = r F = (0.65 m)(1.61)(9.8) = 10.2557 N-m, or, 10.3 N-m.
b. Since the angle is 22.5°, or (90° – 22.5°) from the axis, then the torque is t = r F sin Θ = 10.3 sin 67.5° = 9.52 N.

9. Suppose a torque rotates your body about one of three different axes of rotation: case A, an axis through your spine; case B, an axis through your hips; and case C, an axis through your ankles. Rank these three axes of rotation in increasing order of the angular acceleration produced by the torque. Indicate ties where appropriate.
Solution:
The least torque will be through the spine, the greatest through the ankles.

13. A person holds a ladder horizontally at its center. Treating the ladder as a uniform rod of length 3.15 meter and mass of 8.42 kg, find the torque the person must exert on the ladder to give it an angular acceleration of 0.302 radians/sec2.
Solution:
One of our relationships is t = I a, and to find torque, we need the moment of inertia, I, and the angular acceleration, a. Do we have them? Yes! The moment of inertia around a uniform rod of length 3.15 meter and mass of 8.42 kg, is I = 1/12 m L2, or, I = (0)(8.42)(3.15)2. = (0.083)(8.42)(9.9225) = 6.96 kg m2.

Now we can find the torque: t = I a = (6.96)(0.302) = 2.1 N m.


19. A fish takes the bait and pulls on the line with a force of 2.2 N. The fishing reel, which rotates without friction, is a cylinder of radius 0.055 m and mass of 0.99 kg.
a. What is the angular acceleration of the fishing reel?
b. How much line does the fish pull from the reel in ¼ sec.?
Solution:
a. We are looking for acceleration and can use: t = I a = r F, so a = r F/I. Since I = ½ MR2 for a cylinder, then = 2F/Mr = 4.4/(0.55) = 8 rad/sec2.
b. First we use Θ = Θ 0 + w0 t + ½ a t2 = (in radians), but initial angle and initial angular velocity are both 0.0. So, we rewrite it to be: Θ = ½ a t2 = (0.50)(8)(0.25)2 = 0.25 radian. Since distance traveled is s = r , then s = (0.055m)(0.25) = 0.01375 or about 0.014m.

25. Refer to the person holding the baseball in problem 5 on page 366. Suppose the biceps exert just enough upward force to help keep the system in static equilibrium.
a. is the force exerted by the biceps more, less or same as the combined weight of the forearm, hand and baseball? Explain.
b. Determine the force exerted by the biceps.
Solution:
a. No, but the torque is equal
b. 1000 N >Fbiceps > 1.0 N
lo
29. A remote control gadget of mass 0.0122 kg is 23.0 cm long and rests on a table, as shown in figure 11-28 on page 368. Part of the remote control unit, of length of L, is hanging over the table’s edge. In order to operate the power button on this remote control device requires a force of 0.365 N. How far can the remote control doohickey extend beyond the edge of the table and still not tip over when one presses its power button? Assume the mass of the remote it distributed uniformly, i.e., the density is homogenous and that the power button is 1.41 cm from the overhanging end of the remote.
Solution: We need to find the point between the left end of the remote, which is hanging over the edge at is at x0 = 0.0 cm, and the point to the right of that, at the edge of the table, x = L, which will be the point of axis (or, fulcrum, or center of mass) through which it rotates. The right end of the remote will be at x = 23.0 cm.

In order for this to be in equilibrium, all torques must add up to zero, i.e.,

St = 0. And remember torque is t = r F.

the torque to the left of that point is going to be t1 = ( ½ L) F1 + (L – 1.41)(F2) ccw, such that F1 = the force of gravity on that portion of the remote that is to the left of the fulcrum point. F2 = 0.365 N, the force needed to operate the button. We need to calculate the force of gravity, F1. By the way, ccw = counter clockwise. The distance that gravity acts through has to be halfway between x = 0.0 and x = L, so it's ½ L.

F1 = m1 g where the mass here, m1, is the portion of the remote to the left of the fulcrum. But how much is m1? The entire remote has a mass of M = 0.0122 kg, and is 23.0 cm long, of which L cm is to the left of the fulcrum. Thus, m1 = (L/23)0.0122 kg; and m2 = [(23 – L)/ (23)](0.0122 kg).

