12.7: Exercises
1. A ball is drawn randomly from a jar that contains 6 red balls, 2 white balls, and 5 yellow balls. Find the probability of the given event.
- A red ball is drawn
- A white ball is drawn
2. Suppose you write each letter of the alphabet on a different slip of paper and put the slips into a hat. What is the probability of drawing one slip of paper from the hat at random and getting:
- A consonant
- A vowel
3. A group of people were asked if they had run a red light in the last year. 150 responded "yes", and 185 responded "no". Find the probability that if a person is chosen at random, they have run a red light in the last year.
4. In a survey, 205 people indicated they prefer cats, 160 indicated they prefer dots, and 40 indicated they don’t enjoy either pet. Find the probability that if a person is chosen at random, they prefer cats.
5. Compute the probability of tossing a six-sided die (with sides numbered 1 through 6) and getting a 5.
6. Compute the probability of tossing a six-sided die and getting a 7.
7. Giving a test to a group of students, the grades and gender are summarized below. If one student was chosen at random, find the probability that the student was female.
\(\begin{array}{|l|l|l|l|l|}
\hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \text { Total } \\
\hline \text { Male } & 8 & 18 & 13 & 39 \\
\hline \text { Female } & 10 & 4 & 12 & 26 \\
\hline \text { Total } & 18 & 22 & 25 & 65 \\
\hline
\end{array}\)
8. The table below shows the number of credit cards owned by a group of individuals. If one person was chosen at random, find the probability that the person had no credit cards.
\(\begin{array}{|l|l|l|l|l|}
\hline & \text { Zero } & \text { One } & \text { Two or more } & \text { Total } \\
\hline \text { Male } & 9 & 5 & 19 & 33 \\
\hline \text { Female } & 18 & 10 & 20 & 48 \\
\hline \text { Total } & 27 & 15 & 39 & 81 \\
\hline
\end{array}\)
9. Compute the probability of tossing a six-sided die and getting an even number.
10. Compute the probability of tossing a six-sided die and getting a number less than 3.
11. If you pick one card at random from a standard deck of cards, what is the probability it will be a King?
12. If you pick one card at random from a standard deck of cards, what is the probability it will be a Diamond?
13. Compute the probability of rolling a 12-sided die and getting a number other than 8.
14. If you pick one card at random from a standard deck of cards, what is the probability it is not the Ace of Spades?
15. Referring to the grade table from question #7, what is the probability that a student chosen at random did NOT earn a C?
16. Referring to the credit card table from question #8, what is the probability that a person chosen at random has at least one credit card?
17.  What is the probability of showing a 6 on both rolls?
18. A fair coin is flipped twice. What is the probability of showing heads on both flips?
19.  What is the probability of showing a 5 on the first roll and an even number on the second roll?
20. Suppose that 21% of people own dogs. If you pick two people at random, what is the probability that they both own a dog?
21. Suppose a jar contains 17 red marbles and 32 blue marbles. If you reach in the jar and pull out 2 marbles at random, find the probability that both are red.
22. Suppose you write each letter of the alphabet on a different slip of paper and put the slips into a hat. If you pull out two slips at random, find the probability that both are vowels.
23. Bert and Ernie each have a well-shuffled standard deck of 52 cards. They each draw one card from their own deck. Compute the probability that:
- Bert and Ernie both draw an Ace.
- Bert draws an Ace but Ernie does not.
- neither Bert nor Ernie draws an Ace.
- Bert and Ernie both draw a heart.
- Bert gets a card that is not a Jack and Ernie draws a card that is not a heart.
24. Bert has a well-shuffled standard deck of 52 cards, from which he draws one card; Ernie has a 12-sided die, which he rolls at the same time Bert draws a card. Compute the probability that:
- Bert gets a Jack and Ernie rolls a five.
- Bert gets a heart and Ernie rolls a number less than six.
- Bert gets a face card (Jack, Queen or King) and Ernie rolls an even number.
- Bert gets a red card and Ernie rolls a fifteen.
- Bert gets a card that is not a Jack and Ernie rolls a number that is not twelve.
25. Compute the probability of drawing a King from a deck of cards and then drawing a Queen.
26. Compute the probability of drawing two spades from a deck of cards.
27. A math class consists of 25 students, 14 female and 11 male. Two students are selected at random to participate in a probability experiment. Compute the probability that
- a male is selected, then a female.
- a female is selected, then a male.
- two males are selected.
- two females are selected.
- no males are selected.
28. A math class consists of 25 students, 14 female and 11 male. Three students are selected at random to participate in a probability experiment. Compute the probability that
- a male is selected, then two females.
- a female is selected, then two males.
- two females are selected, then one male.
- three males are selected.
- three females are selected.
29. Giving a test to a group of students, the grades and gender are summarized below. If one student was chosen at random, find the probability that the student was female and earned an A.
\(\begin{array}{|l|l|l|l|l|}
\hline & \mathrm{A} & \mathrm{B} & \mathrm{C} & \text { Total } \\
\hline \text { Male } & 8 & 18 & 13 & 39 \\
\hline \text { Female } & 10 & 4 & 12 & 26 \\
\hline \text { Total } & 18 & 22 & 25 & 65 \\
\hline
\end{array}\)
30. The table below shows the number of credit cards owned by a group of individuals. If one person was chosen at random, find the probability that the person was male and had two or more credit cards.
\(\begin{array}{|l|l|l|l|l|}
\hline & \text { Zero } & \text { One } & \text { Two or more } & \text { Total } \\
\hline \text { Male } & 9 & 5 & 19 & 33 \\
\hline \text { Female } & 18 & 10 & 20 & 48 \\
\hline \text { Total } & 27 & 15 & 39 & 81 \\
\hline
\end{array}\)
31.A jar contains 6 red marbles numbered 1 to 6 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is red or odd-numbered.
32. A jar contains 4 red marbles numbered 1 to 4 and 10 blue marbles numbered 1 to 10. A marble is drawn at random from the jar. Find the probability the marble is blue or even-numbered.
33. Referring to the table from #29, find the probability that a student chosen at random is female or earned a B.
34. Referring to the table from #30, find the probability that a person chosen at random is male or has no credit cards.
35. Compute the probability of drawing the King of hearts or a Queen from a deck of cards.
36. Compute the probability of drawing a King or a heart from a deck of cards.
37. A jar contains 5 red marbles numbered 1 to 5 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is
- Even-numbered given that the marble is red.
- Red given that the marble is even-numbered.
38. A jar contains 4 red marbles numbered 1 to 4 and 8 blue marbles numbered 1 to 8. A marble is drawn at random from the jar. Find the probability the marble is
- Odd-numbered given that the marble is blue.
- Blue given that the marble is odd-numbered.
39. Compute the probability of flipping a coin and getting heads, given that the previous flip was tails.
40. Find the probability of rolling a “1” on a fair die, given that the last 3 rolls were all ones.
41. Suppose a math class contains 25 students, 14 females (three of whom speak French) and 11 males (two of whom speak French). Compute the probability that a randomly selected student speaks French, given that the student is female.
42. Suppose a math class contains 25 students, 14 females (three of whom speak French) and 11 males (two of whom speak French). Compute the probability that a randomly selected student is male, given that the student speaks French.
43. A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus.
- Find the probability that a person has the virus given that they have tested positive, i.e. find \(P(A | B)\).
- Find the probability that a person does not have the virus given that they test negative, i.e. find \(P(\text { not } A | \text { not } B)\).
44. A certain virus infects one in every 2000 people. A test used to detect the virus in a person is positive 96% of the time if the person has the virus and 4% of the time if the person does not have the virus.
- Find the probability that a person has the virus given that they have tested positive, i.e. find \(P(A | B)\).
- Find the probability that a person does not have the virus given that they test negative, i.e. find \(P(\text { not } A | \text { not } B)\).
45. A certain disease has an incidence rate of 0.3%. If the false negative rate is 6% and the false positive rate is 4%, compute the probability that a person who tests positive actually has the disease.
46. A certain disease has an incidence rate of 0.1%. If the false negative rate is 8% and the false positive rate is 3%, compute the probability that a person who tests positive actually has the disease.
47. A certain group of symptom-free women between the ages of 40 and 50 are randomly selected to participate in mammography screening. The incidence rate of breast cancer among such women is 0.8%. The false negative rate for the mammogram is 10%. The false positive rate is 7%. If a the mammogram results for a particular woman are positive (indicating that she has breast cancer), what is the probability that she actually has breast cancer?
48. About 0.01% of men with no known risk behavior are infected with HIV. The false negative rate for the standard HIV test 0.01% and the false positive rate is also 0.01%. If a randomly selected man with no known risk behavior tests positive for HIV, what is the probability that he is actually infected with HIV?
49. A boy owns 2 pairs of pants, 3 shirts, 8 ties, and 2 jackets. How many different outfits can he wear to school if he must wear one of each item?
50. At a restaurant you can choose from 3 appetizers, 8 entrees, and 2 desserts. How many different three-course meals can you have?
51. How many three-letter "words" can be made from 4 letters "FGHI" if
- repetition of letters is allowed
- repetition of letters is not allowed
52. How many four-letter "words" can be made from 6 letters "AEBWDP" if
- repetition of letters is allowed
- repetition of letters is not allowed
53. All of the license plates in a particular state feature three letters followed by three digits (e.g. ABC 123). How many different license plate numbers are available to the state's Department of Motor Vehicles?
54. A computer password must be eight characters long. How many passwords are possible if only the 26 letters of the alphabet are allowed?
55. A pianist plans to play 4 pieces at a recital. In how many ways can she arrange these pieces in the program?
56. In how many ways can first, second, and third prizes be awarded in a contest with 210 contestants?
57. Seven Olympic sprinters are eligible to compete in the 4 x 100 m relay race for the USA Olympic team. How many four-person relay teams can be selected from among the seven athletes?
58. A computer user has downloaded 25 songs using an online file-sharing program and wants to create a CD-R with ten songs to use in his portable CD player. If the order that the songs are placed on the CD-R is important to him, how many different CD-Rs could he make from the 25 songs available to him?
59. In western music, an octave is divided into 12 pitches. For the film Close Encounters of the Third Kind , director Steven Spielberg asked composer John Williams to write a five-note theme, which aliens would use to communicate with people on Earth. Disregarding rhythm and octave changes, how many five-note themes are possible if no note is repeated?
60. In the early twentieth century, proponents of the Second Viennese School of musical composition (including Arnold Schönberg, Anton Webern and Alban Berg) devised the twelve-tone technique, which utilized a tone row consisting of all 12 pitches from the chromatic scale in any order, but with not pitches repeated in the row.
61. In how many ways can 4 pizza toppings be chosen from 12 available toppings?
62. At a baby shower 17 guests are in attendance and 5 of them are randomly selected to receive a door prize. If all 5 prizes are identical, in how many ways can the prizes be awarded?
63. In the 6/50 lottery game, a player picks six numbers from 1 to 50. How many different choices does the player have if order doesn’t matter?
64. In a lottery daily game, a player picks three numbers from 0 to 9. How many different choices does the player have if order doesn’t matter?
65. A jury pool consists of 27 people. How many different ways can 11 people be chosen to serve on a jury and one additional person be chosen to serve as the jury foreman?
