DOCSLIB.ORG

1 Circular Motion by Prof. Massimiliano Galeazzi, University of Miami If You Spin a Bucket Full of Water in a Vertical Circl

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
1 Circular Motion by Prof. Massimiliano Galeazzi, University of Miami If You Spin a Bucket Full of Water in a Vertical Circl Circular Motion By Prof. Massimiliano Galeazzi, University of Miami If you spin a bucket full of water in a vertical circle fast enough, the water will stay in the bucket. How fast do you need to spin it to make sure you don’t get showered? What is the force that makes your car turn when you turn the steering wheel? How big is that force? These are some of the questions we will answer in this chapter. The general topic is that of circular motion, that is, objects moving on a circular trajectory. Along the way we will learn about ferris wheels, cars turning without going off the road, etc. But first, we need to improve our math understanding by reviewing the product between vectors. --------------------------------------------------------------------------------------------------------------------------------------------------- MATH INSERT In chapter 1 we have introduced the concept of vectors and found how vectors can be added and or subtracted. We also mentioned the fact that vectors can be multiplied with each other in two different ways, what are called the scalar (or dot) product and the vector (or cross) product. In this chapter we will make use, for the first time, of the dot-product, so it is appropriate to take a look at it before we proceed. 1. The scalar (dot) product between vectors FIG. 1 Let’s consider two vectors, = + + and = + + . The scalar product between and , denoted (“A dot B”, hence the퐴⃗ term퐴푥 횤“̂ dot퐴 product푦횥̂ 퐴푧”),푘� is a 퐵�scalar⃗ 퐵 푥qu횤̂ antity퐵푦횥̂ equal퐵푧푘� to the product between the magnitudes퐴⃗ 퐵�⃗ of the two vectors, times the cosine of the angle between the two: 퐴⃗ ∙ 퐵�⃗ = cos , 휙 [1] where 퐴the⃗ ∙ 퐵� ⃗angle�퐴⃗ �� 퐵�is⃗� measured휙 by drawing the two vectors with the tails in the same point (see Fig. 1a-b). We notice that, from Eq. 1, the dot product, like the “regular” product between scalars is commutative, i.e., 휙 = [2] Also,퐴⃗ ∙ 퐵�⃗ we퐵� ⃗can∙ 퐴⃗ write Eq. [1] as: = cos , [3] ⃗ �⃗ ⃗ �⃗ That is,퐴 the∙ 퐵 scalar�퐴�� product�퐵� 휙between� and is equal to the magnitude of times the projection of in the direction of , i.e., cos (see Fig. 1c). Similarly, we can write: 퐴⃗ 퐵�⃗ 퐴⃗ 퐵�⃗ 퐴⃗ �퐵�⃗� 휙 1 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. = cos , [4] ⃗ �⃗ �⃗ ⃗ That is,퐴 the∙ 퐵 scalar�퐵�� product�퐴� 휙between� and is equal to the magnitude of times the projection of in the direction of (see Fig. 1d). 퐴⃗ 퐵�⃗ 퐵�⃗ 퐴⃗ 퐵�⃗ From the definition of the scalar product it is possible to derive a few simple properties: • The scalar product between two vectors can be positive, negative, or zero. In particular, if 0 < 90 then the product is positive, while if 90 < 180 then the product is negative. 표 ≤ 휙 • The scalar product of two parallel vectors표 is equal표 to the product of the magnitudes of the two ( = 0). 휙 ≤ • The scalar product of two vectors that are anti-parallel is equal to the negative of the product of the two 휙 magnitudes ( = 180 ). • The scalar product of two표 perpendicular vectors is equal to zero ( = 90 ). 