7.11: Why It Matters- Consumer Behavior
Why learn about consumer behavior?
Please welcome a new arrival
At 1:26 a.m. this morning, in Houston’s Memorial Hermann Hospital, a consumer was born. His name is Finnegan Henry James. By the time he goes home three days later, some of America’s biggest marketers will be pursuing him with samples, coupons, and assorted freebies. Proctor & Gamble hopes its Pampers brand will win the battle for Finn’s bottom, but retailer Target has a lower-priced contender. To welcome Finn’s family, Johnson & Johnson has already sent his mother a sample of its gentle baby wash. Bristol-Myers Squibb Company is sending a free, bulky box of Enfamil baby formula.
Like no generation before, Finn enters a consumer culture surrounded by logos, labels, and messages almost from the moment of birth. As an infant, he may wear Sesame Street diapers and a miniature NBA jersey. Right away, this little boy will begin influencing his parents’ purchasing decisions–that’s what spitting out spoonfuls of baby food is all about. By the time he is twenty months old, he will start to recognize some of the thousands of brands flashed in front of him each day. Around age four, Finn will begin making decisions about how to spend his own money. At age seven, if he is anything like the typical kid, he will see some forty thousand commercials a year. [1] By the time he is twelve, he will have his own entry in the massive data banks of marketers.
Many forces are at work influencing Finn’s consumer choices from a very early age. Some of these forces are social: his parents, cousins, and play group. Some of these forces are cultural: Finn is a Texan and an American. As Finn grows and matures, his age, gender, education, economic status, life stage, and personality all play a role in his decisions as a consumer. Multiply Finn by millions of babies born in the U.S. every year, and you have new, increasingly marketing-savvy generations flooding the market.
This is Finn’s story. And if you’re living in the U.S. today, your story probably sounds a lot like his.
You Are the Target and the Hunter
Setting aside the ethics of marketing to children, the fact remains that you are a consumer living in a highly commercialized, modern society. Marketing artifacts are so woven into the fabric of our lives that many people hardly recognize them. Every year, companies and marketing organizations spend billions of dollars focused on one central goal: to influence consumers’ purchasing decisions.
As a consumer, hopefully your growing understanding of marketing is helping you see the world around you a little differently, with more and better information about the forces that are trying to influence you.
With your increasing skills as a marketer, you recognize how important it is to understand your customers if you are going to reach them effectively. Part of that is understanding the factors that influence their purchasing decisions. Once you’re educated about those influencing factors, they’ll be tools you can use to create effective marketing.
- www.apa.org/pi/families/resources/advertising-children.pdf ↵
Contributors and Attributions
- Chapter 4: Understanding Buyer Behavior, from Introducing Marketing. Authored by : John Burnett. Provided by : Global Marketing. Located at : http://solr.bccampus.ca:8001/bcc/file/ddbe3343-9796-4801-a0cb-7af7b02e3191/1/Core%20Concepts%20of%20Marketing.pdf . License : CC BY: Attributio n


8.2: Dual-Enrollment Writing Classes Should Always be Pursued
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- West Virginia University via Digital Publishing Institute and West Virginia University Libraries
Author: Caroline Wilkinson, English, New Jersey City University
Dual enrollment is a program that allows a high-school student to take a college course and receive credit for both high school and college. According to the most comprehensive study on dual enrollment, in the 2002–2003 academic year, 70% of high schools allowed students to take college courses. Depending on where the student lives, many school districts and states also pay for the student to take the college class. High schools and universities offer dual enrollment because it addresses the growing national interest in college readiness. This course would seem to be a benefit to both the high-school students, who can make an early start on their college careers, and the university that can target potential applicants for enrollment. However, dual-enrollment programs are problematic because the cultures of high school and college are so distinct that a dual-enrollment course cannot provide the institutional context of college (such as the instructor, the classroom, the campus, fellow students, the technology in use, and so on).
Unlike a college composition course, a dual-enrollment course is held at a university campus or at a local high school. A college instructor or high-school instructor will teach it. The constant is that high-school students will be involved. When dual-enrollment courses began to spread in the 1990s, composition scholar David Schwalm explained that dual enrollment implies that a college class can be duplicated in a different environment when, in fact, it cannot because of the disparity between a high-school setting and a college one. Institutional norms at the secondary level, such as shorter class times and a higher number of students per class, impact the pedagogy of a dual-enrollment writing course when it occurs at the site of the high school.
College composition courses usually focus on writing as a process so that a student will go through the actions of drafting and revising one or more texts throughout the semester. Many composition courses focus on peer review, where students read each other’s work and give guided feedback to one another, and on conferences, where the student meets with a professor one-on-one to go over what the student can most improve. This drafting and revision process functions well in the college environment, where class times can be flexible. However, at the high-school level, class times can typically run 45 minutes with seven periods every single day. Thus, there is less time for class discussion in dual-enrollment courses. Since the class meets more during the week than a college class would, the amount of reading or homework required changes because of the institutional environment. In a dual-enrollment course, it is nearly impossible to conference with students one-on-one because the instructor has to be in the classroom at all times to watch over all the students. The instructor cannot cancel class so that conferences can occur; therefore, the way that writing is taught is different because of the context of the classroom.
If the main issues with dual enrollment are institutional, then it would make sense to offer dual-enrollment writing courses at the sponsoring university instead of the high school. This would negate some of the problems with timing and number of students that impact pedagogy. What the change in environment would not account for is the maturity of the students in the class. Composition scholars Kara Taczak and William H. Thelin studied the impact of adding high-school students to a mixed composition course on campus with college students. Although most of these dual-enrollment high-school students performed well academically, they were also considered a distraction to the instructor and college students. The high-school students’ familiarity with each other meant they behaved in ways that potentially disrupted the norms of a first-year college classroom that would otherwise be filled with students who do not know each other. Although the institution of instruction had changed to an on-campus classroom, the culture of high school continued.
There will be cultural issues in dual-enrollment courses because of the varied factors unique to these kinds of courses that impact the teaching of writing. In order to save money, most universities and high schools enter a partnership in a dual-enrollment program so that a high-school instructor teaches the course. Efforts at collaborative partnerships between high schools and colleges frequently leave the dual-credit writing instructor in a precarious space because this instructor will represent the college curriculum and culture to high-school students. This high-school instructor will need to be trained effectively by the sponsoring university. Ideally, this training will ensure that the instructor understands and possesses a specific curriculum to teach, is well versed in the objectives of the specific composition course, and employs the composition assessment policy of the university. If the university fails to provide the instructor with this training, the course will then fail the students because it will not represent the same culture of writing as a course taught on campus.
Additionally, high-school instructors may not have any experience in teaching college-level writing courses; most have experience only in teaching literature, creative writing, and speech or journalism courses. This lack of experience teaching at the university level—compounded by a lack of training from the university— could result in conflicting information being relayed to students about what college-level writing means. To complicate matters even more, students who earn dual enrollment writing credit will typically not need to take the first introductory writing course at a college; as a result, they will remain unfamiliar with the writing expectations of that college until later courses.
Alternatives in Dual-Enrollment Writing Courses
If dual-enrollment programs are here to stay, as indicated by the number of students that participate in them, then there can be more effective ways to prepare students for writing at the college level. There are already collaborative relationships between high schools and colleges throughout the United States in the form of the National Writing Project (NWP). The NWP offers professional development for educators in writing for local schools, districts, and higher education. Faculty from universities and K–12 schools co-direct sites in their local area. This collaboration has been successful in that there are nearly 200 sites in all 50 states. One way to make dual-enrollment programs more effective is for colleges to create better collaborative relationships with local high schools building on the sites of the NWP.
Colleges and high schools need to be in closer conversation with one another about the purposes and curriculum of the dual-enrollment courses. They can do so by the college sharing its composition objectives, curriculum, sample syllabi, sample schedules, and sample assignments with the schools. The National Alliance of Concurrent Enrollment Partnerships (NACEP) serves as a national accrediting body for dual-enrollment partnerships. NACEP works to make sure that the dual-enrollment courses are just as rigorous as the on-campus college courses by applying measurable criteria in five categories: curriculum, faculty, students, assessment, and program evaluation. Currently, NACEP has accredited 98 dual-enrollment programs according to these standards. Dual-enrollment programs that want to be more effective can concentrate on the following standards recommended by NACEP:
- Courses reflect the theoretical and pedagogical orientation of the sponsoring department at the university.
- University faculty members perform classroom observations to demonstrate validity of the class.