66. The United States Senate Committee on Commerce, Science, and Transportation consists of 23 members, 12 Republicans and 11 Democrats. The Surface Transportation and Merchant Marine Subcommittee consists of 8 Republicans and 7 Democrats. How many ways can members of the Subcommittee be chosen from the Committee?
67. You own 16 CDs. You want to randomly arrange 5 of them in a CD rack. What is the probability that the rack ends up in alphabetical order?
68. A jury pool consists of 27 people, 14 men and 13 women. Compute the probability that a randomly selected jury of 12 people is all male.
69. In a lottery game, a player picks six numbers from 1 to 48. If 5 of the 6 numbers match those drawn, they player wins second prize. What is the probability of winning this prize?
70. In a lottery game, a player picks six numbers from 1 to 48. If 4 of the 6 numbers match those drawn, they player wins third prize. What is the probability of winning this prize?
71. Compute the probability that a 5-card poker hand is dealt to you that contains all hearts.
72. Compute the probability that a 5-card poker hand is dealt to you that contains four Aces.
73. A bag contains 3 gold marbles, 6 silver marbles, and 28 black marbles. Someone offers to play this game: You randomly select on marble from the bag. If it is gold, you win $3. If it is silver, you win $2. If it is black, you lose $1.
74.   Compute the expected value for this game. Should you play this game?
75. In a lottery game, a player picks six numbers from 1 to 23. If the player matches all six numbers, they win 30,000 dollars. Otherwise, they lose $1. Find the expected value of this game.
76. A game is played by picking two cards from a deck. If they are the same value, then you win $5, otherwise you lose $1. What is the expected value of this game?
77. A company estimates that 0.7% of their products will fail after the original warranty period but within 2 years of the purchase, with a replacement cost of $350. If they offer a 2 year extended warranty for $48, what is the company's expected value of each warranty sold?
78. An insurance company estimates the probability of an earthquake in the next year to be 0.0013. The average damage done by an earthquake it estimates to be $60,000. If the company offers earthquake insurance for $100, what is their expected value of the policy?
Exploration
Some of these questions were adapted from puzzles at mindyourdecisions.com.
79. A small college has been accused of gender bias in its admissions to graduate programs.
- Out of 500 men who applied, 255 were accepted. Out of 700 women who applied, 240 were accepted. Find the acceptance rate for each gender. Does this suggest bias?
- The college then looked at each of the two departments with graduate programs, and found the data below. Compute the acceptance rate within each department by gender. Does this suggest bias?
- Looking at our results from Parts a and b , what can you conclude? Is there gender bias in this college’s admissions? If so, in which direction?
80. A bet on “black” in Roulette has a probability of 18/38 of winning. If you win, you double your money. You can bet anywhere from $1 to $100 on each spin.
- Suppose you have $10, and are going to play until you go broke or have $20. What is your best strategy for playing?
- Suppose you have $10, and are going to play until you go broke or have $30. What is your best strategy for playing?
81. Your friend proposes a game: You flip a coin. If it’s heads, you win $1. If it’s tails, you lose $1. However, you are worried the coin might not be fair coin. How could you change the game to make the game fair, without replacing the coin?
82. Fifty people are in a line. The first person in the line to have a birthday matching someone in front of them will win a prize. Of course, this means the first person in the line has no chance of winning. Which person has the highest likelihood of winning?
83. Three people put their names in a hat, then each draw a name, as part of a randomized gift exchange. What is the probability that no one draws their own name? What about with four people?
84. How many different “words” can be formed by using all the letters of each of the following words exactly once?
- “ALICE”
- “APPLE”
85. How many different “words” can be formed by using all the letters of each of the following words exactly once?
- “TRUMPS”
- “TEETER”
86. The Monty Hall problem is named for the host of the game show Let’s make a Deal . In this game, there would be three doors, behind one of which there was a prize. The contestant was asked to choose one of the doors. Monty Hall would then open one of the other doors to show there was no prize there. The contestant was then asked if they wanted to stay with their original door, or switch to the other unopened door. Is it better to stay or switch, or does it matter?
87. Suppose you have two coins, where one is a fair coin, and the other coin comes up heads 70% of the time. What is the probability you have the fair coin given each of the following outcomes from a series of flips?
- 5 Heads and 0 Tails
- 8 Heads and 3 Tails
- 10 Heads and 10 Tails
- 3 Heads and 8 Tails
88. Suppose you have six coins, where five are fair coins, and one coin comes up heads 80% of the time. What is the probability you have a fair coin given each of the following outcomes from a series of flips?
- 5 Heads and 0 Tails
- 8 Heads and 3 Tails
- 10 Heads and 10 Tails
- 3 Heads and 8 Tails
89. In this problem, we will explore probabilities from a series of events.
- If you flip 20 coins, how many would you expect to come up “heads”, on average? Would you expect every flip of 20 coins to come up with exactly that many heads?
- If you were to flip 20 coins, what would you consider a “usual” result? An “unusual” result?
- Flip 20 coins (or one coin 20 times) and record how many come up “heads”. Repeat this experiment 9 more times. Collect the data from the entire class.
- When flipping 20 coins, what is the theoretic probability of flipping 20 heads?
- Based on the class’s experimental data, what appears to be the probability of flipping 10 heads out of 20 coins?
- The formula \(_{n} C_{x} p^{x}(1-p)^{n-x}\) will compute the probability of an event with probability \(p\) occurring \(x\) times out of \(n\), such as flipping \(x\) heads out of \(n\) coins where the probability of heads is \(p = \frac{1}{2}\). Use this to compute the theoretic probability of flipping 10 heads out of 20 coins.
- If you were to flip 20 coins, based on the class’s experimental data, what range of values would you consider a “usual” result? What is the combined probability of these results? What would you consider an “unusual” result? What is the combined probability of these results?
- We’ll now consider a simplification of a case from the 1960s. In the area, about 26% of the jury eligible population was black. In the court case, there were 100 men on the juror panel, of which 8 were black. Does this provide evidence of racial bias in jury selection?


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12: Social and Emotional Development in Early Adulthood (Ages 18 to 29)
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12.0: What Does Psychology Say?
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The text describes a moment in the life of Isaac, a college graduate, who is filled with mixed emotions during his graduation ceremony. As the first in his family to graduate, he faces both pride and the anxiety of moving to the city for a job. Supported by family and friends, Isaac is determined and optimistic about his future. The chapter will explore how new graduates develop their sense of self, choose career paths, and navigate relationships and romance in new adult roles and contexts.
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12.1: The Development of Self in Early Adulthood
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This page provides an overview of developmental tasks in emerging and early adulthood, emphasizing the evolving sense of self and personality during this stage. Key themes include cultural influences on self-development, the concept of a social clock, and differences in developmental priorities across cultures. It also discusses how personality traits may shift slightly in response to life events.
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12.2: Identity Development in Context in Early Adulthood
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The page covers identity development in early adulthood through Marcia???s framework, intersectionality, and the growth of ethnoracial, gender and sexual, and religious identities. It highlights Logan Eggleston's journey to embracing her identity as a Black female student athlete, emphasizing the transitions in early adulthood impacting identity. The discussion extends to intersectionality, identity in different domains, and identity reconciliation amid new roles and contexts.
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12.3: Relationships with Friends and Family in Early Adulthood
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This page discusses the role of social contexts in early adulthood, focusing on friendships and family relations. It explores how these relationships are crucial for personal development and achieving independence. Self-determination theory highlights the importance of balancing autonomy, competence, and relatedness. The narrative of emerging adulthood today involves deferring traditional milestones in favor of personal goals.
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12.4: Contexts- School and Work Settings in Early Adulthood
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Young adults face varied education and work decisions post-high school, influenced by personal, familial, and societal contexts. College remains a common pathway, although many pursue vocational training, military service, or immediate employment. Factors like race, gender, and geography further shape these trajectories. Career exploration and work-life balance are crucial during early adulthood, with many seeking transferable skills and the right developmental opportunities.
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12.5: Finding Love, Intimacy, and Romance in Early Adulthood
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This page explores the development of romantic relationships in early adulthood, focusing on identity and intimacy as critical factors.


22.1: Rotational Variables
We have already talked about angular velocity in Chapter 6 , so here we will review it, provide a few more details that we will need, and then move on to the discuss the notion of angular acceleration.
Angular Velocity
Uniform circular motion (discussed previously in Circular Motion ) is motion in a circle at constant speed. Although this is the simplest case of rotational motion, it is very useful for many situations, and we use it here to introduce rotational variables.
In Figure \(\PageIndex{1}\), we show a particle moving in a circle. The coordinate system is fixed and serves as a frame of reference to define the particle’s position. Its position vector from the origin of the circle to the particle sweeps out the angle \(\theta\), which increases in the counterclockwise direction as the particle moves along its circular path. The angle \(\theta\) is called the angular position of the particle. As the particle moves in its circular path, it also traces an arc length s.
The angle is related to the radius of the circle and the arc length by
\[\theta = \frac{s}{r} \ldotp \label{10.1}\]
We can assign vectors to the quantities in Equation \ref{10.1}. The angle \(\vec{\theta}\) is a vector out of the page in Figure \(\PageIndex{1}\). The angular position vector \(\vec{r}\) and the arc length \(\vec{s}\) both lie in the plane of the page. These three vectors are related to each other by
\[\vec{s} = \vec{\theta} \times \vec{r} \ldotp \label{10.2}\]
That is, the arc length is the cross product of the angle vector and the position vector, as shown in Figure \(\PageIndex{2}\).
The magnitude of the angular velocity, denoted by \(\omega\), is the time rate of change of the angle \(\theta\) as the particle moves in its circular path. The instantaneous angular velocity is defined as the limit in which \(\Delta\)t → 0 in the average angular velocity \(\bar{\omega} = \frac{\Delta \theta}{\Delta t}\):
\[\omega = \lim_{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t} = \frac{d \theta}{dt}, \label{10.3}\]
where \(\theta\) is the angle of rotation (Figure \(\PageIndex{2}\)).
We can see how angular velocity is related to the tangential speed of the particle by differentiating Equation \ref{10.1} with respect to time. We rewrite Equation \ref{10.1} as
\[s = r \theta \ldotp\]
Taking the derivative with respect to time and noting that the radius r is a constant, we have
\[\frac{ds}{dt} = \frac{d}{dt} (r \theta) = \theta \frac{dr}{dt} + r \frac{d \theta}{dt} = r \frac{d \theta}{dt}\]
where \(\theta \frac{dr}{dt}\) = 0. Here, \(\frac{ds}{dt}\) is just the tangential speed v t of the particle in Figure \(\PageIndex{1}\). Thus, by using Equation \ref{10.3}, we arrive at an expression we first encounted in Section 6.1 :
\[v_{t} = r \omega \ldotp \label{10.4}\]
That is, the tangential speed of the particle is its angular velocity times the radius of the circle. From Equation \ref{10.4}, we see that the tangential speed of the particle increases with its distance from the axis of rotation for a constant angular velocity. This effect is shown in Figure \(\PageIndex{3}\). Two particles are placed at different radii on a rotating disk with a constant angular velocity. As the disk rotates, the tangential speed increases linearly with the radius from the axis of rotation. In Figure \(\PageIndex{3}\), we see that v 1 = r 1 \(\omega_{1}\) and v 2 = r 2 \(\omega_{2}\). But the disk has a constant angular velocity, so \(\omega_{1} = \omega_{2}\). This means \(\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}\) or v 2 = \(\left(\dfrac{r_{2}}{r_{1}}\right)\)v 1 . Thus, since r 2 > r 1 , v 2 > v 1 .