휙 • From the previous result it follows that, if the scalar product between two표 vectors is zero and their magnitudes 휙 are different from zero, then the two vectors must be perpendicular. This provides a simple, yet very powerful way to determine whether two vectors are perpendicular. While the definition of scalar product in Eq. [1] is very useful, sometimes it is easier to work with vector components, rather than magnitude and direction. It is also possible to calculate the scalar product of two vectors in terms of their components: = + + + + . [5] 푥 푦 푧 푥 푦 푧 To퐴⃗ ∙ 퐵find�⃗ the�퐴 value횤̂ 퐴 of횥̂ the퐴 scalar푘�� ∙ � product퐵 횤̂ 퐵 in횥 ̂ terms퐵 푘� of� components, we simply expand the product in Eq. [5]: = + + + + + + + [6] 퐴⃗ ∙ 퐵�⃗ 퐴푥횤̂ ∙ 퐵푥횤̂ 퐴푥횤̂ ∙ 퐵푦횥̂ 퐴푥횤̂ ∙ 퐵푧푘� 퐴푦횥̂ ∙ 퐵푥횤̂ + + + 퐴푦횥̂ ∙ 퐵푦횥̂ 퐴푦횥̂ ∙ 퐵푧푘� 푧 � 푥 푧 � 푦 푧 � 푧 � And taking퐴 푘into∙ 퐵 account횤̂ 퐴 푘 that,∙ 퐵 횥usinĝ 퐴 the푘 ∙ 퐵properties푘 of the scalar product and of unit vectors = = = 1, and = = = 0, we get: 횤̂ ∙ 횤̂ 횥̂ ∙ 횥̂ 푘� ∙ 푘� � � 횤̂ ∙ 횥̂ 횤̂ ∙=푘 횥̂ ∙ 푘+ + . [7] 푥 푥 푦 푦 푧 푧 END OF퐴⃗ ∙ MATH퐵�⃗ 퐴 퐵INSERT퐴 퐵 퐴 퐵 --------------------------------------------------------------------------------------------------------------------------------------------------- 2. Motion along a circular path Consider a particle with mass m that is constrained to move on a circular path, as shown in Fig. 2. To describe the motion of the particle we can set up a coordinate system with the origin coincident with the center of the circle and then use the x and y coordinate of the particle. We could, however, also use the angle that the particle position makes with the positive x-axis, or the distance of the particle from the positive x-axis along the circular path. Note that in our choice 휃 of coordinates the particle is moving around the z-axis which is perpendicular to the page and passes through the center of 푙 the circle. A general convention is to consider the angle and the distance l as positive if measured counterclockwise. The Cartesian coordinates x and y of the particle and the vector position can then be easily calculated as: 휃 푟⃗ 2 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. FIG. 2 = cos [8] 푥 = 푅 sin 휃 [9] 푦 = 푅 + 휃 = ( cos ) + ( sin ) , [10] where 푟⃗R is푥 the횤̂ 푦radius횥̂ 푅of the휃 circle.횤̂ 푅 Moreover,휃 횥̂ since the distance l traveled by the particle is nothing more than the arc subtended by the angle , we can also write: = . 휃 [11] From푙 푅 휃Eq. [10] we can also easily find an expression for the particle velocity. By using the chain rule for derivatives and the fact that in circular motion R is constant, i.e., the distance from the origin does not change, we get: = = ( sin ) + ( cos ) , [12] 푑푟⃗ 푑휃 푑휃 where 푣⃗we mad푑푡 e use−푅 of the휃 fact푑푡 that,횤̂ indeed,푅 휃 the푑푡 angle횥̂ is the only quantity in Eq. [10] that changes with time. The first derivative of the angle versus time is 휃called the angular velocity of the particle and is indicated with the symbol (the Greek letter omega): 푧 휔= , [13] 푑휃 푧 Where휔 the index푑푡 “z” is used to indicate that, as mentioned before, the rotation is around the z-axis. We note that, in general, the angular velocity is a vector, with each component quantifying the rotation around a different axis. However, in this chapter (and in most of the book) we will only consider rotations around the z-axis, and thus only use the z- component of the vector. We also point out that the angular velocity measures angles per time, or radians per second. However, radians are usually treated as if they are not a real unit, and the SI unit for angular velocity is simply . Using 휔푧 Eq. [13] we can rewrite Eq. [12] as: −1 푠 = sin + cos . [14] 푣⃗ −푅휔푧 휃 횤̂ 푅휔푧 휃 횥̂ We can also use Eq. [12] to find the speed of the particle = | | = + . However, it is easier to calculate it 2 2 using the fact that the speed is the first derivative of the distance 푣l traveled푣⃗ in� Eq.푣푥 [11]:푣푦 ( ) = = = = , or = , [15] 푑푙 푑 푅휃 푑휃 푣 푧 푧 where 푣we used푑푡 the푑푡 fact that푅 푑푡 the 푅radius휔 R휔 is constant.푅 Equation [15] represents a very simple and straightforward relation between the particle speed and its angular velocity. 