Therefore, F1 = m1 g = [(L/23)(0.0122)] g. and its portion of the torque will be ( ½ L) F1 = ( ½ L)([(L/23)(0.0122)] g). ccw.

The torque from the other force will be (L – 1.41)(F2) = (L – 1.41)(0.365 N).ccw. =

Thus, the net torque on the left will be

t1 = ( ½ L)([(L/23)(0.0122)] g) + (L – 1.41)(F2) = (L – 1.41)(0.365 N). ccw.

The torque from the remote to the right of the fulcrum will be t2 = (23.0 – L) F3, such that

F3 = m2 g. = ([( ½ )(23 – L)/ (23)](0.0122 kg) g. =

Thus, t2 = (23.0 – L)[([( ½ )(23 – L)/ (23)](0.0122 kg) g]. cw,of course. =

And for equilibrium, the torques would be equal (but opposite):

t1 = t2 , or,

( ½ L)([(L/23)(0.0122)] g) + (L – 1.41)(0.365 N) = (23.0 – L)[([( ½ )(23 – L)/ (23)](0.0122 kg) g],

Which, as one can obviously see, is quite easy to work out. Yeah right.

(L)([(L/23)(0.0122)] g) + 2(L – 1.41)(0.365 N) = (23.0 – L)[(23 – L)/ (23)](0.0122 kg) g],

(L)(L/23)(0.0122)(g) + 2L(0.365) - (1.41)(0.365 N) = [(23 – L)2/ (23)](0.0122 kg) g],

(L2)(0.0122)(g) + 2L(0.365)(23) - (1.41)(23)(0.365 N) = [(23 – L)2/ (0.0122 kg)(g)],

(L2)(0.0122)2(g) + 2L(0.365)(23)(0.0122 kg) - (0.0122 kg)(1.41)(23)(0.365 N) = [(23 – L)2 ](g)],

(L2)(0.0122)2(9.8) + 2L(0.365)(23)(0.0122) - (0.0122)(1.41)(23)(0.365 N) = [(23 – L)2 ](9.8)],

0.00146 (L2) + 0.205(L) – 0.144 = [232 – 46L + L2 ](9.8)]

0.00146 (L2) + 0.205(L) – 0.144 = (529 – 451L + 9.8L2 ); or

(9.8 – 0.00146)(L2) – 450.8 L + 528.86 = 0. For the sake of sanity, we re-write it as:

9.8(L2) – 451 L + 529 = 0; and divide by 9.8, again for simplicity:

(L2) – 46 L + 54 = 0.

quadratic:

L = [(46) +/- √(2116 – 216)] / 2 = [(46) +/- √(1900)] / 2 = [(46) +/- 43.6] / 2; gives two answers:

a. 2.4/2 = 1.2 cm, or
b. 89.6 / 2 = 44.8 cm.

Clearly, it has to be 1.2 cm. So the Remote can go just slightly less than that, or,

L < 1.2 cm.

There's gotta be a simpler way.

31.Repeat Example 11-4 (on page 342 in the book); this time with a uniform diving board that weighs 225 N. (In that Example, it says: A 5.0-meter diving board of negligible mass is supported by two pillars. One pillar is at the left end of the diving board, as shown in the picture here on page 342. The other is 1.50 meters away. Find the forces exerted by the pillars when a 90.0-kg diver stands at the far [right] end of the board).
Solution: First, in the Example 11-4 on page 342 in the book, F2 = 2940 N, and F1 = – 2060 N. Now that has changed with the addition of a board that has a weight of + 225 N. This is divided up as follows: DF2 + DF1 = 225 N where = +380 N and = - 150 N due to the geometry of the set up. That means there are now new values of F2 = +3320 N and F1 = -2210 N.

39. A uniform crate with a mass of 16.2 kilograms rests on the floor with a coefficient of friction of m = 0.571. The crate is a uniform cube with sides 1.21 meters in length.
a. what horizontal force applied to the top of the crate will start the crate to tip?
b. If the horizontal force is applied halfway to the top of the crate, it will begin to slip, before it tips. Explain.
Solution: The force of friction is: Ff = mN = m m g = (0.571)(16.2)(9.8) = 90.65 N.
a. So, if a force of F > 90.65 N is applied at the top, it will tip over.
b. Pushing at halfway results in a net torque of zero (0.0).