Up until now, we have discussed the magnitude of the angular velocity \(\omega = \frac{d \theta}{dt}\), which is a scalar quantity—the change in angular position with respect to time. The vector \(\vec{\omega}\) is the vector associated with the angular velocity and points along the axis of rotation. This is useful because when a rigid body is rotating, we want to know both the axis of rotation and the direction that the body is rotating about the axis, clockwise or counterclockwise. The angular velocity \(\vec{\omega}\) gives us this information. The angular velocity \(\vec{\omega}\) has a direction determined by what is called the right-hand rule. The right-hand rule is such that if the fingers of your right hand wrap counterclockwise from the x-axis (the direction in which \(\theta\) increases) toward the y-axis, your thumb points in the direction of the positive z-axis (Figure \(\PageIndex{4}\)). An angular velocity \(\vec{\omega}\) that points along the positive z-axis therefore corresponds to a counterclockwise rotation, whereas an angular velocity \(\vec{\omega}\) that points along the negative z-axis corresponds to a clockwise rotation.
We can verify the right-hand-rule using the vector expression for the arc length \(\vec{s} = \vec{\theta} \times \vec{r}\), Equation \ref{10.2}. If we differentiate this equation with respect to time, we find
\[\frac{d \vec{s}}{dt} = \frac{d}{dt}(\vec{\theta} \times \vec{r}) = \left(\dfrac{d \theta}{dt} \times \vec{r}\right) + \left(\vec{\theta} \times \dfrac{d \vec{r}}{dt}\right) = \frac{d \theta}{dt} \times \vec{r} \ldotp\]
Since \(\vec{r}\) is constant, the term \(\vec{\theta} \times \frac{d \vec{r}}{dt}\) = 0. Since \(\vec{v} = \frac{d \vec{s}}{dt}\) is the tangential velocity and \(\omega = \frac{d \vec{\theta}}{dt}\) is the angular velocity, we have
\[\vec{v} = \vec{\omega} \times \vec{r} \ldotp \label{10.5}\]
That is, the tangential velocity is the cross product of the angular velocity and the position vector, as shown in Figure \(\PageIndex{5}\). From part (a) of this figure, we see that with the angular velocity in the positive z-direction, the rotation in the xy-plane is counterclockwise. In part (b), the angular velocity is in the negative z-direction, giving a clockwise rotation in the xy-plane.
A flywheel rotates such that it sweeps out an angle at the rate of \(\theta\) = \(\omega\)t = (45.0 rad/s)t radians. The wheel rotates counterclockwise when viewed in the plane of the page. (a) What is the angular velocity of the flywheel? (b) What direction is the angular velocity? (c) How many radians does the flywheel rotate through in 30 s? (d) What is the tangential speed of a point on the flywheel 10 cm from the axis of rotation?
Strategy
The functional form of the angular position of the flywheel is given in the problem as \(\theta\)(t) = \(\omega\)t, so by taking the derivative with respect to time, we can find the angular velocity. We use the right-hand rule to find the angular velocity. To find the angular displacement of the flywheel during 30 s, we seek the angular displacement \(\Delta \theta\), where the change in angular position is between 0 and 30 s. To find the tangential speed of a point at a distance from the axis of rotation, we multiply its distance times the angular velocity of the flywheel.
Solution
- \(\omega\) = \(\frac{d \theta}{dt}\) = 45 rad/s. We see that the angular velocity is a constant.
- By the right-hand rule, we curl the fingers in the direction of rotation, which is counterclockwise in the plane of the page, and the thumb points in the direction of the angular velocity, which is out of the page.
- \(\Delta \theta\) = \(\theta\)(30 s) − \(\theta\)(0 s) = 45.0(30.0 s) − 45.0(0 s) = 1350.0 rad.
- v t = r\(\omega\) = (0.1 m)(45.0 rad/s) = 4.5 m/s.
Significance
In 30 s, the flywheel has rotated through quite a number of revolutions, about 215 if we divide the angular displacement by 2\(\pi\). A massive flywheel can be used to store energy in this way, if the losses due to friction are minimal. Recent research has considered superconducting bearings on which the flywheel rests, with zero energy loss due to friction.
Angular Acceleration
We have just discussed angular velocity for uniform circular motion, but not all motion is uniform. Envision an ice skater spinning with his arms outstretched—when he pulls his arms inward, his angular velocity increases. Or think about a computer’s hard disk slowing to a halt as the angular velocity decreases. We will explore these situations later, but we can already see a need to define an angular acceleration for describing situations where \(\omega\) changes. The faster the change in \(\omega\), the greater the angular acceleration. We define the instantaneous angular acceleration \(\alpha\) as the derivative of angular velocity with respect to time:
\[\alpha = \lim_{\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t} = \frac{d \omega}{dt} = \frac{d^{2} \theta}{dt^{2}}, \label{10.6}\]
where we have taken the limit of the average angular acceleration, \(\bar{\alpha} = \frac{\Delta \omega}{\Delta t}\) as \(\Delta t → 0\). The units of angular acceleration are (rad/s)/s, or rad/s 2 .
In the same way as we defined the vector associated with angular velocity \(\vec{\omega}\), we can define \(\vec{\alpha}\), the vector associated with angular acceleration (Figure \(\PageIndex{6}\)). If the angular velocity is along the positive z-axis, as in Figure \(\PageIndex{4}\), and \(\frac{d \omega}{dt}\) is positive, then the angular acceleration \(\vec{\alpha}\) is positive and points along the +z- axis. Similarly, if the angular velocity \(\vec{\omega}\) is along the positive z-axis and \(\frac{d \omega}{dt}\) is negative, then the angular acceleration is negative and points along the +z-axis.
We can express the tangential acceleration vector as a cross product of the angular acceleration and the position vector. This expression can be found by taking the time derivative of \(\vec{v} = \vec{\omega} \times \vec{r}\); the result is:
\[\vec{a} = \vec{\alpha} \times \vec{r} \ldotp \label{10.7}\]
The vector relationships for the angular acceleration and tangential acceleration are shown in Figure \(\PageIndex{7}\).
We can relate the tangential acceleration of a point on a rotating body at a distance from the axis of rotation in the same way that we related the tangential speed to the angular velocity. If we differentiate Equation \ref{10.4} with respect to time, noting that the radius r is constant, we obtain
\[a_{t} = r \alpha \ldotp \label{10.8}\]
Thus, the tangential acceleration a t is the radius times the angular acceleration. Equations \ref{10.4} and \ref{10.8} are important for the discussion of rolling motion.
Let’s apply these ideas to the analysis of a few simple fixed-axis rotation scenarios. Before doing so, we present a problem-solving strategy that can be applied to rotational kinematics: the description of rotational motion.
- Examine the situation to determine that rotational kinematics (rotational motion) is involved.
- Identify exactly what needs to be determined in the problem (identify the unknowns). A sketch of the situation is useful.
- Make a complete list of what is given or can be inferred from the problem as stated (identify the knowns).
- Solve the appropriate equation or equations for the quantity to be determined (the unknown). It can be useful to think in terms of a translational analog, because by now you are familiar with the equations of translational motion.
- Substitute the known values along with their units into the appropriate equation and obtain numerical solutions complete with units. Be sure to use units of radians for angles.
- Check your answer to see if it is reasonable: Does your answer make sense?
Now let’s apply this problem-solving strategy to a few specific examples.
A bicycle mechanic mounts a bicycle on the repair stand and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the average angular acceleration in rad/s 2 . (b) If she now hits the brakes, causing an angular acceleration of −87.3 rad/s 2 , how long does it take the wheel to stop?
Strategy
The average angular acceleration can be found directly from its definition \(\bar{\alpha} = \frac{\Delta \omega}{\Delta t}\) because the final angular velocity and time are given. We see that \(\Delta \omega\) = \(\omega_{final}\) − \(\omega_{initial}\) = 250 rev/min and \(\Delta\)t is 5.00 s. For part (b), we know the angular acceleration and the initial angular velocity. We can find the stopping time by using the definition of average angular acceleration and solving for \(\Delta\)t, yielding
\[\Delta t = \frac{\Delta \omega}{\alpha} \ldotp\]
Solution
- Entering known information into the definition of angular acceleration, we get $\bar{\alpha} = \frac{\Delta \omega}{\Delta t} = \frac{250\; rpm}{5.00\; s} \ldotp$Because \(\Delta \omega\) is in revolutions per minute (rpm) and we want the standard units of rad/s 2 for angular acceleration, we need to convert from rpm to rad/s: $\Delta \omega = 250 \frac{rev}{min}\; \cdotp \frac{2 \pi\; rad}{rev}\; \cdotp \frac{1\; min}{60\; s} = 26.2\; rad/s \ldotp$Entering this quantity into the expression for \(\alpha\), we get $bar{\alpha} = \frac{\Delta \omega}{\Delta t} = \frac{26.2\; rpm}{5.00\; s} = 5.24\; rad/s^{2} \ldotp$
- Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that \(\Delta \omega\) is −26.2 rad/s, and \(\alpha\) is given to be –87.3 rad/s 2 . Thus $\Delta t = \frac{-26.2\; rad/s}{-87.3\; rad/s^{2}} = 0.300\; s \ldotp$
Significance
Note that the angular acceleration as the mechanic spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero.
The fan blades on a turbofan jet engine (shown below) accelerate from rest up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in the world, capable of thrusts of 330–510 kN.) (a) What is the average angular acceleration? (b) What is the instantaneous angular acceleration at any time during the first 20 s?
A wind turbine (Figure \(\PageIndex{9}\)) in a wind farm is being shut down for maintenance. It takes 30 s for the turbine to go from its operating angular velocity to a complete stop in which the angular velocity function is \(\omega\)(t) = \(\Big[\frac{(ts^{−1} −30.0)^{2}}{100.0} \Big]\)rad/s. If the turbine is rotating counterclockwise looking into the page, (a) what are the directions of the angular velocity and acceleration vectors? (b) What is the average angular acceleration? (c) What is the instantaneous angular acceleration at t = 0.0, 15.0, 30.0 s?
Strategy
- We are given the rotational sense of the turbine, which is counterclockwise in the plane of the page. Using the right hand rule (Figure 10.5), we can establish the directions of the angular velocity and acceleration vectors.
- We calculate the initial and final angular velocities to get the average angular acceleration. We establish the sign of the angular acceleration from the results in (a).
- We are given the functional form of the angular velocity, so we can find the functional form of the angular acceleration function by taking its derivative with respect to time.
Solution
- Since the turbine is rotating counterclockwise, angular velocity \(\vec{\omega}\) points out of the page. But since the angular velocity is decreasing, the angular acceleration \(\vec{\alpha}\) points into the page, in the opposite sense to the angular velocity.