3 Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden. 3. Position vs. Velocity in circular motion Before continuing with our analysis, let’s take a quick look at the properties of the vectors position and velocity associated with circular motion. The vector position is the vector going from the origin of the coordinate system to the particle position. However, since in our choice the origin of the coordinate system coincides with the center of the circular path, the magnitude of the vector position will always푟⃗ be constant and equal to the radius R of the circle and will always have a “radial” direction, i.e., it will always be oriented with the radius of the circle at the position of the particle. This property can be expressed as = , [16] where 푟⃗ is 푅the푟̂ generic unit vector (magnitude 1) aligned with the radius of the circle at the position of the particle. Using푟̂ Eqs. [10] and [14] we can also demonstrate that, in circular motion, the velocity is perpendicular to the position . The simplest way to demonstrate it is by calculating the dot-product between the two vectors, using the 푣⃗ definition of dot-product with components derived in Eq. [7] and the fact that the two vectors only have x and y components:푟⃗ = ( + ) + = + = ( cos )( sin ) + ( sin )( cos ) , [17] 푥 푦 푥 푦 푧 푧 which 푟⃗simplifies∙ 푣⃗ 푥횤 ̂to: 푦횥̂ ∙ �푣 횤̂ 푣 횥̂� 푥푣 푦푣 푅 휃 −푅휔 휃 푅 휃 푅휔 휃 = cos sin + sin cos = 0. [18] 2 2 Equation푟⃗ ∙ 푣⃗ − 푅[18휔] 푧is valid휃 in 휃the most푅 휔 푧general휃 case휃 of circular motion, and implies that, if a particle can only move on a circular path, then its velocity is always perpendicular to its vector position. Moreover, since the vector position is always aligned with a radius of the circular path, this means that the velocity is always perpendicular to the radius, i.e., it is always tangent to the circular trajectory.
Load more

Recommended publications
  • Velocity-Corrected Area Calculation SCIEX PA 800 Plus Empower
    Velocity-corrected area calculation: SCIEX PA 800 Plus Empower Driver version 1.3 vs. 32 Karat™ Software Firdous Farooqui1, Peter Holper1, Steve Questa1, John D. Walsh2, Handy Yowanto1 1SCIEX, Brea, CA 2Waters Corporation, Milford, MA Since the introduction of commercial capillary electrophoresis (CE) systems over 30 years ago, it has been important to not always use conventional “chromatography thinking” when using CE. This is especially true when processing data, as there are some key differences between electrophoretic and chromatographic data. For instance, in most capillary electrophoresis separations, peak area is not only a function of sample load, but also of an analyte’s velocity past the detection window. In this case, early migrating peaks move past the detection window faster than later migrating peaks. This creates a peak area bias, as any relative difference in an analyte’s migration velocity will lead to an error in peak area determination and relative peak area percent. To help minimize Figure 1: The PA 800 Plus Pharmaceutical Analysis System. this bias, peak areas are normalized by migration velocity. The resulting parameter is commonly referred to as corrected peak The capillary temperature was maintained at 25°C in all area or velocity corrected area. separations. The voltage was applied using reverse polarity. This technical note provides a comparison of velocity corrected The following methods were used with the SCIEX PA 800 Plus area calculations using 32 Karat™ and Empower software. For Empower™ Driver v1.3: both, standard processing methods without manual integration were used to process each result. For 32 Karat™ software, IgG_HR_Conditioning: conditions the capillary Caesar integration1 was turned off.