- High-school instructors meet the standards required of teaching a course at the college.
- Professional development activities occur that offer both the college and high school a way to have conversations on what writing means and how it functions in a dual-enrollment class.
These standards provide college and high-school instructors with strategies that will not only help them to address the institutional and cultural issues of dual enrollment, but also instruct them how to work within this setting to create a beneficial learning environment.
In order to make these partnerships effective, there should be one contact at the university, such as the Writing Program Administrator, that high-school instructors can come to with questions about materials and institutional knowledge of composition. This same contact would also observe the instructors who teach dual-enrollment at the high-school level and provide feedback on the classroom. These liaisons could also provide high-school instructors with representative assignments for college-level writing courses so that high-school instructors do not end up teaching a literature course as a composition course.
Colleges could provide a more collaborative relationship with a high school through program work in the dual-enrollment writing program. Respect for where students are coming from in their writing is central to composition. There should also be respect for other levels of educators by college instructors. One way to create a partnership between a college and high school is to provide funding for high-school instructors to participate. Christine Farris has started a program at Indiana University, where she offers 35-hour seminars in the summer that introduce high-school teachers to current methods in college composition and ways to teach the composition course in a manner that is consistent with the university approach. The high-school teachers are also funded for the summer seminar and participate in the fall and spring colloquia and classroom site visits. This set of support mechanisms establishes a clearer idea of what is expected of college writing in the dual-enrollment course and ensures high-school instructors have a relationship with the college writing program. Dual-enrollment programs are seen as ways to make money by many universities because students or high schools have to pay a discounted tuition in order to take them. It seems only fair that the university should pay for the professionalization of the high-school instructors who are teaching this course.
Although the contexts of high schools and colleges will be different no matter what, dual-enrollment instructors can also reflect on ways to prepare students more in class, such as having group conferences so students have the experience of discussing their revision plans, and taking a field trip to the university campus so students can see where the library and writing center are. These experiences provide another way to show students they are in a college class that differs from high school. Drawing upon the relationships formed in the NWP, following guidelines under NACEP, mandating one person as the contact for dual enrollment in the department at the university, and providing professionalization benefits for dual-enrollment instructors will create more effective dual-enrollment writing programs.
Further Reading
For more about the standards and assessment of dual-enrollment writing programs, see the Council of Writing Program Administrators’ “CWPA Position Statement on Pre-College Credit for Writing.” This statement discusses pedagogy, student readiness, and curriculum for dual enrollment compared to Advanced Placement and other pre-college credit programs. For more information on these programs, see Rob Jenkins’ “Advanced Placement vs. Dual Enrollment.” Additionally, see the National Alliance of
Concurrent Enrollment Partnerships’ “Statement of National Concurrent Enrollment Partnership Standards” for more understanding of how to assess an effective dual-enrollment writing program.
For more about how writing scholars are reflecting on the growing number of dual-enrollment programs, see Kristine Hansen and Christine Farris’ College Credit for Writing in High School: The ‘Taking Care of’ Business (National Council of Teachers of English). This edited collection explains the economics of dual-enrollment writing programs and the institutional and cultural issues that face dual enrollment. Additionally, see Kara Taczak and William H. Thelin’s “(Re)Envisioning the Divide: The Impact of College Courses on High School Students” in Teaching English in the Two Year College and Howard Tinberg and Jean Paul Nadeau’s “Contesting the Space between High School and College in the Era of Dual-Enrollment” in College Composition and Communication .
Keywords
college-level writing, composition/rhetoric, dual enrollment, writing studies, writing-program administration
Author Bio
Caroline Wilkinson is an assistant professor of English at New Jersey City University. She writes about dual-enrollment writing courses and the relationship between high schools and colleges. She also writes about the portrayals of what college-level writing might mean.


8.1: Motor Drives
A frequently-used component in any modern industry are motor drives, sometimes called variable frequency drives or VFDs. In order to understand how VFDs control the speed of a motor, you must first understand what happens inside an AC induction motor.
Below, Jim Pytel explains an example motor drive by Omron. Use this information as an introduction to the topic. There are MANY different types of motor drives, and they all have their own little idiosyncrasies. Even VFDs made by the same manufacturer may have very different programming steps. READ THE MANUAL for the drive you’re working with. As you become more familiar with these, you’ll be able to quickly find the information you need to set-up your particular drive.


1.74: Normal subgroup
A subgroup H of a group G is normal in G (H \(\triangleleft\) G) if gH = Hg for any g ∈G. Equivalently, H ⊂ G is normal if and only if gHg -1 = H for any g ∈G, i.e., if and only if each conjugacy class of G is either entirely inside H or entirely outside H. This is equivalent to say that H is invariant under all inner automorphisms of G.
The property gH = Hg means that left and rights cosets of H in G coincide. From this one sees that the cosets form a group with the operation g 1 H * g 2 H = g 1 g 2 H which is called the factor group or quotient group of G by H , denoted by G/H .
In the special case that a subgroup H has only two cosets in G (namely H and gH for some g not contained in H ), the subgroup H is always normal in G .


15.7: Community, Development, and Broadcast Media
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- Jennifer Hasty, David G. Lewis, & Marjorie M. Snipes
- OpenStax
Learning Objectives
By the end of this section, you will be able to:
- Explain how radio is associated with different themes and audiences than print media.
- Define the concept of community radio.
- Explain how community radio gives expression to local forms of identity and social action.
- Define the notion of Indigenous media.
The media scholar Marshall McLuhan is famous for his aphorism “ the medium is the message ” (1964, 23). What he meant by this is that each genre of media has its own set of features that suggest certain uses and kinds of content. In contrast to print media, radio allows for real-time talk and discussion as well as music. Radio reaches beyond the limited audience of avid readers who have time to focus on text to wider audiences of listeners who may be too busy to read or have not had access to formal education. As an oral medium, radio lends itself more readily to linguistic diversity. In places where many languages are spoken, often the language of state is the only one that circulates in written form, while the rest function as spoken languages only. Print media may therefore be limited to dominant languages, while the oral genre of radio can provide content in alternative and even multiple languages. Finally, while reading print media is largely a solitary and silent experience, radio provides a shared and noisy experience. A personal experience shared by Jennifer Hasty illustrates this.
In Ghana, I could nearly always hear a radio blasting from someone’s compound or kiosk or car. Radio was woven into daily life, a sort of auditory backdrop to everyday work and leisure. News headlines were read out each day on the morning talk shows, generating discussions in households and buses as people made their way to work. On the popular radio talk shows, Ghanaians from all walks of life called in to broadcast their perspectives on the issues of the day. Even during the music shows, listeners participated with heartfelt requests dedicated to friends, lovers, and family members.
Because of its distinctive features, the genre of radio is not as narrowly focused as print media on themes of political economy such as nationalism and democracy. While including attention to current events, radio typically provides listeners with a wider variety of content, including music, talk shows, drama, and quiz shows. In an effort to provide relevant content to the broadest spectrum of listeners in an area, local radio stations design their programming to reflect the tastes and issues of particular communities. Certainly, print media does this to some degree, but the audience for print media constitutes a narrower segment of the community. Radio attempts to address the community as a whole.
Commercial and state radio are dominant forces in the media landscapes of most countries, but an alternative form, community radio , has been growing rapidly in recent decades. Community radio refers to radio stations that are community owned and operated, staffed by groups of professionals and volunteers. The involvement of local volunteers allows for community participation in programming, production, and on-air performance. Community radio stations often focus on local current events, educational programs, and development initiatives. Typically, they are low wattage with minimal range and thus are nonprofit.
In places where people want to start up a community radio station but lack the capital and technological know-how, nongovernmental organizations (NGOs) and development organizations have provided support in partnership with community organizations. Such collaborations between community groups and foreign NGOs have made possible the start-up of community radio stations in many countries, including Nepal, Sri Lanka, and the Philippines. Throughout Africa, community radio stations sponsored by government and/or NGOs have been used to educate rural peoples about farming methods and spread public interest health messages. In Thailand, the global leader in community radio, more than 7,000 independent radio stations have been started since 2001.
Established in 1997, the Nepalese station Radio Sagarmatha was the first independent community radio station in South Asia. The station was started by the Nepal Forum of Environmental Journalists in an effort to break the government monopoly on radio and provide better coverage of community issues. Regulated by government, Radio Sagarmatha is not allowed to address political or economic issues. Focusing instead on community development, the station features daily discussion programs addressing such issues as public health, education, women’s empowerment, and workers’ concerns. Though not explicitly political, the station identifies itself as the defender of democracy and free speech in Nepal, giving voice to the people. In 2005, the army raided the studio, seizing equipment and arresting staff for rebroadcasting a BBC interview with a politician. The station reemerged after the incident and remains on air today. With 2.5 million regular listeners, Radio Sagarmatha is one of the largest and most successful community radio stations in the world.