53. Calculate the angular momentum of the Earth around its own axis due to its daily motion. Assume a sphere with uniform density.

Solution: First, a sphere with uniform density has a moment of inertia of I = (2/5) M r2.
We know M = 6 x 1024 kg; we know r = 6.4 x 106 m; we know w = 2 p radians in 24 hours, or
w = (2 p) / (86,400 sec) = 7.2685 x 10-5 radians/sec. Will want to use this relationship: l = I w.

Then if l = I w = [(2/5) M r2 ] w then l = [(2/5)(6 x 1024 kg)(6.4 x 106 m)2 ] (7.2685 x 10-5 radians/sec) =

[(0.4)(6 x 1024 kg)(4.1 x 1013 m)] (7.2685 x 10-5 radians/sec) = [(0.4)(6)(4.1)(7.2685)][(1024)(1013)(10-5)]

= [7.15 x 101] [1032] = 7.15 x 1033 kg m2 /sec.

63. A puck (like, air puck; or hockey puck, etc.) on a horizontal frictionless surface is attached to a string that passes through a hole in the surface as shown in Figure 11-34 on page 370. As the puck rotates revolves around the hole (the author must be an idiot; the puck is revolving, not rotating, so I changed it), the string is pulled downward by a magical elf, bringing the puck closer to the hole.
a. During the process, does the puck’s linear speed increase, decrease, or remain constant?
b. During the process, does the puck’s angular speed increase, decrease, or remain constant?
c. During the process, does the puck’s angular momentum increase, decrease, or remain constant?
Solution:
a. increase
b. increase
c. decrease.

65.As an ice skater begins a spin, his angular speed is 3.17 rad/s. After pulling in his arms, his angular speed increases to 5.46 rad/s. Find the ratio of the skater’s final moment of inertia to his initial moment of inertia.
Solution: We know that l = I w and that angular momentum must be conserved, i.e., l i = l f.

So, l i = I i w i and also that l f. = I f w f . Plus, l i = l f.

So the ratio of the skater’s final moment of inertia to his initial moment of inertia will be:

(I f ) / (I i) = (w i ) / (w f ) = (3.17 rad/s) / (5.46 rad/s) = 0.58 = 5.8 x 10-1 .

77. How much work must be done to accelerate a baton from rest to an angular speed of 7.4 rad/s about its center? Consider the baton to be a uniform rod of length 0.53 m and a mass of 0.44 kg.
Solution: We know that K.E. = ½ I w2.
The initial KEi = 0.0 Joules, as nothing is moving. The final KEf = ½ I w f2. So the work required to do what would be DKE = ½ I w f2 since initial radial velocity is zero (at rest).

So, what is the moment of inertial for this object? From the chart, this is like a slender rod of length L, being rotated about its center. The moment of inertia for such an object is I = (1/12) M L2. =

I = (0.085)(0.44 kg)(0.53 m)2 = (0.083)(0.44 kg)(0.281 m2) = 0.0103 kg m2.

And w f = (7.4 rad/s)2 = 54.76 rad/s2. So, to find the final K.E., which is the work required:

Work = KEf = ½ I w f2. = (0.5)(0.0103 kg m2)(54.76 rad/s2) = 0.282 Joules = 2.82 x 10-1 Joule.


Chapter 12: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53 and 63 on pp 409-412

1. System A has masses of m & m candies. Separated by a distance, r. System B has masses of m and “2m” separated by a distance “2r.” System C has masses of 2m and 3m separated by a distance of 2r. And, finally, System D has masses of 4m and 5m separated by a distance of 3r. Rank these systems in order of increasing gravitational force. Indicate “ties” where appropriate.
Solution: System A has a Force divided by Gravitation constant of FA/G. In the same fashion, Systems B, C, and D will be FB/G, FC/G, and FD/G. The values of each of these will give us the answer, so let's start: Remember, the force of gravity is: F = G mM/r2.

System A: FA/G = mm/r2; = (m/r)2 ;

System B: FB/G = (m)(2m)/(2r)2; = (2 m2)/(4r2) = ½ (m/r)2; thus System B is half System A;

System C: FC/G = (2m)(3m)/(2r)2; = (5 m2)/(4r2) = (5/4)(m/r)2; System C is larger than A or B

System D: FD/G = (4m)(5m)/(3r)2; = (20 m2) /(9r2) = (20/9)(m/r)2; System D is larger than C;

So, D, C, A, B.