    [Show full text]
  • The Origins of Velocity Functions
    The Origins of Velocity Functions Thomas M. Humphrey ike any practical, policy-oriented discipline, monetary economics em- ploys useful concepts long after their prototypes and originators are L forgotten. A case in point is the notion of a velocity function relating money’s rate of turnover to its independent determining variables. Most economists recognize Milton Friedman’s influential 1956 version of the function. Written v = Y/M = v(rb, re,1/PdP/dt, w, Y/P, u), it expresses in- come velocity as a function of bond interest rates, equity yields, expected inflation, wealth, real income, and a catch-all taste-and-technology variable that captures the impact of a myriad of influences on velocity, including degree of monetization, spread of banking, proliferation of money substitutes, devel- opment of cash management practices, confidence in the future stability of the economy and the like. Many also are aware of Irving Fisher’s 1911 transactions velocity func- tion, although few realize that it incorporates most of the same variables as Friedman’s.1 On velocity’s interest rate determinant, Fisher writes: “Each per- son regulates his turnover” to avoid “waste of interest” (1963, p. 152). When rates rise, cashholders “will avoid carrying too much” money thus prompting a rise in velocity. On expected inflation, he says: “When...depreciation is anticipated, there is a tendency among owners of money to spend it speedily . the result being to raise prices by increasing the velocity of circulation” (p. 263). And on real income: “The rich have a higher rate of turnover than the poor. They spend money faster, not only absolutely but relatively to the money they keep on hand.
    [Show full text]
  • VELOCITY from Our Establishment in 1957, We Have Become One of the Oldest Exclusive Manufacturers of Commercial Flooring in the United States
    VELOCITY From our establishment in 1957, we have become one of the oldest exclusive manufacturers of commercial flooring in the United States. As one of the largest privately held mills, our FAMILY-OWNERSHIP provides a heritage of proven performance and expansive industry knowledge. Most importantly, our focus has always been on people... ensuring them that our products deliver the highest levels of BEAUTY, PERFORMANCE and DEPENDABILITY. (cover) Velocity Move, quarter turn. (right) Velocity Move with Pop Rojo and Azul, quarter turn. VELOCITY 3 velocity 1814 style 1814 style 1814 style 1814 color 1603 color 1604 color 1605 position direction magnitude style 1814 style 1814 style 1814 color 1607 color 1608 color 1609 reaction move constant style 1814 color 1610 vector Velocity Vector, quarter turn. VELOCITY 5 where to use kinetex Healthcare Fitness Centers kinetex overview Acute care hospitals, medical Health Clubs/Gyms office buildings, urgent care • Cardio Centers clinics, outpatient surgery • Stationary Weight Centers centers, outpatient physical • Dry Locker Room Areas therapy/rehab centers, • Snack Bars outpatient imaging centers, etc. • Offices Kinetex® is an advanced textile composite flooring that combines key attributes of • Cafeteria, dining areas soft-surface floor covering with the long-wearing performance characteristics of • Chapel Retail / Mercantile hard-surface flooring. Created as a unique floor covering alternative to hard-surface Wholesale / Retail merchants • Computer room products, J+J Flooring’s Kinetex encompasses an unprecedented range of • Corridors • Checkout / cash wrap performance attributes for retail, healthcare, education and institutional environments. • Diagnostic imaging suites • Dressing rooms In addition to its human-centered qualities and highly functional design, Kinetex • Dry physical therapy • Sales floor offers a reduced environmental footprint compared to traditional hard-surface options.