Community radio stations in Brazil have faced similar forms of government regulation and harassment. Anthropologist Derek Pardue (2011) describes the expansion of community radio in the wake of political liberalization in the 1980s. As of 2013, there were 4,700 community radio stations operating in Brazil, an increase of 70 percent since 2002. Moreover, approximately 5,000 such stations have been shut down by government, their equipment confiscated and management prosecuted as felons. Associated with free speech and political activism, community radio attracts involvement by counterculture artists and performers such as the hip-hop communities of impoverished favela neighborhoods in São Paulo. Through community radio, local hip-hop artists narrate their stories of hardship and heroism, defining their spatially marginal neighborhoods as economically and politically marginalized periferias (peripheries). Pardue describes how community radio gives hip-hop artists and other community members a platform for demonstrating their awareness of social issues and command of information. Using slang that signals racial and class identities, they publicize otherwise unreported events and perspectives such as police violence and gang activities, providing a much more inclusive public sphere than commercial media.
In Australia, more than 400 independent radio stations broadcast in 70 different community languages. Many of these community radio stations have been started by Indigenous Australian communities as a means of cultural survival and language preservation. Indigenous media refers to the use of media by Indigenous communities for community identity, cultural representation, and activism. In the 1990s, some Indigenous broadcasters developed the ability to link community radio stations together in regional and national networks. As many Indigenous community stations featured call-in request programs, the linking of stations allowed a person in one community to publicly greet a relative in a faraway community with a song dedication. Anthropologist Daniel Fisher (2009) describes how radio has become a way for Indigenous Australian people to celebrate kinship connections in the context of kin dispersal due to government policy, travel, work, and incarceration. Throughout the 20th century, Indigenous children were seized from their families and sent to state institutions and foster homes in order to assimilate them into White Australian settler culture. In the present, Indigenous family ties are further troubled by the disproportionate numbers of young men incarcerated in Australian prisons. In this context, call-in request shows have become wildly popular on Indigenous radio networks, as relatives phone in to dedicate emotionally charged songs about love, separation, and loss to relatives in distant places.
Inspired by a wide variety of social issues, community radio is catching hold in the United States as well. In response to the domination of American radio by large media corporations, the US Congress passed the Local Community Radio Act in 2010, authorizing the Federal Communications Commission (FCC) to provide licenses to low-powered community radio stations. A group of community organizers in Madrid, New Mexico, just outside of Albuquerque, was awarded a license and began broadcasting KMRD 96.9 in 2015. As an alternative to commercial radio programming, KMRD, like many community stations, features more diverse and locally relevant content. Local DJs host call-in and talk programs about community issues and spin a wide variety of music, including alternative, pop, techno, garage, folk, and western. Local bands get frequent airplay, stimulating the local music scene. On Monday nights, you can hear a program devoted to African music, hosted by the author of this chapter. Those not within range of the station can listen to KRMD online . Over a thousand new community radio stations have emerged as a result of the Local Community Radio Act.


16.15: Endocrine System Worksheet Answers
1. Fill in the gaps in the sentences below using the words in the list.
- target; blood system; ducts; hormones
- a. Endocrine glands release their secretions directly into the blood. In other words they have no ducts .
- b. Endocrine glands secrete chemicals called hormones .
- c. Hormones are transported from the endocrine glands to all parts of the body by the blood system.
- d. Although hormones are carried throughout the body they only affect specific target organs and tissues
2. The position of endocrine organs have been indicated in red on the diagram of a composite male and female dog shown below. Add the labels in the list to the diagram.
- Ovary; Pancreas; Thyroid gland; Pituitary gland; Testis; Adrenal gland; Pineal gland; Parathyroid gland
3. On the diagram of the brain below indicate the position of the Hypothalamus and Pituitary gland.
4. In the table below list 3 hormones produced by the pituitary gland and state the function of each.
| Hormone | Function |
|---|---|
| 1. Growth hormone. | Stimulates growth of the body by increase in length of the long bones |
| 2. Oxytocin | Stimulates milk "let down" |
| 3.Follicle stimulating hormone (FSH) | Stimulates the development of the ovarian follicle. |
| Plus: Luteinising hormone(LH) | Stimulates development of the corpus luteum of the ovary |
| Plus: Antidiuretic hormone (ADH) | Stimulates the production of concentrated urine |
| Plus some others |
5. Fill in the following table with the endocrine organ the hormones are produced by.
| Hormone | Produced by: |
|---|---|
| Insulin | Pancreas |
| Progesterone | Corpus luteum |
| Oestrogen | Ovarian follicle |
| Growth hormone | (Anterior) pituitary gland |
| Adrenaline | Adrenal medulla |
| Antidiuretic hormone | (Posterior) pituitary gland |
| Testosterone | Testis |
| Aldosterone | Adrenal cortex |
| Melatonin | Pineal gland |
| Oxytocin | (Posterior) pituitary gland |
| Thyroxine | Thyroid gland |
6. Match the hormones in the list below with their functions.
- Oxytocin; Insulin; Oestrogen; Growth hormone; Antidiuretic hormone; Testosterone; Adrenaline; Cortisone; Melatonin; Progesterone; Thyroxine; Luteinising hormone; Follicle stimulating hormone
| Hormone | Function |
|---|---|
| Follicle stimulating hormone (FSH) | 1. Stimulates development of the ovarian follicle. |
| Oxytocin | 2. Stimulates milk “let down”. |
| Insulin | 3. Controls blood glucose levels. |
| Thyroxine | 4. Influences the rate of growth and development of young animals. |
| Growth hormone | 5. Stimulates the growth of long bones. |
| Antidiuretic hormone (ADH) | 6. Stimulates absorption of water from the kidney tubule. |
| Melatonin | 7. Influences the development of sexual maturity. |
| Luteinising hormone (LH) | 8. Stimulates the development of the corpus luteum. |
| Oestrogen | 9. Stimulates the development of female sexual characteristics. |
| Testosterone | 10.  Affect glucose, protein and fat metabolism. |
| Progesterone | 12. Prepares the lining of the uterus for pregnancy. |
| Adrenaline | 13. Prepares the body for emergency situations. |
7. Circle the odd one out.
- 1. melatonin ; oxytocin; growth hormone; antidiuretic hormone; follicle stimulating hormone. Melatonin is the only hormone in the list not produced by the pituitary gland.
- 2. progesterone; oestrogen; luteinising hormone; cortisone ; follicle stimulating. Cortisone is the only hormone in the list not involved in any major way with reproduction .
- 3. adrenaline; cortisone; aldosterone, oestrogen, insulin . Insulin is the only hormone in the list not produced by the adrenal gland.


6.20: Discrete Random Variables (5 of 5)
Learning Objectives
- Use probability distributions for discrete and continuous random variables to estimate probabilities and identify unusual events.
Here is another example of how to use the mean and standard deviation of a discrete random variable to identify unusual values for a random variable.
Example
Changing Majors
Here we have again the probability distribution of the number of changes in major.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P ( X ) | 0.135 | 0.271 | 0.271 | 0.180 | 0.090 | 0.036 | 0.012 | 0.003 | 0.002 |
How often do we expect a college student to change majors?
This question is asking for the expected value, which is the mean of the probability distribution. So we calculate the weighted average, as before:
What is the standard deviation of the probability distribution?
We have drawn lines to show the mean and 1 standard deviation above and below the mean.
Recall that earlier, we discussed what would be considered an unusual (and not unusual) number of changes in major, and we used probability calculations to assess that. For example, we found that changing majors 5 or more times occurs only about 5% of the time and therefore can be considered unusual.
Another way to think about defining “unusual” is to look at outcomes relative to the mean. We might consider outcomes more than 2 standard deviations above the mean as unusual.
What values are more than 2 standard deviations above the mean of 2?
Mean + 2 (standard deviation) = , which rounds to 5.
We conclude from this line of reasoning that a college student who changes majors 5 or more times is “unusual.”
In Summarizing Data Graphically and Numerically , we used the standard deviation to identify usual, or typical, values. We said that a typical range of values falls within 1 standard deviation of the mean. We can use a similar idea here.