3.A 6.1-kg bowling ball and a 7.2-kg bowling sphere rest on a rack 0.75 m apart.
a. What is the force of gravity on each of the spheres by the other one?
b. At what distance is the force of gravity between the spheres equal to 2.0 x 10-9 N?
Solution: The force of gravity is: F = G mM/r2. = [(6.67 x 10-11)(6.1)(7.2)] / (0.75)2 =
(292.95)(10-11)/(0.5625) = (2.9295)(10-9)/(0.5625) = [(2.9295)/(0.5625)] (10-9) = 5.208 x 10-9 N ~
5.2 x 10-9 N.

9.When the Earth, Moon, and Sun are aligned in such a way as to make a right triangle, with the Moon at the right angle junction as shown in Figure 12-22 on page 409 in the book, the Moon is in a phase called “Third Quarter.” The arrangement in Figure 12-22 is as if from above Earth's North Pole. If you look carefully, you may see Santa Claus. Find the magnitude and direction of the net force exerted on the Moon. Give the direction relative to the line connecting the Moon and the Sun.
Solution: The force of gravity on the Moon from the Sun is: Fs = G m M / R2. Where the mass of the Moon is 'm.' The Sun's mass is M. The distance between the Sun and the Moon is R. The Sun's force is the + x direction.
The force of gravity on the Moon from the Earth is: Fm = G m M / r2. Where the mass of the Moon is 'm.' The Earth's mass is M. The distance between the Earth and the Moon is r. The Earth's force is the - y direction (the same as 270°
.
We thus know that the net force on the Moon will be in the 4th Quadrant. But in order to get the net force we must find the magnitude and the exact angle. We also must know the masses and distances, which are: M = 2 x 1027 kg for the Sun; M = 6 x 1024 kg for the Earth; and m = 7.22 x 1022 kg for the Moon. The distance from the Sun to the Moon (on average) is R = 1.5 x 1011 meters; and from Earth to the Moon (on average) is r = 3.85 x 108 meters. Remember that G = 6.67 x 10-11.

So, now we have to plug and chug to find each Force:

Fs = [(6.67 x 10-11)(7.22 x 1022)(2 x 1027)] / (1.5 x 1011)2 =

(6.67)(7.22)(2)(10-11)(1022)(1027) / (1.5)2 (1011)2 = (96.31)(1038) / (2.25)(1022) =

(42.8)(1038/1022) = 42.8 x 1016= 4.3 x 1017 N.. And now for Earth's force:

Fe = [(6.67 x 10-11)(7.22 x 1022)(6 x 1024)] / (3.85 x 108)2 =

(6.67)(7.22)(6)(10-11)(1022)(1024) / (3.85)2 (108)2 = (288.94)(1035) / (14.82)(1016) =

(14.82)(1035/1016) = 14.82 x 1019= 1.48 x 1020 N. Now let's find the Net Force:

Fnet = Fs + Fe= (4.3 x 1017 N @ 0°) + (1.48 x 1020 N @ 270°); now let's pause and think. A right triangle with one side Fs and another side of Fe will have hypotenuse of Fnet. A similar triangle will exist if we divide both forces by 1017 to get one side of 4.3 and the other side of 1.48 x 103 = 1480. So, the length squared of the hypotenuse of that similar triangle will be l 2 = (4.3)2 + (1480)2 = (18.49) + (2,190,400) = 2,190,418.49; so the square root of that is the hypotenuse, or l = 1480.006247, or, 1.48 x 103 .

Multiply that back by 1017 we get the net force to be Fnet = 1.48 x 1020 N. This gives us the conclusion that the force from the Sun is negligible compared to Earth's. And if we were to compute the vanishingly small angle with trigonometry, it would probably be 270° (or directly to the -y direction).


13. Suppose that three astronomical objects, 1, 2, & 3, are observed to lie in a straight line, and that the distance from 1 to 3 is D. Given that object 1 has four times the mass of object 3 and seven times the mass of object 2, find the distance between objects 1 and 2 for which the net force on object 2 is zero (0.0 N).
Solution: Sounds like fun.