- The initial angular velocity of the turbine, setting t = 0, is \(\omega\) = 9.0 rad/s. The final angular velocity is zero, so the average angular acceleration is $\bar{\alpha} \frac{\Delta \omega}{\Delta t} = \frac{\omega - \omega_{0}}{t - t_{0}} = \frac{0 - 9.0\; rad/s}{30.0 - 0\; s} = -0.3\; rad/s^{2} \ldotp$
- Taking the derivative of the angular velocity with respect to time gives \(\alpha = \frac{d \omega}{dt} = \frac{(t − 30.0)}{50.0}\) rad/s 2 $\alpha (0.0; s) = -0.6\; rad/s^{2}, \alpha (15.0\; s) = -0.3\; rad/s^{2}, and\; \alpha (30.0\; s) = 0\; rad/s \ldotp$
Significance
We found from the calculations in (a) and (b) that the angular acceleration α and the average angular acceleration \(\bar{\alpha}\) are negative. The turbine has an angular acceleration in the opposite sense to its angular velocity.
We now have a basic vocabulary for discussing fixed-axis rotational kinematics and relationships between rotational variables. We discuss more definitions and connections in the next section.


8.3: Preventing Foodborne Illness
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- BC Cook Articulation Committee
- BCcampus
Food-handling and Storage Procedures
Proper food handling and storage can prevent most foodborne illnesses. In order for pathogens to grow in food, certain conditions must be present. By controlling the environment and conditions, even if potentially harmful bacteria are present in the unprepared or raw food, they will not be able to survive, grow, and multiply, causing illness.
There are six factors that affect bacterial growth, which can be referred to by the mnemonic FATTOM :
- F ood
- A cid
- T emperature
- T ime
- O xygen
- M oisture
Each of these factors contributes to bacterial growth in the following ways:
- Food: Bacteria require food to survive. For this reason, moist, protein-rich foods are good potential sources of bacterial growth.
- Acid: Bacteria do not grow in acidic environments. This is why acidic foods like lemon juice and vinegar do not support the growth of bacteria and can be used as preservatives
- Temperature: Most bacteria will grow rapidly between 40°F and 140°F. This is referred to as the danger zone (see the section below for more information on the danger zone).
- Time: Bacteria require time to multiply. When small numbers of bacteria are present, the risk is usually low, but extended time with the right conditions will allow the bacteria to multiply and increase the risk of contamination
- Oxygen: There are two types of bacteria. Aerobic bacteria require oxygen to grow, so will not multiply in an oxygen-free environment such as a vacuum-packaged container. Anaerobic bacteria will only grow in oxygen-free environments. Food that has been improperly processed and then stored at room temperature can be at risk from anaerobic bacteria. A common example is a product containing harmful Clostridium botulinum (botulism-causing) bacteria that has been improperly processed during canning, and then is consumed without any further cooking or reheating.
- Moisture: Bacteria need moisture to survive and will grow rapidly in moist foods. This is why dry and salted foods are at lower risk of being hazardous.
Identifying Time/Temperature Control for Safety Foods (TCS Foods)
Foods that have the FATTOM conditions are considered Time/Temperature Control for Safety Foods (TCS Foods). TCS foods are those foods that are considered perishable. That is, they will spoil or “go bad” if left at room temperature. TCS foods support the growth or survival of disease-causing bacteria (pathogens) or may be contaminated by pathogens.
Generally, a food is a TCS if it is:
- Of animal origin such as meat, milk, eggs, fish, shellfish, poultry (or if it contains any of these products)
- Of plant origin (vegetables, beans, fruit, etc.) that has been heat-treated or cooked
- Any of the raw sprouts (bean, alfalfa, radish, etc.)
- Any cooked starch (rice, pasta, etc.)
- Any type of soya protein (soya milk, tofu, etc.)
Table 2 identifies common foods as either TCS or non-TCS:
| TCS | Non-TCS |
|---|---|
| Chicken, beef, pork, and other meats | Beef jerky |
| Pastries filled with meat, cheese, or cream | Bread |
| Cooked rice | Uncooked rice |
| Tofu | Uncooked beans |
| Fresh garlic in oil | Fresh garlic |
| Fresh or cooked eggs | Powdered eggs |
| Gravy | Flour |
| Dry soup mix with water added | Dry soup mix |
The Danger Zone
One of the most important factors to consider when handling food properly is temperature. Table 3 lists the most temperatures to be aware of when handling food.
| Fahrenheit | What happens? |
|---|---|
| 212° | Water boils |
| 140° | Most pathogenic bacteria are destroyed. Keep hot foods above this temperature. |
| 68° | Food must be cooled from 140°F to 70°F(60°C to 20°C) within two hours or less |
| 40° | Food must be cooled from 70°F to 40°F (20°C to 4°C) within four hours or less |
| 32° | Water freezes |
| 0° | Frozen food must be stored at 0°F (−18°C) or below |
Figure 1. Danger Zone Chart, Used with permission from BC Centre for Disease Control (BCCDC). [Image description]
The range of temperature from 40°F and 140°F (4°C and 60°C )is known as the danger zone , or the range at which most pathogenic bacteria will grow and multiply.
Time-temperature Control
Pathogen growth is controlled by a time-temperature relationship. To kill micro-organisms, food must be held at a sufficient temperature for a sufficient time. Cooking is a scheduled process in which each of a series of continuous temperature combinations can be equally effective.
Table 4 show the minimum time-temperature requirements to keep food safe.
| Critical control point | Type of food | Temperature |
|---|---|---|
| Refrigeration | Cold food storage, all foods. | 40°F (4°C) or less |
| Freezing | Frozen food storage, all foods. | 0°F (−18°C) or less |
| Cooking | Food mixtures containing poultry, eggs, meat, fish, or other potentially hazardous foods | Internal temperature of 165°F (74°C )for at least 15 seconds |
| Cooking | Pork, lamb, veal, beef | Internal temperature of 145°F (63°C) |
| Cooking | Poultry | Internal temperature of 74°C (165°F) for 15 seconds |
| Cooking | Ground meat (Includes chopped, ground, flaked, or minced beef, pork, or fish) | 160°F (71°C) |
| Cooking | Eggs [1] | 160°F (71°C) |
| Cooking | Fish [2] | Internal temperature of 145°F (63°C) |
| Holding | Hot foods | 140°F (60°C) |
| Reheating | All foods | 165°F (74°C) for at least 15 seconds |
The Top 10 List: Do’s and Don’ts
Figure 1 illustrates the top 10 improper food-handling methods and the percentage of foodborne illnesses they cause.
This section describes each food-handling practice outlined in the top 10 list and the ways to prevent each problem.
1. Improper cooling
Many people think that once a food has been properly cooked, all disease-causing organisms (pathogens) have been killed. This is not true. Some pathogens can form heat-resistant spores, which can survive cooking temperatures. When the food begins cooling down and enters the danger zone, these spores begin to grow and multiply. If the food spends too much time in the danger zone, the pathogens will increase in number to a point where the food will make people sick. That is why the cooling process is crucial. Cooked food must be cooled from 140°F to 70°F (60°C to 20°C) in two hours or less, AND then from 70°F to 40°F (60°C to 20°C) in four hours or less.
Image by: Lizz Daniels, RDN
Even in modern walk-in coolers, large cuts of meat will not cool down properly. Neither will whole poultry. Even large pots (4 L/1 gal. or more) of soup, stews, gravy, etc., can take a day or more to cool to 40°F . However, you can cool these foods down quickly by using one or more of the following methods depending on the type of food being cooled:
- Place the food in shallow pans (with the food no deeper than 5 cm/2 in.) and put the pans in the cooler.
- When the food is cooling, do not tightly cover. Doing so only seals in the heat.
- Do not stack the shallow cooling pans during the cooling step. This will defeat the purpose of shallow panning by preventing cold air from reaching the food. You may need to add more shelves to your cooler.
- Cut large cuts of meat or whole poultry into smaller or thinner portions. Then place these portions into shallow pans for cooling.
- Use cooling wands or cooling sticks to cool foods quickly.
- Use rapid cooling equipment such as walk-in coolers with wire shelving and good air flow. Home-style refrigerators or reach-ins do not cool food well.
- Stir the food in a container placed in an ice-water bath.
- Use containers that help heat transfer, such as stainless steel or aluminum. Plastic does not transfer heat well.
- Use ice as an ingredient (e.g., in stews or soups).
- For large pots of cooked desserts (e.g., custard), divide it into serving sizes and then cool.
2. Advance preparation
Advance preparation is the cause of many food-poisoning outbreaks, usually because food has been improperly cooled. Often, foods that are prepared well before serving spend too much time in the danger zone. This may happen for one or more of the following reasons:
- The food is left out at room temperature too long.
- The food is not heated or reheated properly (to a high enough temperature), or not cooled properly.
- The food is brought in and out of the danger zone too many times (e.g., cooked, hot held, cooled, reheated, hot held, cooled, reheated again).
To prevent problems of advance preparation:
- Try to prepare all foods for same-day use and as close to serving time as possible.
- To prevent outside contamination of foods prepared in advance, cover them tightly after they have been properly cooled.
- Reheat leftovers only once. If leftovers are not consumed after being reheated, throw them out.
- For foods prepared and held refrigerated in the cooler for more than 24 hours, mark the date of preparation and a “serve by” date. Generally, TCS foods should be thrown out if not used within three days from date they were made.
- If you must prepare foods in advance, be sure you properly cool and refrigerate them.
3. Infected person
Many people carry pathogens somewhere on or in their bodies without knowing it—in their gut, in their nose, on their hands, in their mouth, and in other warm, moist places. People who are carrying pathogens often have no outward signs of illness. However, people with symptoms of illness (diarrhea, fever, vomiting, jaundice, sore throat with a fever, hand infections, etc.) are much more likely to spread pathogens to food.
Another problem is that pathogens can be present in the cooked and cooled food that, if given enough time, can still grow. These pathogens multiply slowly but they can eventually reach numbers where they can make people sick. This means that foods that are prepared improperly, many days before serving, yet stored properly the entire time can make people sick.
Some pathogens are more dangerous than others (e.g., salmonella, E. coli, campylobacter). Even if they are only present in low numbers, they can make people very sick. A food handler who is carrying these kinds of pathogens can easily spread them to foods – usually from their hands. Ready-to-eat food is extra dangerous. Ready-to-eat food gets no further cooking after being prepared, so any pathogens will not be killed or controlled by cooking.
To prevent problems:
- Make sure all food handlers wash their hands properly after any job that could dirty their hands (e.g., using the toilet, eating, handling raw meats, blowing their nose, smoking).
- Food handlers with infected cuts on their hands or arms (including sores, burns, lesions, etc.) must not handle food or utensils unless the cuts are properly covered (e.g., waterproof bandage covered with a latex glove or finger cot).
- When using gloves or finger cots , food handlers must still wash their hands. As well, gloves or cots must be replaced if they are soiled, have a hole, and at the end of each day.
- Food handlers with infection symptoms must not handle utensils or food and should be sent home.
- Where possible, avoid direct hand contact with food – especially ready-to-eat foods (e.g., use plastic utensils plastic or latex gloves).
4. Inadequate reheating for hot holding
Many restaurants prepare some of menu items in advance or use leftovers in their hot hold units the next day. In both cases, the foods travel through the danger zone when they are cooled for storage and again when they are reheated.