Example
Detecting Fraud
Legitimate records often display a surprising pattern that is not present in faked tax returns or other fraudulent accounting records. In legitimate records, the distribution of first digits can be modeled using Benford’s law. For example, suppose the total income recorded on a tax return is $20,712. The first digit is 2. Now we examine a very large number of tax returns and record the first digit of total income for all of the returns. The relative frequency of each first digit will behave according to Benford’s law.
Benford’s law can also be described using a mathematical formula, but we will not go into that here. Instead, let’s double-check that this distribution meets the criteria for a probability distribution of a discrete random variable. For a randomly selected tax return, we cannot predict what the first digit will be, but the first digits behave according to a predictable pattern described by Benford’s law. The model assigns probabilities to all possible values for a first digit (notice that the first digit cannot be zero). All possible outcomes taken together have a probability of 1. You can verify this by adding together the probabilities in the table.
Here is the probability distribution for first digits based on Benford’s law shown in a histogram. The mean is approximately 3.4, with a standard deviation of about 2.5 (calculations not shown).
Now, let’s compare this distribution to real data.
The second line in the following table is the probability distribution for the first significant digit in true tax data collected by Mark Nigrini from 169,662 IRS model files. You can see that relative frequencies of first digits in the legitimate tax records follow Benford’s law very closely.
By comparison, here is the probability distribution for first digits in fraudulent tax records from a study of fraudulent cash disbursement and payroll expenditures conducted in 1995 by the district attorney’s office in Kings County, New York. For fraudulent data, the mean is approximately 5.2, with a standard deviation of about 0.9.
Obviously, the relative frequencies of first digits from the fraudulent data do not follow Benford’s law (shown again below). The distributions have very different shapes, means, and standard deviations. Compared to legitimate data, in fraudulent data, we are much more likely to see numbers with a first digit of 5 and much less likely to see numbers with a first digit of 1, 2, or 3.
Comment
When we compare the two distributions above, we can get a better understanding of the standard deviation of a random variable. The distribution in which it is more likely to find values that are further from the mean will have a larger standard deviation.
Likewise, the distribution in which it is less likely to find values that are further from the mean will have a smaller standard deviation.
In the fraudulent distribution, values like 1 or 2 that are far from the mean are very unlikely. On the other hand, in the Benford’s law distribution, the values 1 and 2 are quite likely. Indeed, the standard deviation of the Benford law is 2.5, which is larger than the standard deviation of 0.9 in the fraudulent distribution.
Try It
Use the following histograms to answer the activity question:
Let’s Summarize
- The probability of an event is a measure of the likelihood that the event occurs.
- Probabilities are always between 0 and 1. The closer the probability is to 0, the less likely the event is to occur. The closer the probability is to 1, the more likely the event is to occur.
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The two ways of determining probabilities are empirical and theoretical.
- Empirical methods use a series of trials that produce outcomes that cannot be predicted in advance (hence the uncertainty). The probability of an event is approximated by the relative frequency of the event.
- Theoretical methods use the nature of the situation to determine probabilities. Probability rules allow us to calculate theoretical probabilities.
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Some common probability rules:
- The probability of the complement of an event A can be found by subtracting the probability of A from 1: P (not A ) = 1 – P ( A )
- Events are called disjoint or mutually exclusive if they have no events in common. If A and B are disjoint events, then P ( A or B ) = P ( A ) + P ( B ).
- When the knowledge of the occurrence of one event A does not affect the probability of another event B, we say the events are independent. If A and B are independent events, then P ( A and B ) = P ( A ) · P ( B ).
- When we have a quantitative variable with outcomes that occur as a result of some random process (e.g., rolling a die, choosing a person at random), we call it a random variable.
-
There are two types of random variables:
- Discrete random variables have numeric values that can be listed and often can be counted.
- Continuous random variables can take any value in an interval and are often measurements. This type of random variable will be discussed in section 6.2.
- A probability distribution of a random variable tells us the probabilities of all the possible outcomes (for discrete random variables) of the variable or ranges of values (for continuous random variables). A probability distribution shows us the regular, predictable distribution of outcomes in a large number of repetitions of a random variable.
- For a discrete random variable, the probabilities of values are areas of the corresponding regions of the probability histogram for the variable.
Contributors and Attributions
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


4.2: The VSEPR Model
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Learning Objectives
- To use the VSEPR model to predict molecular geometries.
- To predict whether a molecule has a dipole moment.
The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the valence-shell electron-pair repulsion (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds.
The VSEPR Model
The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach.
We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the central atom , ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form groups , which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(\PageIndex{1}\) and \(\PageIndex{2}\).
In the VSEPR model, the molecule or polyatomic ion is given an AX m E n designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the bonded atoms in a molecule or polyatomic ion.
VESPR Produce to predict Molecular geometry
This VESPR procedure is summarized as follows:
- Draw the Lewis electron structure of the molecule or polyatomic ion.
- Determine the electron group arrangement around the central atom that minimizes repulsions.
- Assign an AX m E n designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations from ideal bond angles.
- Describe the molecular geometry.
We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(\PageIndex{2}\) and Figure \(\PageIndex{3}\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups.
Two Electron Groups
Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\).
AX 2 Molecules: BeH 2
1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is
3. Both groups around the central atom are bonding pairs (BP). Thus BeH 2 is designated as AX 2 .
4. From Figure \(\PageIndex{3}\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH 2 is linear .
AX 2 Molecules: CO 2
1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is
2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH 2 , the arrangement that minimizes repulsions places the groups 180° apart.
3. Once again, both groups around the central atom are bonding pairs (BP), so CO 2 is designated as AX 2 .
4. VSEPR only recognizes groups around the central atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO 2 is linear (Figure \(\PageIndex{3}\)). The structure of \(\ce{CO2}\) is shown in Figure \(\PageIndex{1}\).
Three Electron Groups
AX 3 Molecules: BCl 3
1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is
3. All electron groups are bonding pairs (BP), so the structure is designated as AX 3 .
4. From Figure \(\PageIndex{3}\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl 3 is trigonal planar , as shown in Figure \(\PageIndex{2}\).
AX 3 Molecules: CO 3 2 −
1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as
3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX 3 .
4. We see from Figure \(\PageIndex{3}\) that the molecular geometry of CO 3 2 − is trigonal planar with bond angles of 120°.
In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time.
AX 2 E Molecules: SO 2
1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below.
3. There are two bonding pairs and one lone pair, so the structure is designated as AX 2 E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(\PageIndex{4}\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO 2 , we have one BP–BP interaction and two LP–BP interactions.
4. The molecular geometry is described only by the positions of the nuclei, not by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is bent , or V shaped , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(\PageIndex{2}\) and \(\PageIndex{3}\)). The O-S-O bond angle is expected to be less than 120° because of the extra space taken up by the lone pair.
As with SO 2 , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom.
Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH 2 O (AX 3 ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°).
Four Electron Groups
One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions.
AX 4 Molecules: CH 4
1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is
2. There are four electron groups around the central atom. As shown in Figure \(\PageIndex{2}\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°.
3. All electron groups are bonding pairs, so the structure is designated as AX 4 .
4. With four bonding pairs, the molecular geometry of methane is tetrahedral (Figure \(\PageIndex{3}\)).
AX 3 E Molecules: NH 3
1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure
2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
3. With three bonding pairs and one lone pair, the structure is designated as AX 3 E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
4. There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal . In essence, this is a tetrahedron with a vertex missing (Figure \(\PageIndex{3}\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(\PageIndex{3}\) and Figure \(\PageIndex{4}\)).
AX 2 E 2 Molecules: H 2 O
1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure
3. With two bonding pairs and two lone pairs, the structure is designated as AX 2 E 2 with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles.
4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is bent , or V shaped , with an H–O–H angle that is even less than the H–N–H angles in NH 3 , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices.
Five Electron Groups
In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups.
AX 5 Molecules: PCl 5
1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl 5 is
3. All electron groups are bonding pairs, so the structure is designated as AX 5 . There are no lone pair interactions.
4. The molecular geometry of PCl 5 is trigonal bipyramidal , as shown in Figure \(\PageIndex{3}\). The molecule has three atoms in a plane in equatorial positions and two atoms above and below the plane in axial positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example.
AX 4 E Molecules: SF 4
1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(\PageIndex{2}\).
3. We designate SF 4 as AX 4 E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the equatorial position is more stable than the one with the lone pair in the axial position . We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair.
At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions.
4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a seesaw . The F axial –S–F axial angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane.