1 2 3
* * *
7m m 7/4 m

|----x----|------(D – x) --|

|<-------------D---------->|

F12 = G(7m)(m)/(x)2

F23 = - G(7/4 m)(m)/(D – x)2

(this force starts negative coming from another direction

And F12 + (- F23 ) = 0.0 N. Or,

Fnet = G(7m)(m)/(x)2 - G(7/4 m)(m)/(D – x)2 = 0

(7 m2 ) / (x)2 - (7m2 /4) / (D – x)2 = 0

(7 ) / (x)2 - (7 /4) / (D – x)2 = 0

(7 ) / (x)2 = (7 /4) / (D – x)2

(7)(D – x)2 = (7 /4)(x)2

(D – x)2 = (1 /4)(x)2

4(D – x)2 = (x)2

4(D2 – 2xD + x2) = (x)2

(4D2 – 8xD + 4x2) = (x)2

3x2 – 8xD + 4D2 = 0

Using the quadratic equation, then ax2 + bx + c = 0; and then

x = [- b +/- √(b2 – 4ac)] / 2a, where a = 3, b = - 8D, c = 4D2.

So, x = [+ 8D +/- √(64D2 – 48D2)] / 10, = ( 8D +/- √(16D2) / 10 = ( 8D +/- 4D) / 10.

the two mathematical solutions are x = {1.2 D, 0.4D} Since in this framework x has to be shorter then D, then the answer is x = 0.4 D.

19. At a certain distance from the center of Earth, a 4.6-kg object has a weight of 2.2 N.
a. Find the distance.
b. If the object is released at this location and allowed to fall toward the Earth, what is its initial acceleration?
c. If the object is now moved twice as far from Earth by what factor does its weight change? Explain.
d. by what factor does its initial acceleration change? Explain.
Solution: The force of gravity is: F = G mM/r2. We have everything but the distance, r for the first part, so, re-write:
(a) r2 = G mM/F, so taking the square root gives us the distance: r = √(G mM/F) =
√[(6.67 x 10-11)(4.6)(6 x 1024)/(2.2)] = √[(184 x 1013)/(2.2)] = √[(83.678 x 1013)] =
√[(8.3678 x 1014)] = √[(8.3678)(1014)] = 2.892 x 107 m ~ 2.9 x 107 m from the Earth's center. How far above the surface would be another question.
(b) At 6.4 x 106 meters (surface) the acceleration is – 9.8 m/s2. At 2.9 x 107 m, it is (29/6.4) times (4.53) farther, and since it's (gravity) an inverse square law, the new acceleration would be
(-9.8)/(4.53)2 = - 0.477 m/s2..
(c)It's an inverse square law, so 2.0 times further would be ( ½ )2 = ¼ of what it was.
(d) It's ( ¼ )(- 0.477 m/s2) = - 0.119 m/s2


25. A satellite orbits the Earth in a circular orbit of radius r. At some point its rocket engine his fired in such a way that its speed increases rapidly by a small amount. As a result,
a. does the apogee distance increase, decrease, or stay the same?
b. does the perigee distance increase, decrease, or stay the same?
Solution: Since v = w r, then radius increases, and it moves further out, but eventually comes back. The apogee gets longer, the perigee gets shorter.

29. Find the orbital speed of a satellite in a geosynchronous circular orbit 3.58 x 107 m above Earth's surface.
Solution: Using Kepler's 3rd law, P2 (m + M) = (4p2 / G) r3. The third law has the term (m + M) which is the mass of the very light satellite, m, and Earth, M. So we can ignore little 'm'. Now the formula becomes:P2 = [(4p2 / GM) r3.]

The speed or velocity will be the circumference, c, of its orbit divided by the period, or v = c/P. For convenience, let's find v2 first, or, v2 = (c/P)2 and (c/P)2 = (2 p r)2 / [(4p2 / GM) r3.] =
(2 p r)2 / [(2pr)2 / GM) r.] = GM / (r) = [(6.67 x 10-11)(6 x 1024)] / [(4.22 x 107)] =
(40 x 1013) / [(4.22 x 107)] = 9.5 x 106 . and this is v2, so v is the square root of that number, or

v = √(9.5 x 106 ) = 3.1 x 103 m/s.