Foods that are hot held before serving are particularly vulnerable to pathogens. In addition to traveling through the danger zone twice, even in properly operating hot hold units, the food is close to the temperature that will allow pathogens to grow.
To prevent problems:
- Do not use hot hold units to reheat food. They are not designed for this purpose. Instead, rapidly reheat to 165° (and hold the food at that temperature for at least 15 seconds before putting it in the hot hold unit. This will kill any pathogens that may have grown during the cool-down step and the reheat step.
- If using direct heat (stove top, oven, etc.), the temperature of the reheated food must reach at least 165°F for at least 15 seconds within two hours. Keep a thermometer handy to check the temperature of the food.
- If using a microwave, rotate or stir the food at least once during the reheat step, as microwaves heat unevenly. As well, the food must be heated to at least 165°F and then stand covered for two minutes after reheating before adding to the hot hold unit. The snapping and crackling sounds coming from food being reheated in a microwave do not mean the food is hot.
5. Improper hot holding
Hot hold units are meant to keep hot foods at 140°F or hotter. At or above this temperature, pathogens will not grow.
6. Contaminated raw food or ingredient
We know that many raw foods often contain pathogens, yet certain foods are often served raw.
- Raw oysters served in the shell
- Raw eggs in certain recipes (e.g., Caesar salad, eggnog made from raw eggs)
- Rare hamburger
- Sushi/sashimi
- Steak tartare
These foods have caused many food-poisoning outbreaks. Always remember: you cannot tell if a food contains pathogens just by look, taste, or smell.
To prevent problems:
- Buy all your foods or ingredients from approved suppliers.
- If available, buy foods or ingredients from suppliers who also have food safety plans for their operations.
- Where possible, use processed or pasteurized alternatives (e.g., pasteurized liquid eggs).
- Never serve these types of foods to high-risk customers (e.g., seniors, young children, people in poor health, people in hospitals or nursing homes).
7. Unsafe source
Foods from approved sources are less likely to contain high levels of pathogens or other forms of contamination. Approved sources are those suppliers that are inspected for cleanliness and safety by a government food inspector. Foods supplied from unreliable or disreputable sources, while being cheaper, may contain high levels of pathogens that can cause many food-poisoning outbreaks.
Fly-by-night suppliers (trunk sales) often do not care if the product is safe to sell to you, but approved suppliers do! As well, many fly-by-night suppliers have obtained their product illegally (e.g., closed shellfish fisheries, rustled cattle, poached game and fish) and often do not have the equipment to properly process, handle, store, and transport the food safely.
Of particular concern is seafood from unapproved sources. Seafood, especially shellfish, from unapproved sources can be heavily contaminated with pathogens or poisons if they have been harvested from closed areas.
8. Use of leftovers
Using leftovers has been the cause of many outbreaks of food poisoning because of improper cooling and reheating (of “hot” leftovers). Leftovers that are intended to be served hot pass through the danger zone twice (during the initial cooling of the hot food and when reheating). Those leftovers intended to be served without reheating, or as an ingredient in other foods (e.g., sandwich filler), go through the danger zone during cooling and then, when being prepared and portioned, often stay in the danger zone for another long period. The time in the danger zone adds up unless the food is quickly cooled and then quickly reheated (if being served hot), or kept cold until serving (if not being served hot).
Contamination can also occur with leftover foods when they are stored in the cooler. Improperly stored leftovers can accidentally be contaminated by raw foods (e.g., blood dripping from a higher shelf).
To prevent problems:
- Reheat leftovers only once. Throw out any leftovers that have already been reheated once.
- Do not mix leftover foods with fresh foods.
- Be sure to follow the proper cooling and reheating procedures when handling leftovers. These are critical control points.
- Cool leftovers in uncovered containers separate from any raw foods. After they are cooled, cover them tightly.
9. Cross-contamination
You can expect certain foods to contain pathogens, especially raw meat, raw poultry, and raw seafood. Use extreme caution when you bring these foods into your kitchen. Cross-contamination happens when something that can cause illness (pathogens or chemicals) is accidentally put into a food where not previously found. This can include, for example, pathogens from raw meats getting into ready-to-eat foods like deli meats.
A term used to describe when this occurs with an allergen, such as nuts or dairy, is cross contact . Cross contact can be avoided using the same prevention steps as cross contamination..
To prevent problems:
- Use separate cutting boards, separate cleaning cloths, knives/utensils, sinks, preparation areas, etc., for raw and for ready-to-eat foods. Otherwise, wash all of these items with detergent and sanitize them with bleach between use.
- Use separate storage areas for raw and ready-to-eat foods. Always store ready-to-eat foods on separate shelves and above raw foods. Store dry foods above wet foods.
- After handling raw foods, always wash your hands properly before doing anything else.
- Use clean utensils, not your hands, to handle cooked or ready-to-eat foods.
- If a customer indicates a food allergy, follow all the same steps to avoid cross contact and use separate or freshly sanitized tools and utensils to prepare food for the individual with the allergy.
10. Inadequate cooking
Proper cooking is one of the best means of making sure your operation does not cause a food-poisoning outbreak. Proper cooking kills all pathogens (except spores) or at least reduces their numbers to a point where they cannot make people sick. Inadequate cooking is often done by accident: for example, cooking a hamburger until the meat is no longer pink instead of checking the internal temperature to ensure it has reached the proper endpoint cooking temperature.
To prevent problems:
- Don’t rely on cooking times alone. Check the internal temperature of the food being cooked.
- For large cuts of meat or large batches of food, check the temperature in several spots.
- Be extra careful when cooking partially frozen foods. There can be cold spots in the food that are not properly cooked. The normal cooking time will have to be increased.
- When grilling or frying meat, cook until the juices run clear. Cooked fish until it flakes easily. Make thin, not thick, hamburgers.
Figure 2 image description:
- Improper cooling, 30%.
- Advance preparation, 15%.
- Infected person, 12%.
- Inadequate reheating, 10%.
- Improper hot holding, 8%.
- Contaminated raw food, 4%.
- Unsafe source, 3%.
- Use of leftovers, 2%.
- Cross-contamination, 2%.
- Inadequate cooking, 1%.
- All other causes, 3%.


Sulfur
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Learning Objectives
Sulfur is incorporated into protein structures in the body. Amino acids, methionine and cysteine contain sulfur which are essential for the antioxidant enzyme glutathione peroxidase. Some vitamins like thiamin and biotin also contain sulfur which are important in regulating acidity in the body. Sulfur is a major mineral with no recommended intake or deficiencies when protein needs are met. Sulfur is mostly consumed as a part of dietary proteins and sulfur containing vitamins.
University of Hawai’i at Mānoa Food Science and Human Nutrition Program : Allison Calabrese, Cheryl Gibby, Billy Meinke, Marie Kainoa Fialkowski Revilla, and Alan Titchenal


22.5: Far-Field Radiation from a Half-Wave Dipole
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A simple and important current distribution is that of the thin half-wave dipole (HWD), shown in Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): Current distribution of the half-wave dipole (HWD). ( CC BY-SA 4.0 ; C. Wang)This is the distribution expected on a thin straight wire having length \(L=\lambda/2\), where \(\lambda\) is wavelength. This distribution is described mathematically as follows:
\[\widetilde{I}(z) \approx I_0 \cos \left( \pi \frac{z}{L} \right) ~~~ \mbox{for} ~ \left|z\right|\le\frac{L}{2} \label{m0200_eHWD1} \]
where \(I_0\) (SI base units of A) is a complex-valued constant indicating the maximum magnitude of the current and its phase. Note that the current is zero at the ends of the dipole; i.e., \(\widetilde{I}(z) = 0\) for \(\left|z\right|= L/2\). Note also that this “cosine pulse” distribution is very similar to the triangular distribution of the ESD, and is reminiscent of the sinusoidal variation of current in a standing wave.
Since \(L=\lambda/2\) for the HWD, Equation \ref{m0200_eHWD1} may equivalently be written:
\[\widetilde{I}(z) \approx I_0 \cos \left( 2\pi \frac{z}{\lambda} \right) \label{m0200_eHWD2} \]
The electromagnetic field radiated by this distribution of current may be calculated using the method described in Section 9.6, in particular:
\[ \tilde{\mathbf{E}}(\mathbf{r}) \approx \hat{\theta} j \frac{\eta}{2} \frac{e^{-j \beta r}}{r}(\sin \theta) \cdot \left[ \frac{1}{\lambda} \int_{-L / 2}^{+L / 2} \tilde{I}\left(z^{\prime}\right) e^{+j \beta z^{\prime} \cos \theta} d z^{\prime}\right] \label{m0200_eESDE} \]
which is valid for field points \({\bf r}\) far from the dipole; i.e., for \(r\gg L\) and \(r\gg \lambda\). For the HWD, the quantity in square brackets is
\[\frac{I_0}{\lambda}\int_{-\lambda/4}^{+\lambda/4} \cos \left( 2\pi \frac{z'}{\lambda} \right) e^{+j\beta z'\cos{\theta}} dz' \nonumber \]
The evaluation of this integral is straightforward, but tedious. The integral reduces to
\[\frac{I_0}{\pi} \frac{\cos\left[\left(\pi/2\right)\cos\theta\right]}{\sin^2\theta} \nonumber \]
Substitution into Equation \ref{m0200_eESDE} yields
\[\widetilde{\bf E}({\bf r}) \approx \hat{\bf \theta} j \frac{\eta I_0}{2\pi} \frac{\cos\left[\left(\pi/2\right)\cos\theta\right]}{\sin\theta} \frac{e^{-j\beta r}}{r} \nonumber \]
The magnetic field may be determined from this result using Ampere’s law. However, a simpler method is to use the fact that the electric field, magnetic field, and direction of propagation \(\hat{\bf r}\) are mutually perpendicular and related by:
\[\widetilde{\bf H} = \frac{1}{\eta} \hat{\bf r} \times \widetilde{\bf E} \nonumber \]
This relationship indicates that the magnetic field will be \(+\hat{\bf \phi}\)-directed.
The magnitude and polarization of the radiated field is similar to that of the electrically-short dipole (ESD; Section 9.5). A comparison of the magnitudes in any radial plane containing the \(z\)-axis is shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\): Comparison of the magnitude of the radiated field of the HWD to that of an electrically-short dipole also oriented along the z-axis. This result is for any radial plane that includes the z-axis. ( CC BY-SA 4.0 ; C. Wang)For either current distribution, the maximum magnitude of the fields occurs in the \(z=0\) plane. For a given terminal current \(I_0\), the maximum magnitude is greater for the HWD than for the ESD. Both current distributions yield zero magnitude along the axis of the dipole. The polarization characteristics of the fields of both current distributions are identical.
Additional Reading:
- “Dipole antenna” (section entitled “Half-wave dipole”) on Wikipedia.


PROBLEM SET: EXPONENTIAL GROWTH AND DECAY FUNCTIONS
Identify each as an exponential, linear, or polynomial function.