AX 3 E 2 Molecules: BrF 3
1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is
Once again, we have a compound that is an exception to the octet rule.
2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid.
3. With three bonding pairs and two lone pairs, the structural designation is AX 3 E 2 with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°:
Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons.
4. The three nuclei in BrF 3 determine its molecular structure, which is described as T shaped . This is essentially a trigonal bipyramid that is missing two equatorial vertices. The F axial –Br–F axial angle is 172°, less than 180° because of LP–BP repulsions (Figure \(\PageIndex{2}\).1).
Because lone pairs occupy more space around the central atom than bonding pairs, electrostatic repulsions are more important for lone pairs than for bonding pairs.
AX 2 E 3 Molecules: I 3 −
1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is
2. There are five electron groups about the central atom in I 3 − , two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid.
3. With two bonding pairs and three lone pairs, I 3 − has a total of five electron pairs and is designated as AX 2 E 3 . We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions.
The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles.
4. With three nuclei and three lone pairs of electrons, the molecular geometry of I 3 − is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected.
Six Electron Groups
Six electron groups form an octahedron , a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(\PageIndex{2}\).)
AX 6 Molecules: SF 6
1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is
With an expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(\PageIndex{2}\) that the geometry that minimizes repulsions is octahedral .
3. With only bonding pairs, SF 6 is designated as AX 6 . All positions are chemically equivalent, so all electronic interactions are equivalent.
4. There are six nuclei, so the molecular geometry of SF 6 is octahedral.
AX 5 E Molecules: BrF 5
1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is
With its expanded valence, this species is an exception to the octet rule.
2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure:
3. With five bonding pairs and one lone pair, BrF 5 is designated as AX 5 E; it has a total of six electron pairs. The BrF 5 structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all F axial –Br–F equatorial angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs.
4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is square pyramidal . The F axial –B–F equatorial angles are 85.1°, less than 90° because of LP–BP repulsions.
AX 4 E 2 Molecules: ICl 4 −
1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is
2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is
3. ICl 4 − is designated as AX 4 E 2 and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles.
4. With five nuclei, the ICl4− ion forms a molecular structure that is square planar , an octahedron with two opposite vertices missing.
The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(\PageIndex{6}\).
Figure \(\PageIndex{6}\): Overview of Molecular Geometries
Example \(\PageIndex{1}\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
- PF 5 (phosphorus pentafluoride, a catalyst used in certain organic reactions)
- H 3 O + (hydronium ion)
Given: two chemical species
Asked for: molecular geometry
Strategy:
- Draw the Lewis electron structure of the molecule or polyatomic ion.
- Determine the electron group arrangement around the central atom that minimizes repulsions.
- Assign an AX m E n designation; then identify the LP–LP, LP–BP, or BP–BP interactions and predict deviations in bond angles.
- Describe the molecular geometry.
Solution:
-
A
The central atom, P, has five valence electrons and each fluorine has seven valence electrons, so the Lewis structure of PF
5
is
C All electron groups are bonding pairs, so PF 5 is designated as AX 5 . Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal.
D The PF 5 molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal.
-
A
The central atom, O, has six valence electrons, and each H atom contributes one valence electron. Subtracting one electron for the positive charge gives a total of eight valence electrons, so the Lewis electron structure is
B There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH 3 , repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron.
C With three bonding pairs and one lone pair, the structure is designated as AX 3 E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron.
D There are three nuclei and one lone pair, so the molecular geometry is trigonal pyramidal , in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions:
Exercise \(\PageIndex{1}\)
Using the VSEPR model, predict the molecular geometry of each molecule or ion.
- XeO 3
- PF 6 −
- NO 2 +
- Answer a
-
trigonal pyramidal
- Answer b
-
octahedral
- Answer c
-
linear
Example \(\PageIndex{2}\)
Predict the molecular geometry of each molecule.
- XeF 2
- SnCl 2
Given: two chemical compounds
Asked for: molecular geometry
Strategy:
Use the strategy given in Example\(\PageIndex{1}\).
Solution:
-
A
Xenon contributes eight electrons and each fluorine seven valence electrons, so the Lewis electron structure is
B There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid.
C From B, XeF 2 is designated as AX 2 E 3 and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial:
The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I 3 − . All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle.
D With two nuclei about the central atom, the molecular geometry of XeF 2 is linear. It is a trigonal bipyramid with three missing equatorial vertices.
-
A
The tin atom donates 4 valence electrons and each chlorine atom donates 7 valence electrons. With 18 valence electrons, the Lewis electron structure is
B There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other.
C From B we designate SnCl 2 as AX 2 E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions.
D With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl 2 is bent, like SO 2 , but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing.
Exercise \(\PageIndex{2}\)
Predict the molecular geometry of each molecule.
- SO 3
- XeF 4
- Answer a
-
trigonal planar
- Answer b
-
square planar
Molecules with No Single Central Atom
The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AX m E n fragments. We will demonstrate with methyl isocyanate (CH 3 –N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops.
We can treat methyl isocyanate as linked AX m E n fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX 4 . We can therefore predict the CH 3 –N portion of the molecule to be roughly tetrahedral, similar to methane:
The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair:
Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°.
The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure:
Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO 2 , so its geometry, like that of CO 2 , is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules.
Example \(\PageIndex{3}\)
Use the VSEPR model to predict the molecular geometry of propyne (H 3 C–C≡CH), a gas with some anesthetic properties.
Given: chemical compound
Asked for: molecular geometry
Strategy:
Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(\PageIndex{3}\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole.
Solution:
Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°.
Exercise \(\PageIndex{3}\)
Predict the geometry of allene (H 2 C=C=CH 2 ), a compound with narcotic properties that is used to make more complex organic molecules.
- Answer
-
The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°.
Molecular Dipole Moments
You previously learned how to calculate the dipole moments of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are vectors ; they possess both a magnitude and a direction . The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO 2 , a linear molecule (Figure \(\PageIndex{8a}\)). Each C–O bond in CO 2 is polar, yet experiments show that the CO 2 molecule has no dipole moment. Because the two C–O bond dipoles in CO 2 are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO 2 molecule has no net dipole moment even though it has a substantial separation of charge. In contrast, the H 2 O molecule is not linear (Figure \(\PageIndex{8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H 2 O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H 2 O to hydrogen-bond to other polarized or charged species, including other water molecules.
Other examples of molecules with polar bonds are shown in Figure \(\PageIndex{9}\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl 3 is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment.
Example \(\PageIndex{4}\)
Which molecule(s) has a net dipole moment?
- \(\ce{H2S}\)
- \(\ce{NHF2}\)
- \(\ce{BF3}\)
Given: three chemical compounds
Asked for: net dipole moment
Strategy:
For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment.
Solution:
-
The total number of electrons around the central atom, S, is eight, which gives four electron pairs. Two of these electron pairs are bonding pairs and two are lone pairs, so the molecular geometry of \(\ce{H2S}\) is bent (Figure \(\PageIndex{6}\)). The bond dipoles cannot cancel one another, so the molecule has a net dipole moment.
-
Difluoroamine has a trigonal pyramidal molecular geometry. Because there is one hydrogen and two fluorines, and because of the lone pair of electrons on nitrogen, the molecule is not symmetrical, and the bond dipoles of NHF
2
cannot cancel one another. This means that NHF
2
has a net dipole moment. We expect polarization from the two fluorine atoms, the most electronegative atoms in the periodic table, to have a greater affect on the net dipole moment than polarization from the lone pair of electrons on nitrogen.
- The molecular geometry of BF 3 is trigonal planar. Because all the B–F bonds are equal and the molecule is highly symmetrical, the dipoles cancel one another in three-dimensional space. Thus BF 3 has a net dipole moment of zero:
Exercise \(\PageIndex{4}\)
Which molecule(s) has a net dipole moment?
- \(\ce{CH3Cl}\)
- \(\ce{SO3}\)
- \(\ce{XeO3}\)
- Answer
-
\(\ce{CH3Cl}\) and \(\ce{XeO3}\)
Summary
Lewis electron structures give no information about molecular geometry , the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The valence-shell electron-pair repulsion (VSEPR) model allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AX m E n designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and m and n are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles . From this we can describe the molecular geometry . The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds.
Molecules with polar covalent bonds can have a dipole moment , an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero.