31.Phobos, a moon of Mars, orbits at a distance of 9378 km from the center of Mars. What's its orbital period?
Solution: Using Kepler's 3rd law, P2 (m + M) = (4p2 / G) r3. The third law has the term (m + M) which is the mass of the very light satellite, m, and Mars, M. So we can ignore little 'm'. Now the formula becomes:P2 = [(4p2 / GM) r3.] = [(40) / (6.67 x 10-11)(7 x 1023)] (9.378 x 106 m)3.] =
[(40)(9.378 x 106 m)3] / [(6.67 x 10-11)(7 x 1023)] = {[(40)(8.25 x 1020)] / (47 x 1012)} =

{[(330 x 1020)] / (47 x 1012)} = 7.0 x 108. And this is P2, so the period is the square root of this. P = √(7.0 x 108.) = 2.65 x 104 sec (this is about 7.3 hours).

39. Find the speed of the binary stars Centauri A and Centauri B. They are separated by a distance of 3.45 x 1012 m and they have an orbital period of 2.52 x 109 seconds (that's about 80 years). They have the same mass. (Binary stars are near each other and orbit each other).
Solution: Using Kepler's 3rd law, P2 (M + M) = (4p2 / G) r3. The third law has the term (M + M) which is the mass of each of the binary stars. So we can now the formula becomes:P2 = [(4p2 / G2M) r3.]

The speed or velocity will be the circumference, c, of its orbit divided by the period, or v = c/P =
(2 p r / P) = [(2)(3.14)(3.45 x 1012)] / (2.52 x 109)] = [(21.666)(1012)] / [(2.52)(109)] = 8.6 x 103 m/s.


53. Find the Escape Velocity, vesc, for the planet Mercury. Do the same for the planet Venus. (How fast must they go to escape the Sun's gravity?)
Solution: Let's use the formula for escape velocity: vesc = √(2 G M/ r)..
So, for Mercury, M = 3.3 x 1023 kg and r = 2.4 x 103 m.

Thus, vesc = √[(2)(6.67 x 10-11)(3.3 x 1023) / (2.4 x 103)].= √[(1.834 x 1010) ]..= 1.35 x 105 m/s.

For Venus, M = 4.8 x 1024 kg; and r = 6 x 103 m.; so

Thus, vesc = √[(2)(6.67 x 10-11)(4.8 x 1024) / (6 x 103)].= √[(10.67 x 1010) ]..= 3.27 x 105 m/s.

For the Sun, M = 2 x 1030 kg; and r = 1.44 x 109 m.; so

Thus, vesc = √[(2)(6.67 x 10-11)(2 x 1030) / (1.44 x 109)].= √[(18.53 x 1010) ]..= 4.3 x 105 m/s.


63. A “dumbbell” has a weight, m, on either end of a rod of length 2a. The center of the dumbbell is at a distance, r, from Earth's center. The dumbbell is aligned radially (pointing towards Earth's center and away from it, i.e., vertical. If r >>a, show that the difference in the gravitation force exerted on the two masses by Earth is approximately equal to 4GmMa/r3. [The difference in the gravitational pull between the two weights creates a stress tension in the bar known as the tidal force. For r >>a, use 1/(r - a)2 – 1/(r + a) 2 – 4a/r3. )
Solution: This is an interesting problem, but let's skip it.


Chapter 13: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 445-447.

1. A small cart on a 5.0-m air-track moves at a speed of 0.85 m/s. (This is like the air hockey table that has no friction while two players keep hitting a round puck back and forth; however, an air track is one-dimensional, and instead of a puck, there is a small “cart.”). Bumpers at either end of the track cause the cart to reverse direction and maintain the same speed. Find the period and frequency of this motion.
Solution: First, some unknown source caused this cart to move at a constant speed. There is no acceleration. Second, there is no friction. Third, the collisions or impacts at the bumpers are 100% elastic so no energy is lost to heat, mechanical deformation, sound, or whatnot. In other words, this is fantasy land. But let's go with the flow. How much time does it take for the cart to travel from one end to the other end, and back (one cycle)? Since v = x/t, and we are looking for “t,” then t = x/v. The length, x = 5.0 meters, so the entire cycle is “2x” since it takes time to travel one length, and the same to travel back. No matter, t = (5.0 meters) / (0.85 m/s) = 5.88 seconds. The entire period would be double that, or P = 11.76 seconds. The frequency is the reciprocal of the period, so, n = (11.76 s)-1 = 0.085 cycles /sec = 8.5 x 10-2 Hz.