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\(y=640\left(1.25^{x}\right)\)
|
-
\(y=640\left(x^{1.25}\right)\)
|
-
\(y=640(1.25 x)\)
|
-
\(y=1.05 x-2.5\)
|
-
\(y=90-(4 / 5) x\)
|
-
\(y=42\left(0.92^{x}\right)\)
|
-
\(y=37\left(x^{0.25}\right)\)
|
-
\(y=4(1 / 3)^{x}\)
|
Indicate if the function represents exponential growth or exponential decay.
-
\(y=127 \mathrm{e}^{-0.35 \mathrm{t}}\)
|
-
\(y=70\left(0.8^{t}\right)\)
|
-
\(y=453\left(1.2^{t}\right)\)
|
-
\(y=16 e^{0.2 t}\)
|
In each of the following, \(y\) is an exponential function of \(t\) stated in the form \(y = ae^{kt}\) where \(t\) represents time measured in years.. For each:
-
re-express each function in the form \(y = ab^t\) (state the value of \(b\) accurate to 4 decimal places)
-
state the annual growth rate or annual decay rate as a percent, accurate to 2 decimal places
-
\(y=127 \mathrm{e}^{-0.35 \mathrm{t}}\)
|
-
\(y=16 e^{0.4 t}\)
|
-
\(y=17250 \mathrm{e}^{0.24 \mathrm{t}}\)
|
-
\(y=4700 \mathrm{e}^{-0.07 t}\)
|
Identify if the function represents exponential growth, exponential decay, linear growth, or linear decay. In each case write the function and find the value at the indicated time.
-
A house was purchased for $350,000 in the year 2010. The value has been increasing by $7,000 per year. Write the function and find the value of the house after 5 years.
|
-
A house was purchased for $350,000 in the year 2010. The value has been increasing at the rate of 2% per year. Write the function and find the value of the house after 5 years.
|
-
A lab purchases new equipment for $50,000. Its value depreciates over time. The value decreases at the rate of 6% annually. Write the function and find the value after 10 years.
|
-
A lab purchases new equipment for $50,000. Its value depreciates over time. The value decreases by $3000 annually. Write the function and find the value after 10 years
|
-
A population of bats in a cave has 200 bats. The population is increasing by 10 bats annually. Write the function. How many bats live in the cave after 7 years?
|
-
A population of bats in a cave has 200 bats. The population is increasing at the rate of 5% annually. Write the function. How many bats live in the cave after 7 years?
|
-
A population of a certain species of bird in a state park has 300 birds. The population is decreasing at the rate of 7% year. Write the function. How many birds are in the population after 6 years?
|
-
A population of a certain species of bird in a state park has 300 birds. The population decreases by 20 birds per year. Write the function. How many birds are in the population after 6 years?
|
In problems 25-28, the problem represents exponential growth or decay and states the CONTINUOUS growth rate or continuous decay rate. Write the exponential growth or decay function and find the value at the indicated time.
Hint: Use the form of the exponential function that is appropriate when the CONTINUOUS growth or decay rate is given.
-
A population of 400 microbes increases at the
continuous
growth rate of 26% per day. Write the function and find the number of microbes in the population at the end of 7 days.
|
-
The price of a machine needed by a production factory is $28,000. The business expects to replace the machine in 4 years. Due to inflation the price of the machine is increasing at the
continuous
rate of 3.5% per year. Write the function and find the value of the machine 4 years from now.
|
-
A population of an endangered species consists of 4000 animals of that species. The population is decreasing at the continuous rate of 12% per year. Write the function and find the size of the population at the end of 10 years.
|
-
A business buys a computer system for $12000. The value of the system is depreciating and decreases at the continuous rate of 20% per year. Write the function and find the value at the end of 3 years.
|


33.E: The Animal Body - Basic Form and Function (Exercises)
33.1: Animal Form and Function
Animals vary in form and function. From a sponge to a worm to a goat, an organism has a distinct body plan that limits its size and shape. Animals’ bodies are also designed to interact with their environments, whether in the deep sea, a rainforest canopy, or the desert. Therefore, a large amount of information about the structure of an organism's body (anatomy) and the function of its cells, tissues and organs (physiology) can be learned by studying that organism's environment.
Review Questions
Which type of animal maintains a constant internal body temperature?
- endotherm
- ectotherm
- coelomate
- mesoderm
- Answer
-
A
The symmetry found in animals that move swiftly is ________.
- radial
- bilateral
- sequential
- interrupted
- Answer
-
B
What term describes the condition of a desert mouse that lowers its metabolic rate and “sleeps” during the hot day?
- turgid
- hibernation
- estivation
- normal sleep pattern
- Answer
-
C
A plane that divides an animal into equal right and left portions is ________.
- diagonal
- midsagittal
- coronal
- transverse
- Answer
-
B
A plane that divides an animal into dorsal and ventral portions is ________.
- sagittal
- midsagittal
- coronal
- transverse
- Answer
-
D
The pleural cavity is a part of which cavity?
- dorsal cavity
- thoracic cavity
- abdominal cavity
- pericardial cavity
- Answer
-
B
Free Response
How does diffusion limit the size of an organism? How is this counteracted?
- Answer
-
Diffusion is effective over a very short distance. If a cell exceeds this distance in its size, the center of the cell cannot get adequate nutrients nor can it expel enough waste to survive. To compensate for this, cells can loosely adhere to each other in a liquid medium, or develop into multi-celled organisms that use circulatory and respiratory systems to deliver nutrients and remove wastes.
What is the relationship between BMR and body size? Why?
- Answer
-
Basal Metabolic Rate is an expression of the metabolic processes that occur to maintain an individual’s functioning and body temperature. Smaller bodied animals have a relatively large surface area compared to a much larger animal. The large animal’s large surface area leads to increased heat loss that the animal must compensate for, resulting in a higher BMR. A small animal, having less relative surface area, does not lose as much heat and has a correspondingly lower BMR.
33.2: Animal Primary Tissues
The tissues of multicellular, complex animals are four primary types: epithelial, connective, muscle, and nervous. Recall that tissues are groups of similar cells group of similar cells carrying out related functions. These tissues combine to form organs—like the skin or kidney—that have specific, specialized functions within the body. Organs are organized into organ systems to perform functions.
Review Questions
Which type of epithelial cell is best adapted to aid diffusion?
- squamous
- cuboidal
- columnar
- transitional
- Answer
-
C
Which type of epithelial cell is found in glands?
- squamous
- cuboidal
- columnar
- transitional
- Answer
-
B
Which type of epithelial cell is found in the urinary bladder?
- squamous
- cuboidal
- columnar
- transitional
- Answer
-
D
Which type of connective tissue has the most fibers?
- loose connective tissue
- fibrous connective tissue
- cartilage
- bone
- Answer
-
B
Which type of connective tissue has a mineralized different matrix?
- loose connective tissue
- fibrous connective tissue
- cartilage
- bone
- Answer
-
D
The cell found in bone that breaks it down is called an ________.
- osteoblast
- osteocyte
- osteoclast
- osteon
- Answer
-
C
The cell found in bone that makes the bone is called an ________.
- osteoblast
- osteocyte
- osteoclast
- osteon
- Answer
-
A
Plasma is the ________.
- fibers in blood
- matrix of blood
- cell that phagocytizes bacteria
- cell fragment found in the tissue
- Answer
-
B
The type of muscle cell under voluntary control is the ________.
- smooth muscle
- skeletal muscle
- cardiac muscle
- visceral muscle
- Answer
-
B
The part of a neuron that contains the nucleus is the
- cell body
- dendrite
- axon
- glial
- Answer
-
B
Free Response
How can squamous epithelia both facilitate diffusion and prevent damage from abrasion?
- Answer
-
Squamous epithelia can be either simple or stratified. As a single layer of cells, it presents a very thin epithelia that minimally inhibits diffusion. As a stratified epithelia, the surface cells can be sloughed off and the cells in deeper layers protect the underlying tissues from damage.
What are the similarities between cartilage and bone?
- Answer
-
Both contain cells other than the traditional fibroblast. Both have cells that lodge in spaces within the tissue called lacunae. Both collagen and elastic fibers are found in bone and cartilage. Both tissues participate in vertebrate skeletal development and formation.
33.3: Homeostasis
Animal organs and organ systems constantly adjust to internal and external changes through a process called homeostasis (“steady state”). These changes might be in the level of glucose or calcium in blood or in external temperatures. Homeostasis means to maintain dynamic equilibrium in the body. It is dynamic because it is constantly adjusting to the changes that the body’s systems encounter. It is equilibrium because body functions are kept within specific ranges.
Review Questions
When faced with a sudden drop in environmental temperature, an endothermic animal will:
- experience a drop in its body temperature
- wait to see if it goes lower
- increase muscle activity to generate heat
- add fur or fat to increase insulation
- Answer
-
C
Which is an example of negative feedback?
- lowering of blood glucose after a meal
- blood clotting after an injury
- lactation during nursing
- uterine contractions during labor
- Answer
-
A
Which method of heat exchange occurs during direct contact between the source and animal?
- radiation
- evaporation
- convection
- conduction
- Answer
-
D
The body’s thermostat is located in the ________.
- homeostatic receptor
- hypothalamus
- medulla
- vasodilation center
- Answer
-
B
Free Response
Why are negative feedback loops used to control body homeostasis?
- Answer
-
An adjustment to a change in the internal or external environment requires a change in the direction of the stimulus. A negative feedback loop accomplishes this, while a positive feedback loop would continue the stimulus and result in harm to the animal.
Why is a fever a “good thing” during a bacterial infection?
- Answer
-
Mammalian enzymes increase activity to the point of denaturation, increasing the chemical activity of the cells involved. Bacterial enzymes have a specific temperature for their most efficient activity and are inhibited at either higher or lower temperatures. Fever results in an increase in the destruction of the invading bacteria by increasing the effectiveness of body defenses and an inhibiting bacterial metabolism.
How is a condition such as diabetes a good example of the failure of a set point in humans?
- Answer
-
Diabetes is often associated with a lack in production of insulin. Without insulin, blood glucose levels go up after a meal, but never go back down to normal levels.


6.3: The Schwarzschild Metric (Part 2)
Geodetic Effect
As promised in section 5.5 , we now calculate the geodetic effect on Gravity Probe B, including all the niggling factors of 3 and \(\pi\). To make the physics clear, we approach the actual calculation through a series of warmups.
Flat Space
As a first warmup, consider two spatial dimensions, represented by Euclidean polar coordinates (r, \(\phi\)). Parallel-transport of a gyroscope’s angular momentum around a circle of constant r gives
\[\begin{split} \nabla_{\phi} L^{\phi} &= 0 \\ \nabla_{\phi} L^{r} &= 0 \ldotp \end{split}\]
Computing the covariant derivatives, we have
\[\begin{split} 0 &= \partial_{\phi} L^{\phi} + \Gamma^{\phi}_{\phi r} L^{r} \\ 0 &= \partial_{\phi} L^{r} + \Gamma^{r}_{\phi \phi} L^{\phi} \ldotp \end{split}\]
The Christoffel symbols are \(\Gamma^{\phi}_{\phi r} = \frac{1}{r}\) and \(\Gamma^{r}_{\phi \phi}\) = −r. This is all made to look needlessly complicated because L \(\phi\) and L r are expressed in different units. Essentially the vector is staying the same, but we’re expressing it in terms of basis vectors in the r and \(\phi\) directions that are rotating. To see this more transparently, let r = 1, and write P for L \(\phi\) and Q for L r , so that
\[\begin{split} P' &= -Q \\ Q' &= P, \end{split}\]
which have solutions such as P = sin \(\phi\), Q = cos \(\phi\). For each orbit (2π\(\pi\)change in \(\phi\)), the basis vectors rotate by 2\(\phi\), so the angular momentum vector once again has the same components. In other words, it hasn’t really changed at all.