14.3: Self-Inductance and Inductors
By the end of this section, you will be able to:
- Correlate the rate of change of current to the induced emf created by that current in the same circuit
- Derive the self-inductance for a cylindrical solenoid
- Derive the self-inductance for a rectangular toroid
Mutual inductance arises when a current in one circuit produces a changing magnetic field that induces an emf in another circuit. But can the magnetic field affect the current in the original circuit that produced the field? The answer is yes, and this is the phenomenon called self-inductance .
Inductors
Figure \(\PageIndex{1}\) shows some of the magnetic field lines due to the current in a circular loop of wire. If the current is constant, the magnetic flux through the loop is also constant. However, if the current I were to vary with time—say, immediately after switch S is closed—then the magnetic flux \(\Phi_m\) would correspondingly change. Then Faraday’s law tells us that an emf \(\epsilon\) would be induced in the circuit, where
\[\epsilon = - \frac{d\Phi_m}{dt} \label{14.6}. \]
Since the magnetic field due to a current-carrying wire is directly proportional to the current, the flux due to this field is also proportional to the current; that is,
\[\Phi_m \propto I. \label{14.7} \]
This can also be written as
\[\Phi_m = LI \label{14.8} \]
where the constant of proportionality L is known as the self-inductance of the wire loop. If the loop has N turns, this equation becomes
\[\boxed{N\Phi_m = LI} \label{14.9} \]
By convention, the positive sense of the normal to the loop is related to the current by the right-hand rule, so in Figure \(\PageIndex{1}\), the normal points downward. With this convention, \(\Phi_m\) is positive in Equation \ref{14.9}, so L always has a positive value .
For a loop with N turns, \(\epsilon = - Nd\Phi_m/dt\), so the induced emf may be written in terms of the self-inductance as
\[\boxed{\epsilon = - L\frac{dI}{dt}.} \label{14.10} \]
When using this equation to determine L , it is easiest to ignore the signs of \(\epsilon\) and \(dI/dt\), and calculate L as
\[L = \frac{|\epsilon|}{|dI/dt|}. \nonumber \]
Since self-inductance is associated with the magnetic field produced by a current, any configuration of conductors possesses self-inductance. For example, besides the wire loop, a long, straight wire has self-inductance, as does a coaxial cable. A coaxial cable is most commonly used by the cable television industry and may also be found connecting to your cable modem. Coaxial cables are used due to their ability to transmit electrical signals with minimal distortions. Coaxial cables have two long cylindrical conductors that possess current and a self-inductance that may have undesirable effects.
A circuit element used to provide self-inductance is known as an inductor . It is represented by the symbol shown in Figure \(\PageIndex{2}\), which resembles a coil of wire, the basic form of the inductor. Figure \(\PageIndex{3}\) shows several types of inductors commonly used in circuits.
In accordance with Lenz’s law, the negative sign in Equation \ref{14.10} indicates that the induced emf across an inductor always has a polarity that opposes the change in the current. For example, if the current flowing from A to B in Figure \(\PageIndex{4a}\) were increasing, the induced emf (represented by the imaginary battery) would have the polarity shown in order to oppose the increase. If the current from A to B were decreasing, then the induced emf would have the opposite polarity, again to oppose the change in current (Figure \(\PageIndex{4b}\)). Finally, if the current through the inductor were constant, no emf would be induced in the coil.
One common application of inductance is to allow traffic signals to sense when vehicles are waiting at a street intersection. An electrical circuit with an inductor is placed in the road underneath the location where a waiting car will stop. The body of the car increases the inductance and the circuit changes, sending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and a receiver. The pulsed signal from the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path (Figure \(\PageIndex{5}\)). Metal detectors can be adjusted for sensitivity and can also sense the presence of metal on a person.
Large induced voltages are found in camera flashes . Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. Recall from Oscillations on oscillations that “oscillation” is defined as the fluctuation of a quantity, or repeated regular fluctuations of a quantity, between two extreme values around an average value. Also recall (from Electromagnetic Induction on electromagnetic induction) that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil. The oscillator system does this many times as the battery voltage is boosted to over 1000 volts. (You may hear the high-pitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash.
An induced emf of 2.0 V is measured across a coil of 50 closely wound turns while the current through it increases uniformly from 0.0 to 5.0 A in 0.10 s. (a) What is the self-inductance of the coil? (b) With the current at 5.0 A, what is the flux through each turn of the coil?
Strategy
Both parts of this problem give all the information needed to solve for the self-inductance in part (a) or the flux through each turn of the coil in part (b). The equations needed are Equation \ref{14.10} for part (a) and Equation \ref{14.9} for part (b).
Solution
- Ignoring the negative sign and using magnitudes, we have, from Equation \ref{14.10}, \[L = \frac{\epsilon}{dI/dt} = \frac{2.0 \, V}{5.0 \, A/0.10 \, s} = 4.0 \times 10^{-2} H. \nonumber \]
- From Equation \ref{14.9}, the flux is given in terms of the current by \(\Phi_m = LI/N\), so \[\Phi_m = \frac{(4.0 \times 10^{-2} H)(5.0 \, A)}{50 \, turns)} = 4.0 \times 10^{-3} Wb. \nonumber \]
Significance
The self-inductance and flux calculated in parts (a) and (b) are typical values for coils found in contemporary devices. If the current is not changing over time, the flux is not changing in time, so no emf is induced.
Current flows through the inductor in Figure \(\PageIndex{4}\) from B to A instead of from A to B as shown. Is the current increasing or decreasing in order to produce the emf given in diagram (a)? In diagram (b)?
- Answer
-
a. decreasing; b. increasing; Since the current flows in the opposite direction of the diagram, in order to get a positive emf on the left-hand side of diagram (a), we need to decrease the current to the left, which creates a reinforced emf where the positive end is on the left-hand side. To get a positive emf on the right-hand side of diagram (b), we need to increase the current to the left, which creates a reinforced emf where the positive end is on the right-hand side.
A changing current induces an emf of 10 V across a 0.25-H inductor. What is the rate at which the current is changing?
- Answer
-
40 A/s
A good approach for calculating the self-inductance of an inductor consists of the following steps:
- Assume a current I is flowing through the inductor.
- Determine the magnetic field \(\vec{B}\) produced by the current. If there is appropriate symmetry, you may be able to do this with Ampère’s law.
- Obtain the magnetic flux, \(\Phi_m\).
- With the flux known, the self-inductance can be found from Equation \ref{14.9}, \(L = N\Phi_m/I\).
To demonstrate this procedure, we now calculate the self-inductances of two inductors.
Cylindrical Solenoid
Consider a long, cylindrical solenoid with length l , cross-sectional area A , and N turns of wire. We assume that the length of the solenoid is so much larger than its diameter that we can take the magnetic field to be \(B = \mu_0nI\) throughout the interior of the solenoid, that is, we ignore end effects in the solenoid. With a current I flowing through the coils, the magnetic field produced within the solenoid is
\[B = \mu_0 \left(\frac{N}{l} \right) I, \label{14.11} \]
so the magnetic flux through one turn is
\[\Phi_m = BA = \frac{\mu_0 NA}{l}I. \label{14.12} \]
Using Equation \ref{14.9}, we find for the self-inductance of the solenoid,
\[L_{solenoid} = \frac{N\Phi_m}{I} = \frac{\mu_0N^2A}{l}. \label{14.13} \]
If \(n = N/l\) is the number of turns per unit length of the solenoid, we may write Equation \ref{14.13} as
\[L = \mu_0 \left(\frac{N}{l}\right)^2 Al = \mu_0 n^2 Al = \mu_0 n^2 (V), \label{14.14} \]
where \(V = Al\) is the volume of the solenoid. Notice that the self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current. This is true for inductors in general.
Rectangular Toroid
A toroid with a rectangular cross-section is shown in Figure \(\PageIndex{6}\). The inner and outer radii of the toroid are \(R_1\) and \(R_2\), and \(h\) is the height of the toroid. Applying Ampère’s law in the same manner as we did in Example 13.5.2 for a toroid with a circular cross-section, we find the magnetic field inside a rectangular toroid is also given by
\[B = \frac{\mu_0NI}{2\pi r}, \nonumber \]
where r is the distance from the central axis of the toroid. Because the field changes within the toroid, we must calculate the flux by integrating over the toroid’s cross-section. Using the infinitesimal cross-sectional area element \(da = h \, dr\) shown in Figure \(\PageIndex{6}\), we obtain
\[\Phi_m = \int B \, da = \int_{R_1}^{R_2} \left(\frac{\mu_0 NI}{2\pi r}\right) (hdr) = \frac{\mu_0NhI}{2\pi}ln \frac{R_2}{R_1}. \label{14.16} \]
Now from Equation \ref{14.16}, we obtain for the self-inductance of a rectangular toroid
\[L = \frac{N\Phi_m}{I} = \frac{\mu_0N^2h}{2\pi}ln \frac{R_2}{R_1}. \label{14.17} \]
As expected, the self-inductance is a constant determined by only the physical properties of the toroid.