3. While fishing for a catfish, a fisherman suddenly notices that the bobber (a floating device on the fishing line) attached to his line is bobbing up and down with a frequency of 2.6 Hz. Find the period of the bobber's motion.
Solution: We know that the period is the reciprocal of the frequency. So, if n = 2.6 Hz, then P = (2.6)-1 = 0.3846 second = 3.8 x 10-1 s.

9. A mass moves back and forth with simple harmonic motion (SHM). The amplitude is A (this is how high the top of the crest of the wave is above the average). The period is P (this is how long it takes for one full cycle of the wave to go from one crest to another crest).
(a) How long does it take for the mass to move through a distance of x = 2A ?
(b) How long does I take for the mass to move through a total distance of 3A ?
Solution: Like the swinging pendulum that starts at maximum distance (45), the period of one complete cycle would be A + A + A + A = 4 A.
a. So to travel 2 A, it would be ½ P.
b. By the same reasoning, this would be (¾) P.

13. A mass on a spring oscillates with SHM with amplitude, A about the equilibrium point where x = 0. Its maximum speed is vmax and its maximum acceleration is amax.
(a). Find the speed of the mass at x = 0.
(b) Find the acceleration of the mass at x = A.
Solution: Like in the swinging pendulum, the acceleration will be greatest at the maximum angle of 45,
a. so amax is at x = A.
b. The bob reaches maximum speed (vmax) at the “bottom” of the swing, or at x = 0.

19. An object moves with SHM with a period T, and amplitude, A. During one complete cycle, for what length of time is the position of the object greater than A/2?
Solution: p = T for x = 4 A, so for (1/8) of 4 A = A/2, it will be 1/8 T.

25. A 30.0-gram gold finch (that's a bird, not the gold fish).lands on a slender tree branch where it oscillates up and down with SHM with A = 0.0335 m and period of T = 1.65 seconds.
(a) Find the maximum acceleration of the finch, amax. Divide this by Earth's acceleration, g = 9.8 m/s2.
(b) Find maximum speed of the gold finch, vmax.
(c) At the time the goldfinch experiences its maximum acceleration, amax, is the speed, v, maximum or minimum?
Solution: The average acceleration is defined as a = Dv / Dt, or the change of velocity over time. And the average velocity is v = Dx / Dt. So it travels x = 4 A in T, or, 4(0.0335 m) per 1.65 seconds. Thus,

v = 4(0.0335 m) / (1.65 seconds) = 4(0.0335 m) / (1.65 seconds) = (0.134 m) / (1.65 s); or,

v = 0.081 m/s. The maximum speed in the cycle is twice the average, or vmax = 0.162 m/s. Save this answer to use for (b).
a. amax = (vmax - 0.0) / ( ¼ )(1.65 s) = (0.162 m/s - 0.0 m/s) / ( ¼ )(1.65 s) =

(0.162 m/s - 0.0 m/s) / (0.4125 s) = (0.162 m/s) / (0.4125 s) = 0.393 m/s2, or 3.93 x 10-1 m/s2.

b. We found this already above: vmax = 0.162 m/s.
c. Minimum

29. The pistons in an internal combustion engine undergo a motion that approximates SHM. If A = 3.4 cm and the frequency is w = 1700 rpm, then find
(a) the maximum acceleration of the pistons, .amax ; and
(b) their maximum speed, vmax.
Solution: We know that v = w r, and if the maximum “r” is 3.4 cm, then that's easy. And we are given w = 1700 rpm = 28.3 rev/sec. But in this problem, the author uses the wrong symbol for frequency; because n = 28.3 rev/s, and w = 2 p n = 2 p (28.3) = 180 rad/s. And vmax = w rmax =

(180)(3.4 x 10- 2 m) = 6.05 m/s.

a. Since a = Dv / Dt = (6.05 – 0) / ( ¼ ) (P); however, P = (n)-1, or (P)-1 = n. So we can rewrite this as:

amax = (6.05 – 0)n / ( ¼ ) = (6.05 )(28.3) / ( ¼ ) = 4 (6.05 )(28.3) = 685 m/s2, or

amax = 6.85 x 102 m/s2.

b. We found this: vmax = w rmax = (180)(3.4 x 10- 2 m) = 6.05 m/s.