Spatial Curvature Only
The flat-space calculation above differs in two ways from the actual result for an orbiting gyroscope: (1) it uses a flat spatial geometry, and (2) it is purely spatial. The purely spatial nature of the calculation is manifested in the fact that there is nothing in the result relating to how quickly we’ve moved the vector around the circle. We know that if we whip a gyroscope around in a circle on the end of a rope, there will be a Thomas precession ( section 2.5 ), which depends on the speed.
As our next warmup, let’s curve the spatial geometry, but continue to omit the time dimension. Using the Schwarzschild metric, we replace the flat-space Christoffel symbol \(\Gamma^{r}_{\phi \phi}\) = −r with −r+2m. The differential equations for the components of the L vector, again evaluated at r = 1 for convenience, are now
\[\begin{split} P' &= -Q \\ Q' &= (1 - \epsilon) P, \end{split}\]
where \(\epsilon\) = 2m. The solutions rotate with frequency \(\omega' = \sqrt{1 − \epsilon}\). The result is that when the basis vectors rotate by 2\(\pi\), the components no longer return to their original values; they lag by a factor of \(\sqrt{1 − \epsilon}\) ≈ 1 − m. Putting the factors of r back in, this is 1 − \(\frac{m}{r}\). The deviation from unity shows that after one full revolution, the L vector no longer has quite the same components expressed in terms of the (r, \(\phi\)) basis vectors.
To understand the sign of the effect, let’s imagine a counterclockwise rotation. The (r, \(\phi\)) rotate counterclockwise, so relative to them, the L vector rotates clockwise. After one revolution, it has not rotated clockwise by a full 2\(\pi\), so its orientation is now slightly counterclockwise compared to what it was. Thus the contribution to the geodetic effect arising from spatial curvature is in the same direction as the orbit.
Comparing with the actual results from Gravity Probe B, we see that the direction of the effect is correct. The magnitude, however, is off. The precession accumulated over n periods is \(\frac{2 \pi nm}{r}\), or, in SI units, \(\frac{2 \pi nGm}{c^{2} r}\). Using the data from section 2.5 , we find \(\Delta \theta\) = 2 × 10 −5 radians, which is too small compared to the data shown in Figure 5.5.2 .
2+1 Dimensions
To reproduce the experimental results correctly, we need to include the time dimension. The angular momentum vector now has components (L \(\phi\) ,L r ,L t ). The physical interpretation of the L t component is obscure at this point; we’ll return to this question later.
Writing down the total derivatives of the three components, and notating \(\frac{dt}{d \phi}\) as \(\omega^{−1}\), we have
\[\frac{dL^{\phi}}{d \phi} = \partial_{\phi} L^{\phi} + \omega^{-1} \partial_{t} L^{\phi}\]
\[\frac{dL^{r}}{d \phi} = \partial_{\phi} L^{r} + \omega^{-1} \partial_{t} L^{r}\]
\[\frac{dL^{t}}{d \phi} = \partial_{\phi} L^{t} + \omega^{-1} \partial_{t} L^{t}\]
Setting the covariant derivatives equal to zero gives
\[\begin{split} 0 &= \partial_{\phi} L^{\phi} + \Gamma^{\phi}_{\phi r} L^{r} \\ 0 &= \partial_{\phi} L^{r} + \Gamma^{r}_{\phi \phi} L^{\phi} \\ 0 &= \partial_{t} L^{r} + \Gamma^{r}_{tt} L^{t} \\ 0 &= \partial_{t} L^{t} + \Gamma^{t}_{tr} L^{r} \ldotp \end{split}\]
Exercise \(\PageIndex{4}\)
Self-check: There are not just four but six covariant derivatives that could in principle have occurred, and in these six covariant derivatives we could have had a total of 18 Christoffel symbols. Of these 18, only four are nonvanishing. Explain based on symmetry arguments why the following Christoffel symbols must vanish: \(\Gamma^{\phi}_{\phi t}, \Gamma^{t}_{tt}\).
Putting all this together in matrix form, we have L' = ML, where
\[M = \begin{pmatrix} 0 & -1 & 0 \\ 1 - \epsilon & 0 & - \frac{\epsilon (1 - \epsilon)}{2 \omega} \\ 0 & - \frac{\epsilon}{2 \omega (1 - \epsilon)} & 0 \end{pmatrix} \ldotp\]
The solutions of this differential equation oscillate like e i\(\Omega\)t , where i\(\Omega\) is an eigenvalue of the matrix.
Exercise \(\PageIndex{5}\)
Self-check: The frequency in the purely spatial calculation was found by inspection. Verify the result by applying the eigenvalue technique to the relevant 2 × 2 submatrix.
To lowest order, we can use the Newtonian relation \(\omega^{2} r = \frac{Gm}{r}\) and neglect terms of order \(\epsilon^{2}\), so that the two new off-diagonal matrix elements are both approximated as \(\sqrt{\frac{\epsilon}{2}}\). The three resulting eigenfrequencies are zero and \(\Omega\) = ±[1 − \(\frac{3}{2}\))m/r].
The presence of the mysterious zero-frequency solution can now be understood by recalling the earlier mystery of the physical interpretation of the angular momentum’s L t component. Our results come from calculating parallel transport, and parallel transport is a purely geometric process, so it gives the same result regardless of the physical nature of the four-vector. Suppose that we had instead chosen the velocity four-vector as our guinea pig. The definition of a geodesic is that it parallel-transports its own tangent vector, so the velocity vector has to stay constant. If we inspect the eigenvector corresponding to the zero-frequency eigenfrequency, we find a timelike vector that is parallel to the velocity four-vector. In our 2+1-dimensional space, the other two eigenvectors, which are spacelike, span the subspace of spacelike vectors, which are the ones that can physically be realized as the angular momentum of a gyroscope. These two eigenvectors, which vary as e ±i\(\Omega\) , can be superposed to make real-valued spacelike solutions that match the initial conditions, and these lag the rotation of the basis vectors by \(\Delta \Omega\) = \(\frac{3}{2}\)mr. This is greater than the purely spatial result by a factor of \(\frac{3}{2}\). The resulting precession angle, over n orbits of Gravity Probe B, is \(\frac{3 \pi nGm}{c^{2} r}\) = 3 × 10 −5 radians, in excellent agreement with experiment.
One will see apparently contradictory statements in the literature about whether Thomas precession occurs for a satellite: “The Thomas precession comes into play for a gyroscope on the surface of the Earth . . . , but not for a gyroscope in a freely moving satellite.” 6 But: “The total effect, geometrical and Thomas, gives the well-known Fokker-de Sitter precession of \(\frac{3 \pi m}{r}\), in the same sense as the orbit.” 7 The second statement arises from subtracting the purely spatial result from the 2+1-dimensional result, and noting that the absolute value of this difference is the same as the Thomas precession that would have been obtained if the gyroscope had been whirled at the end of a rope. In my opinion this is an unnatural way of looking at the physics, for two reasons. (1) The signs don’t match, so one is forced to say that the Thomas precession has a different sign depending on whether the rotation is the result of gravitational or nongravitational forces. (2) Referring to observation, it is clearly artificial to treat the spatial curvature and Thomas effects separately, since neither one can be disentangled from the other by varying the quantities n, m, and r. For more discussion, see tinyurl.com/me3qf8o.
Orbits
The main event of Newton’s Principia Mathematica is his proof of Kepler’s laws. Similarly, Einstein’s first important application in general relativity, which he began before he even had the exact form of the Schwarzschild metric in hand, was to find the non-Newtonian behavior of the planet Mercury. The planets deviate from Keplerian behavior for a variety of Newtonian reasons, and in particular there is a long list of reasons why the major axis of a planet’s elliptical orbit is expected to gradually rotate. When all of these were taken into account, however, there was a remaining discrepancy of about 40 seconds of arc per century, or 6.6 × 10 −7 radians per orbit. The direction of the effect was in the forward direction, in the sense that if we view Mercury’s orbit from above the ecliptic, so that it orbits in the counterclockwise direction, then the gradual rotation of the major axis is also counterclockwise.
As a very rough hand-wavy explanation for this effect, consider the spatial part of the curvature of the spacetime surrounding the sun. This spatial curvature is positive, so a circle’s circumference is less than 2\(\pi\) times its radius. We could imagine that this would cause Mercury to get back to a previously visited angular position before it has had time to complete its Newtonian cycle of radial motion. Arguments such as this one, however, should not be taken too seriously. A mathematical analysis is required.
Based on the examples in section 5.5 , we expect that the effect will be of order \(\frac{m}{r}\), where m is the mass of the sun and r is the radius of Mercury’s orbit. This works out to be 2.5 × 10 −8 , which is smaller than the observed precession by a factor of about 26.
Conserved Quantities
If Einstein had had a computer on his desk, he probably would simply have integrated the motion numerically using the geodesic equation. But it is possible to simplify the problem enough to attack it with pencil and paper, if we can find the relevant conserved quantities of the motion. Nonrelativistically, these are energy and angular momentum.
Consider a rock falling directly toward the sun. The Schwarzschild metric is of the special form
\[ds^{2} = h(r) dt^{2} - k(r) dr^{2} - \ldots\]
The rock’s trajectory is a geodesic, so it extremizes the proper time s between any two events fixed in spacetime, just as a piece of string stretched across a curved surface extremizes its length. Let the rock pass through distance r 1 in coordinate time t 1 , and then through r 2 in t 2 . (These should really be notated as \(\Delta\)r 1 , . . . or dr 1 , . . . , but we avoid the \(\Delta\)’s or d’s for convenience.) Approximating the geodesic using two line segments, the proper time is
\[\begin{split} s &= s_{1} + s_{2} \\ &= \sqrt{h_{1} t_{1}^{2} - k_{1} r^{2}_{1}} + \sqrt{h_{2} t_{2}^{2} - k_{2} r^{2}_{2}} \\ &= \sqrt{h_{1} t_{1}^{2} - k_{1} r^{2}_{1}} + \sqrt{h_{2} (T - t_{1})^{2} - k_{2} r^{2}_{2}}, \end{split}\]
where T = t 1 + t 2 is fixed. If this is to be extremized with respect to t 1 , then \(\frac{ds}{dt_{1}}\) = 0, which leads to
\[0 = \frac{h_{1} t_{1}}{s_{1}} - \frac{h_{2} t_{2}}{s_{2}},\]
which means that
\[h \frac{dt}{ds} = g_{tt} \frac{dx^{t}}{ds} = \frac{dx_{t}}{ds}\]
is a constant of the motion. Except for an irrelevant factor of m, this is the same as p t , the timelike component of the covariant momentum vector. We’ve already seen that in special relativity, the timelike component of the momentum four-vector is interpreted as the mass-energy E, and the quantity p t has a similar interpretation here. Note that no special assumption was made about the form of the functions h and k. In addition, it turns out that the assumption of purely radial motion was unnecessary. All that really mattered was that h and k were independent of t. Therefore we will have a similar conserved quantity p \(\mu\) any time the metric’s components, expressed in a particular coordinate system, are independent of x \(\mu\) . (This is generalized in section 7.1 .) In particular, the Schwarzschild metric’s components are independent of \(\phi\) as well as t, so we have a second conserved quantity p \(\phi\) , which is interpreted as angular momentum.