(a) Calculate the self-inductance of a solenoid that is tightly wound with wire of diameter 0.10 cm, has a cross-sectional area of \(0.90 \, cm^2\), and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid?
- Answer
-
a. \(4.5 \times 10^{-5} H\); b. \(4.5 \times 10^{-3} V\)
(a) What is the magnetic flux through one turn of a solenoid of self-inductance \(8.0 \times 10^{-5} H\) when a current of 3.0 A flows through it? Assume that the solenoid has 1000 turns and is wound from wire of diameter 1.0 mm. (b) What is the cross-sectional area of the solenoid?
- Answer
-
a. \(2.4 \times 10^{-7} Wb\); b. \(6.4 \times 10^{-5} m^2\)


8.5: Collisions
An important application of the law of momentum conservation is the dynamic analysis of collisions between two objects, like the collision between two billiard balls (see Figure 8.3). We note that methods from previous chapters often cannot be easily applied because the forces and accelerations during a collision are often not accurately known. This inaccuracy is caused by the fast dynamics and high forces during collisions, that are hard to measure. During such a short collision, other forces can usually be neglected, because they are much smaller than the very high collision forces that provide the high accelerations which change the velocity vectors in a short time. As a consequence of the absence of substantial external forces, the law of conservation of momentum Equation 8.12 holds for the system containing both balls. We will now discuss step by step how the dynamics of the collision of two balls can be analysed using momentum conservation and the CoM-frame.
8.5.1 Plane of contact and line of impact
Let us consider two smooth spherical balls \(A, B\) with identical radius, with masses \(m_{A}, m_{B}\) and velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A}\) and \(\overrightarrow{\boldsymbol{v}}_{B}\) that move in the \(x y\)-plane and collide, as shown in Figure 8.3. The motion of the balls consists of 3 phases: 1. before the collision at time \(t_{0}, 2\). during the collision at time \(t_{1}\) and 3 . after the collision at time \(t_{2}\).
We can then choose our Cartesian coordinate system such that the origin \(O\) is at the point of contact. The \(y\)-axis is chosen to be tangential to the surface of the circular disks at the point where they make contact, this surface is called the plane of contact. The \(x\)-axis is chosen perpendicular to the plane of contact and is called the line of impact. The balls \(A\) and \(B\) and velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A}\), \(\overrightarrow{\boldsymbol{v}}_{B}\) can be drawn as shown in Figure 8.3).
The analysis of collisions becomes easier if it is performed in a CoM-frame.
A CoM-frame is a reference frame that is chosen such that the velocity and acceleration of the CoM of the objects is zero.
The analysis process then proceeds along the following steps:
- Determining the velocity vector \(\overrightarrow{\boldsymbol{v}}_{G}\) of the CoM.
- Transforming to the CoM-frame by subtracting the velocity of the centre of mass \(\overrightarrow{\boldsymbol{v}}_{G}\) from all velocity vectors.
- Analysing the system in the centre of mass system.
- Transforming back to the original system by adding \(\overrightarrow{\boldsymbol{v}}_{G}\) to all velocity vectors.
8.5.2 Transformation to the CoM-frame
Since there are no external forces acting on the balls (we neglect gravity), it is a closed system in which momentum conservation holds, such that the velocity \(\overrightarrow{\boldsymbol{v}}_{G}\) of the centre of mass \((\mathrm{CoM})\) is constant. The analysis can be significantly simplified by choosing a coordinate system \(x^{\prime}, y^{\prime}, z^{\prime}\) that moves along with the CoM-frame, such that, measured in these coordinates, the velocity of the CoM is zero \(\overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\overrightarrow{\mathbf{0}}\). A transformation to a different coordinate system like shown in Figure 8.5 was already discussed in Equation 6.24 and is often useful for simplifying the analysis of problems, and therefore an important technique to become familiar with (see Sec. 6.12). The position and velocity vectors in the CoM-frame are found by first determining \(\overrightarrow{\boldsymbol{v}}_{G}\) (using Equation 7.41) and then calculating the velocities \(\overrightarrow{\boldsymbol{v}}^{\prime}\) in the CoM-frame by subtracting \(\overrightarrow{\boldsymbol{v}}_{G}\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{G} & =\frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}}{m_{A}+m_{B}} \tag{8.16} \label{8.16}\\[4pt] \overrightarrow{\boldsymbol{v}}_{A 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{A 0}-\overrightarrow{\boldsymbol{v}}_{G} \tag{8.17} \label{8.17}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{B 0}-\overrightarrow{\boldsymbol{v}}_{G} \tag{8.18} \label{8.18}\\[4pt] \overrightarrow{\boldsymbol{v}}_{G}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{G}-\overrightarrow{\boldsymbol{v}}_{G}=\overrightarrow{\mathbf{0}} \tag{8.19} \label{8.19}\end{align}\]
The result of this transformation to the CoM-frame can be seen in Figure 8.4.
8.5.3 Analysing the collision in the CoM-frame
First we note that in the CoM-frame the velocity vectors of both balls are always parallel, pointing in opposite directions, as can be derived from \(\overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\overrightarrow{\mathbf{0}}\) :
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{G}^{\prime}=\frac{m_{A} \overrightarrow{\boldsymbol{v}}_{A}^{\prime}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}^{\prime}}{m_{A}+m_{B}} & =\overrightarrow{\mathbf{0}} \tag{8.20} \label{8.20}\\[4pt] m_{A} \overrightarrow{\boldsymbol{v}}_{A}^{\prime}+m_{B} \overrightarrow{\boldsymbol{v}}_{B}^{\prime} & =\overrightarrow{\mathbf{0}} \tag{8.21} \label{8.21}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B}^{\prime} & =-\frac{m_{A}}{m_{B}} \overrightarrow{\boldsymbol{v}}_{A}^{\prime} \tag{8.22} \label{8.22}\end{align}\]
Now we can analyse the collision in the CoM-frame in three steps that are illustrated in Figs. 8.4):
- Before the impact, at time \(t=t_{0}\) the velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A 0}^{\prime}, \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime}\) point toward each other.
- During the impact both balls exert forces \(\overrightarrow{\boldsymbol{F}}_{A B}=-\overrightarrow{\boldsymbol{F}}_{B A}=F_{A B} \hat{\boldsymbol{\imath}}\) on each other along the \(x\)-axis and slightly deform at the contact point.
- After the collision, the energy that was stored in the deformation is converted back to kinetic energy, at \(t=t_{2}\) the velocity vectors \(\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}, \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}\) point away from each other.