31. A bronco rider is on a mechanical bucking horse that oscillates up and down with SHM, with a period of T = 0.74 seconds with slowly increasing amplitude, A. At a certain amplitude, Ac, the rider must hang on to prevent being tossed off the horse.
(a) Explain how you'd calculate the amplitude, Ac.
(b) Carry out your method and find the amplitude, Ac.
Solution: We have been doing this for several problems, so
a. The same way we did the previous problems, looking for amax.
b. blah blah blah. S.O.S.

39. When a 0.5-kg mass is attached to a vertical spring, the spring stretches by y = - 15.0 cm. How much mass, in kilograms, must be attached to the vertical spring to stretch it so that it has a period of oscillation, T = 0.75 seconds?
Solution: Well, this is funner. We invoke Robert Hooke: F = - k y = m g. From this we can get the spring constant, k:

- k y = m g
k = - m g / y = - (0.5kg)(- 9.8m/s2) / (0.15 m) = 32.67 N/m. The spring constant is always positive (+) so if you get a negative, take the | ABSOLUTE VALUE | .

We also know that there is a relationship between the spring constant and the frequency:

P = 2 p √(m/k), so we re-write this to have “m” all alone:

P2 = (2 p)2 [(m) / (k)]

[(P) / (2 p)]2 = [(m) / (k)] so,

m = k [(P) / (2 p)]2 = (32.67 N/m)[(0.75) / (6.28)]2 = (32.67 N/m)(0.1194)2 =

(32.67 N/m)(0.0143) = 0.4659 kg.

Or, m = 4.7 x 10-1 kg.

Chapter 14: Problems 1, 3, 9, 13, 19, 25, 29, 31, 39, 53, 63, 67, 69, and 77 on pp 491-495

1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm.
a. Find the amplitude, A.
b. Find the wavelength, l.
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.

3. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.

9. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½

13. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.

19. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]

25. Four waves are described by the following equations (all of them are expressions for harmonic waves), in which all distances are in centimeters and all time is in seconds:

yA = 10 Cos(3x – 4t)
yB = 10 Cos(5x + 4t)
yC = 20 Cos(-10x + 60t)
yD = 20 Cos(-4x – 20t)

a. which wave(s) travel in the +x direction?
b. which wave(s) travel in the - x direction?
c. which wave has the highest frequency?
d. which wave has the longest wavelength?
e. which wave has the fastest speed?
Solution: We immediately realize that all four of these relationships are “expressions for a harmonic wave .” But let's do this first:
a. (+x direction when a neg x is in the equation)
b. (-x direction when a pos x is in the equation)

For the next part, we plug and chug to find the truth, and the truth shall set you free.
The expression for yA would be:

yA = 10 cos(2px/l – 2pt/P), but that has to equal the exact equation that we got for yA above: yA = 10 Cos(3x – 4t). But if yA = yA, then 10 cos(2px/l – 2pt/P) = 10 Cos(3x – 4t), or (2px/l – 2pt/P) = (3x – 4t). In the first one, the coefficient of the variable “x” is 2p/l, while in the second one, the coefficient is “3”. Thus, 2p/l = 3, or lA = 2p/3= 2.27.

Repeating these steps for yB, yC, and yD, we get the following wavelengths:

l B = 1.26,
l C = 0.628,
l D = 1.57,

Thus, the shortest wavelength, and the highest frequency, would have to be from the formula:
yC = 20 Cos(-10x + 60t).

d. yd

The fastest one would have the shortest period, and using the same method as to get the wavelength, we find:

e. for ya, use 2p/T = 4, so the period, T, is 2p/4 = 1.57 s
for yb, use 2p/T = 4, so the period, T, is 2p/4 = 1.57 s (same as ya)
for yc, use 2p/T = 4, so the period, T, is 2p/60 = 0.105 s
for yd, use 2p/T = 20, so the period, T, is 2p/20 = 0.314 s.

Shortest perioid: yC = 20 Cos(-10x + 60t).

29. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged?
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.

a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.


31. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).

Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:

vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:

vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?

What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.

We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi

We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:

t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.

We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.

We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.

t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s

Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s

and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:

vi = 15 m/s.

39. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.

END