Writing these two quantities out explicitly in terms of the contravariant coordinates, in the case of the Schwarzschild spacetime, we have
\[E = \left(1 - \dfrac{2m}{r}\right) \frac{dt}{ds}\]
and
\[L = r^{2} \frac{d \phi}{ds}\]
for the conserved energy per unit mass and angular momentum per unit mass.
In interpreting the energy per unit mass E, it is important to understand that in the general-relativistic context, there is no useful way of separating the rest mass, kinetic energy, and potential energy into separate terms, as we could in Newtonian mechanics. E includes contributions from all of these, and turns out to be less than the contribution due to the rest mass (i.e., less than 1) for a planet orbiting the sun. It turns out that E can be interpreted as a measure of the additional gravitational mass that the solar system possesses as measured by a distant observer, due to the presence of the planet. It then makes sense that E is conserved; by analogy with Newtonian mechanics, we would expect that any gravitational effects that depended on the detailed arrangement of the masses within the solar system would decrease as \(\frac{1}{r^{4}}\), becoming negligible at large distances and leaving a constant field varying as \(\frac{1}{r^{2}}\).
One way of seeing that it doesn’t make sense to split E into parts is that although the equation given above for E involves a specific set of coordinates, E can actually be expressed as a Lorentz-invariant scalar (see section 7.1 ). This property makes E especially interesting and useful (and different from the energy in Newtonian mechanics, which is conserved but not frame-independent). On the other hand, the kinetic and potential energies depend on the velocity and position. These are completely dependent on the coordinate system, and there is nothing physically special about the coordinate system we’ve used here. Suppose a particle is falling directly toward the earth, and an astronaut in a space-suit is free-falling along with it and monitoring its progress. The astronaut judges the particle’s kinetic energy to be zero, but other observers say it’s nonzero, so it’s clearly not a Lorentz scalar. And suppose the astronaut insists on defining a potential energy to go along with this kinetic energy. The potential energy must be decreasing, since the particle is getting closer to the earth, but then there is no way that the sum of the kinetic and potential energies could be constant.
Perihelion Advance
For convenience, let the mass of the orbiting rock be 1, while m stands for the mass of the gravitating body.
The unit mass of the rock is a third conserved quantity, and since the magnitude of the momentum vector equals the square of the mass, we have for an orbit in the plane \(\theta = \frac{\pi}{2}\),
\[\begin{split} 1 &= g^{tt} p_{t}^{2} - g^{rr} p^{2}_{r} - g^{\phi \phi} p_{\phi}^{2} \\ &= g^{tt} p_{t}^{2} g_{rr} (p^{r})^{2} - g^{\phi \phi} p_{\phi}^{2} \\ &= \frac{1}{1 - \frac{2m}{r}} E^{2} - \frac{1}{1 - \frac{2m}{r}} \left(\dfrac{dr}{ds}\right)^{2} - \frac{1}{r^{2}} L^{2} \ldotp \end{split}\]
Rearranging terms and writing \(\dot{r}\) for \(\frac{dr}{ds}\), this becomes
\[\dot{r}^{2} = E^{2} - \left(1 - \dfrac{2m}{r}\right) \left(d\frac{1 + L^{2}}{r^{2}}\right)\]
or
\[\dot{r}^{2} = E^{2} - U^{2}\]
where
\[U^{2} = \left(1 - \frac{2m}{r}\right) \left(1 + \dfrac{L^{2}}{r^{2}}\right) \ldotp\]
There is a varied and strange family of orbits in the Schwarzschild field, including bizarre knife-edge trajectories that take several nearly circular turns before suddenly flying off. We turn our attention instead to the case of an orbit such as Mercury’s which is nearly Newtonian and nearly circular.
Nonrelativistically, a circular orbit has radius \(r = \frac{L^{2}}{m}\) and period \(T = \frac{2 \pi L^{3}}{m^{2}}\).
Relativistically, a circular orbit occurs when there is only one turning point at which \(\dot{r}\) = 0. This requires that E 2 equal the minimum value of U 2 , which occurs at
\[\begin{split} r &= \frac{L^{2}}{2m} \left(1 + \sqrt{1 - \dfrac{12m^{2}}{L^{2}}}\right) \\ &\approx \frac{L^{2}}{m} (1 - \epsilon), \end{split}\]
where \(\epsilon = 3(\frac{m}{L})^{2}\). A planet in a nearly circular orbit oscillates between perihelion and aphelion with a period that depends on the curvature of U 2 at its minimum. We have
\[\begin{split} k &= \frac{d^{2} (U^{2})}{dr^{2}} \\ &= \frac{d^{2}}{dr^{2}} \left(1 - \dfrac{2m}{r} + \dfrac{L^{2}}{r^{2}} - \dfrac{2mL^{2}}{r^{3}}\right) \\ &= - \frac{4m}{r^{3}} + \frac{6L^{2}}{r^{4}} - \frac{24 mL^{2}}{r^{5}} \\ &= 2L^{-6} m^{4} (1 + 2 \epsilon) \end{split}\]
The period of the oscillations is
\[\begin{split} \Delta s_{osc} &= 2 \pi \sqrt{\frac{2}{k}} \\ &= 2 \pi L^{3} m^{-2} (1 - 2 \epsilon) \ldotp \end{split}\]
The period of the azimuthal motion is
\[\begin{split} \Delta s_{az} &= \frac{2 \pi r^{2}}{L} \\ &= 2 \pi L^{3} m^{-2} (1 - 2 \epsilon) \ldotp \end{split}\]
The periods are slightly mismatched because of the relativistic correction terms. The period of the radial oscillations is longer, so that, as expected, the perihelion shift is in the forward direction. The mismatch is \(\epsilon \Delta\)s, and because of it each orbit rotates the major axis by an angle \(2 \pi \epsilon = 6 \pi (\frac{m}{L})^{2} = \frac{6 \pi m}{r}\). Plugging in the data for Mercury, we obtain 5.8 × 10 −7 radians per orbit, which agrees with the observed value to within about 10%. Eliminating some of the approximations we’ve made brings the results in agreement to within the experimental error bars, and Einstein recalled that when the calculation came out right, “for a few days, I was beside myself with joyous excitement.”
Further attempts were made to improve on the precision of this historically crucial test of general relativity. Radar now gives the most precise orbital data for Mercury. At the level of about one part per thousand, however, an effect creeps in due to the oblateness of the sun, which is difficult to measure precisely.
In 1974, astronomers J.H. Taylor and R.A. Hulse of Princeton, working at the Arecibo radio telescope, discovered a binary star system whose members are both neutron stars. The detection of the system was made possible because one of the neutron stars is a pulsar: a neutron star that emits a strong radio pulse in the direction of the earth once per rotational period. The orbit is highly elliptical, and the minimum separation between the two stars is very small, about the same as the radius of our sun. Both because the r is small and because the period is short (about 8 hours), the rate of perihelion advance per unit time is very large, about 4.2 degrees per year. The system has been compared in great detail with the predictions of general relativity,8 giving extremely good agreement, and as a result astronomers have been confident enough to reason in the opposite direction and infer properties of the system, such as its total mass, from the general-relativistic analysis. The system’s orbit is decaying due to the radiation of energy in the form of gravitational waves, which are predicted to exist by relativity.
Deflection of Light
As discussed in section 5.5 , one of the first tests of general relativity was Eddington’s measurement of the deflection of rays of light by the sun’s gravitational field. The deflection measured by Eddington was 1.6 seconds of arc. For a light ray that grazes the sun’s surface, the only physically relevant parameters are the sun’s mass m and radius r. Since the deflection is unitless, it can only depend on \(\frac{m}{r}\), the unitless ratio of the sun’s mass to its radius. Expressed in SI units, this is \(\frac{Gm}{c^{2} r}\), which comes out to be about 10 −6 . Roughly speaking, then, we expect the order of magnitude of the effect to be about this big, and indeed 10 −6 radians comes out to be in the same ball-park as a second of arc. We get a similar estimate in Newtonian physics by treating a photon as a (massive) particle moving at speed c.
It is possible to calculate a precise value for the deflection using methods very much like those used to determine the perihelion advance from earlier . However, some of the details would have to be changed. For example, it is no longer possible to parametrize the trajectory using the proper time s, since a light ray has ds = 0; we must use an affine parameter. Let us instead use this an an example of the numerical technique for solving the geodesic equation, first demonstrated in section 5.9 . Modifying our earlier program, we have the following:
At line 14, we take the mass to be 1000 times greater than the mass of the sun. This helps to make the deflection easier to calculate accurately without running into problems with rounding errors. Lines 17-25 set up the initial conditions to be at the point of closest approach, as the photon is grazing the sun. This is easier to set up than initial conditions in which the photon approaches from far away. Because of this, the deflection angle calculated by the program is cut in half. Combining the factors of 1000 and one half, the final result from the program is to be interpreted as 500 times the actual deflection angle.
The result is that the deflection angle is predicted to be 870 seconds of arc. As a check, we can run the program again with m = 0; the result is a deflection of −8 seconds, which is a measure of the accumulated error due to rounding and the finite increment used for \(\lambda\).
Dividing by 500, we find that the predicted deflection angle is 1.74 seconds, which, expressed in radians, is exactly \(\frac{4Gm}{c^{2} r}\). The unitless factor of 4 is in fact the correct result in the case of small deflections, i.e., for \(\frac{m}{r}\) << 1.
Although the numerical technique has the disadvantage that it doesn’t let us directly prove a nice formula, it has some advantages as well. For one thing, we can use it to investigate cases for which the approximation \(\frac{m}{r}\) << 1 fails. For \(\frac{m}{r}\) = 0.3, the numerical technique gives a deflection of 222 degrees, whereas the weak-field approximation \(\frac{4Gm}{c^{2} r}\) gives only 69 degrees. What is happening here is that we’re getting closer and closer to the event horizon of a black hole. Black holes are the topic of section 6.3 , but it should be intuitively reasonable that something wildly nonlinear has to happen as we get close to the point where the light wouldn’t even be able to escape.
The precision of Eddington’s original test was only about ± 30%, and has never been improved on significantly with visible-light astronomy. A better technique is radio astronomy, which allows measurements to be carried out without waiting for an eclipse. One merely has to wait for the sun to pass in front of a strong, compact radio source such as a quasar. These techniques have now verified the deflection of light predicted by general relativity to a relative precision of about 10 −5 . 9
References
6 Misner, Thorne, and Wheeler, Gravitation , p. 1118
7 Rindler, Essential Relativity , 1969, p. 141
9 For a review article on this topic, see Clifford Will, “The Confrontation between General Relativity and Experiment,” relativity.livingreviews.org/...es/lrr-2006-3/.