 To analyse the change in momentum during the impact, we apply the principle of impulse and momentum Equation 8.9 along the \(y\) and \(x\) axes. Because the balls are smooth and frictionless, there are no forces in the \(y\)-direction tangential to the plane of contact, such that all forces act along the line of impact and are parallel to the \(x\)-axis. Since the forces and impulses along the \(y\)-axis are zero, the momentum and velocity components of both balls are conserved \(\Delta p_{y}^{\prime}=m \Delta v_{y}^{\prime}=0\) along that axis:
\[\begin{align} v_{A 2, y}^{\prime} & =v_{A 0, y}^{\prime} \tag{8.23} \label{8.23}\\[4pt] v_{B 2, y}^{\prime} & =v_{B 0, y}^{\prime} \tag{8.24} \label{8.24}\end{align}\]
Along the \(x\)-axis it is more difficult to analyse the collision, since the contact forces along this axis are unknown, and can depend on the material properties of the balls and other conditions that affect the collision. These properties are conveniently captured by the coefficient of restitution \(e\), that relates the velocity of the ball in the CoM at time \(t_{2}\) to that at \(t_{0}\) :
\[\begin{align} v_{A 2, x}^{\prime} & =-e v_{A 0, x}^{\prime} \tag{8.25} \label{8.25}\\[4pt] v_{B 2, x}^{\prime} & =-e v_{B 0, x}^{\prime} \tag{8.26} \label{8.26}\end{align}\]
Note that these equations follow from momentum conservation, ensuring that Equation 8.21 still holds after the collision. The coefficient \(e\), that can have any value between 0 and 1 , is closely related to the energy loss during the collision as we will discuss later. Combining Eqs. (8.23)-(8.26) we have:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2}^{\prime} & =-e v_{A 0, x}^{\prime} \hat{\boldsymbol{\imath}}+v_{A 0, y}^{\prime} \hat{\boldsymbol{\jmath}} \tag{8.27} \label{8.27}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime} & =-e v_{B 0, x}^{\prime} \hat{\boldsymbol{\imath}}+v_{B 0, y}^{\prime} \hat{\boldsymbol{\jmath}} \tag{8.28} \label{8.28}\end{align}\]
Finally, we transform back to the original coordinate system (Figure 8.3 at \(t_{2}\) ) by adding the CoM velocity \(\overrightarrow{\boldsymbol{v}}_{G}\) to all velocity vectors:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G} \tag{8.29} \label{8.29}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G} \tag{8.30} \label{8.30}\end{align}\]
The whole outlined procedure can also be captured in a single set of equations:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\left[-e\left(v_{A 0, x}-v_{G, x}\right)+v_{G, x}\right] \hat{\boldsymbol{\imath}}+v_{A 0, y} \hat{\boldsymbol{\jmath}} \tag{8.31} \label{8.31}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\left[-e\left(v_{B 0, x}-v_{G, x}\right)+v_{G, x}\right] \hat{\boldsymbol{\imath}}+v_{B 0, y} \hat{\boldsymbol{\jmath}} \tag{8.32} \label{8.32}\end{align}\]
Thus we have analysed the dynamics of collisions with oblique impact (impact at an angle) by only using the conservation of momentum principle, without knowing any details about the forces, except the restitution coefficient \(e\). In the special case that the balls both move parallel to the line of impact ( \(x\)-axis) they experience central impact, where all previous equations still hold with all velocity components along the \(y\)-axis being zero \(\left(v_{A 0, y}=v_{B 0, y}=0\right)\). Note that since we are dealing with spherical balls, the centres of mass of both objects always coincide with the line of impact. For non-circular objects, this is not always the case, and the impact can become eccentric such that the objects can start rotating after the collision. Also if the surface of the objects is not smooth, friction forces might generate rotations under oblique impact. Impact with rotations will be discussed in chapter Ch. 12.
Let us analyse a numerical example of the situation sketched in Figure 8.3 and Figure 8.4 with \(e=0.6, m_{A}=3 \mathrm{~kg}\) and \(m_{B}=5 \mathrm{~kg}\). First we determine the velocity vector \(\overrightarrow{\boldsymbol{v}}_{G}\) of the centre of mass:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 0} & =[3.5 \hat{\boldsymbol{\imath}}-1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.33} \label{8.33}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0} & =[-0.5 \hat{\imath}+2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.34} \label{8.34}\\[4pt] \overrightarrow{\boldsymbol{v}}_{G} & =\left[\frac{3 \cdot 3.5+5 \cdot-0.5}{3+5} \hat{\boldsymbol{\imath}}+\frac{3 \cdot-1.5+5 \cdot 2.5}{3+5} \hat{\boldsymbol{\jmath}}\right] \mathrm{m} / \mathrm{s} \tag{8.35} \label{8.35}\\[4pt] & =[1 \hat{\imath}+1 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.36} \label{8.36}\end{align}\]
Then we transform to the centre of mass system:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{A 0}-\overrightarrow{\boldsymbol{v}}_{G}=[(3.5-1) \hat{\boldsymbol{\imath}}+(-1.5-1) \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.37} \label{8.37}\\[4pt] & =[2.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.38} \label{8.38}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 0}^{\prime} & =\overrightarrow{\boldsymbol{v}}_{B 0}-\overrightarrow{\boldsymbol{v}}_{G}=[(-0.5-1) \hat{\boldsymbol{\imath}}+(2.5-1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.39} \label{8.39}\\[4pt] & =[-1.5 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.40} \label{8.40}\end{align}\]
We multiply the \(x\)-components by \(-e=-0.6\) to obtain the velocities in the CoM-frame after the collision:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}=[-e \cdot 2.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s}=[-1.5 \hat{\boldsymbol{\imath}}-2.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.41} \label{8.41}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}=[-e \cdot-1.5 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s}=[0.9 \hat{\boldsymbol{\imath}}+1.5 \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.42} \label{8.42}\end{align}\]
Finally we add again the centre of mass velocity vector to obtain the velocities in the original system after the impact:
\[\begin{align} \overrightarrow{\boldsymbol{v}}_{A 2} & =\overrightarrow{\boldsymbol{v}}_{A 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G}=[(-1.5+1) \hat{\boldsymbol{\imath}}+(-2.5+1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.43} \label{8.43}\\[4pt] & =[-0.5 \hat{\boldsymbol{\imath}}-1.5 \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.44} \label{8.44}\\[4pt] \overrightarrow{\boldsymbol{v}}_{B 2} & =\overrightarrow{\boldsymbol{v}}_{B 2}^{\prime}+\overrightarrow{\boldsymbol{v}}_{G}=[(0.9+1) \hat{\boldsymbol{\imath}}+(1.5+1) \hat{\boldsymbol{\jmath}}] \mathrm{m} / \mathrm{s} \tag{8.45} \label{8.45}\\[4pt] & =[1.9 \hat{\boldsymbol{\imath}}+2.5 \hat{\jmath}] \mathrm{m} / \mathrm{s} \tag{8.46} \label{8.46}\end{align}\]
As an exercise, you can check if there is momentum conservation during this collision.
8.5.4 Coefficient of restitution
The coefficient of restitution \(e\), that was introduced to capture the effect of the forces during impact, can be used as a measure for the kinetic energy that is lost during the collision in the CoM-frame. To analyse this energy loss, we use that the kinetic energy of ball \(A\) is given by:
\[T_{A}^{\prime}=\frac{1}{2} m_{A}\left(v_{A, x}^{\prime}{ }^{2}+v_{A, y}^{\prime}{ }^{2}\right) \tag{8.47} \label{8.47}\]
If we only consider the kinetic energy \(T_{A, x}^{\prime}\) contributed by the \(x\)-component of the velocity vector, we find, using Equation 8.25 that the kinetic energy after the collision is:
\[T_{A 2, x}^{\prime}=\frac{1}{2} m_{A} v_{A 2, x}^{\prime}{ }^{2}=\frac{1}{2} m_{A} e^{2} v_{A 0, x}^{\prime}{ }^{2}=e^{2} T_{A 0, x}^{\prime} \tag{8.48} \label{8.48}\]
And the same equation holds for ball \(B\). So, the \(x\)-axis kinetic energy \(T_{A, x}^{\prime}\)
reduces by a factor \(e^{2}\) during the collision, while the \(y\)-axis contribution remains constant \(T_{A 2, y}^{\prime}=T_{A 0, y}^{\prime}\). The largest energy reduction occurs for \(e=0\), which is called a plastic or perfectly inelastic collision. Kinetic energy is conserved \(\left(T_{A 2, x}^{\prime}=T_{A 0, x}^{\prime}\right)\) for \(e=1\), which is called an elastic collision. The restitution coefficient can have any value between 0 and \(1(0 \leq e \leq 1)\). Thus the coefficient of restitution provides information on energy conservation during collisions.


18.2E: Arterioles
An arteriole is a small diameter blood vessel in the microcirculation system that branches out from an artery and leads to capillaries.
- Explain the function of arterioles
Key Points
- Arterioles have muscular walls and are the primary site of vascular resistance, which reduces the pressure and velocity of flow for gas and nutrient exchange to occur within the capillaries.
- Arterioles are innervated and can also respond to other circulating factors to regulate their caliber.
Key Terms
- microcirculation : The flow of blood through the smallest vessels: arterioles, capillaries, and venules.
- arteriole : One of the small branches of an artery, especially one that connects with capillaries.
An arteriole is a small-diameter blood vessel which forms part of the microcirculation that extends from an artery and leads to capillaries.
Capillary : Arterioles are part of the microcirculation system, along with capillaries, arteries, veins, venules, and tissue cells.
The microcirculation involves the flow of blood in the smallest blood vessels, including arterioles, capillaries, and venules.
Arterioles have muscular walls that usually consist of one or two layers of smooth muscle. They are the primary site of vascular resistance. This reduces the pressure and velocity of blood flow to enable gas and nutrient exchange to occur within the capillaries. Arterioles are innervated and also respond to various circulating hormones and other factors such as pH in order to regulate their caliber, thus modulating the amount of blood flow into the capillary network and tissues.