    [Show full text]
  • Chapter 3 Motion in Two and Three Dimensions
    Chapter 3 Motion in Two and Three Dimensions 3.1 The Important Stuff 3.1.1 Position In three dimensions, the location of a particle is specified by its location vector, r: r = xi + yj + zk (3.1) If during a time interval ∆t the position vector of the particle changes from r1 to r2, the displacement ∆r for that time interval is ∆r = r1 − r2 (3.2) = (x2 − x1)i +(y2 − y1)j +(z2 − z1)k (3.3) 3.1.2 Velocity If a particle moves through a displacement ∆r in a time interval ∆t then its average velocity for that interval is ∆r ∆x ∆y ∆z v = = i + j + k (3.4) ∆t ∆t ∆t ∆t As before, a more interesting quantity is the instantaneous velocity v, which is the limit of the average velocity when we shrink the time interval ∆t to zero. It is the time derivative of the position vector r: dr v = (3.5) dt d = (xi + yj + zk) (3.6) dt dx dy dz = i + j + k (3.7) dt dt dt can be written: v = vxi + vyj + vzk (3.8) 51 52 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS where dx dy dz v = v = v = (3.9) x dt y dt z dt The instantaneous velocity v of a particle is always tangent to the path of the particle. 3.1.3 Acceleration If a particle’s velocity changes by ∆v in a time period ∆t, the average acceleration a for that period is ∆v ∆v ∆v ∆v a = = x i + y j + z k (3.10) ∆t ∆t ∆t ∆t but a much more interesting quantity is the result of shrinking the period ∆t to zero, which gives us the instantaneous acceleration, a.
    [Show full text]
  • Moment of Inertia
    MOMENT OF INERTIA The moment of inertia, also known as the mass moment of inertia, angular mass or rotational inertia, of a rigid body is a quantity that determines the torque needed for a desired angular acceleration about a rotational axis; similar to how mass determines the force needed for a desired acceleration. It depends on the body's mass distribution and the axis chosen, with larger moments requiring more torque to change the body's rotation rate. It is an extensive (additive) property: for a point mass the moment of inertia is simply the mass times the square of the perpendicular distance to the rotation axis. The moment of inertia of a rigid composite system is the sum of the moments of inertia of its component subsystems (all taken about the same axis). Its simplest definition is the second moment of mass with respect to distance from an axis. For bodies constrained to rotate in a plane, only their moment of inertia about an axis perpendicular to the plane, a scalar value, and matters. For bodies free to rotate in three dimensions, their moments can be described by a symmetric 3 × 3 matrix, with a set of mutually perpendicular principal axes for which this matrix is diagonal and torques around the axes act independently of each other. When a body is free to rotate around an axis, torque must be applied to change its angular momentum. The amount of torque needed to cause any given angular acceleration (the rate of change in angular velocity) is proportional to the moment of inertia of the body.
    [Show full text]
  • Rotation: Moment of Inertia and Torque
    Rotation: Moment of Inertia and Torque Every time we push a door open or tighten a bolt using a wrench, we apply a force that results in a rotational motion about a fixed axis. Through experience we learn that where the force is applied and how the force is applied is just as important as how much force is applied when we want to make something rotate. This tutorial discusses the dynamics of an object rotating about a fixed axis and introduces the concepts of torque and moment of inertia. These concepts allows us to get a better understanding of why pushing a door towards its hinges is not very a very effective way to make it open, why using a longer wrench makes it easier to loosen a tight bolt, etc. This module begins by looking at the kinetic energy of rotation and by defining a quantity known as the moment of inertia which is the rotational analog of mass. Then it proceeds to discuss the quantity called torque which is the rotational analog of force and is the physical quantity that is required to changed an object's state of rotational motion. Moment of Inertia Kinetic Energy of Rotation Consider a rigid object rotating about a fixed axis at a certain angular velocity. Since every particle in the object is moving, every particle has kinetic energy. To find the total kinetic energy related to the rotation of the body, the sum of the kinetic energy of every particle due to the rotational motion is taken. The total kinetic energy can be expressed as ..
    [Show full text]
  • Circular Motion Angular Velocity
    PHY131H1F - Class 8 Quiz time… – Angular Notation: it’s all Today, finishing off Chapter 4: Greek to me! d • Circular Motion dt • Rotation θ is an angle, and the S.I. unit of angle is rad. The time derivative of θ is ω. What are the S.I. units of ω ? A. m/s2 B. rad / s C. N/m D. rad E. rad /s2 Last day I asked at the end of class: Quiz time… – Angular Notation: it’s all • You are driving North Highway Greek to me! d 427, on the smoothly curving part that will join to the Westbound 401. v dt Your speedometer is constant at 115 km/hr. Your steering wheel is The time derivative of ω is α. not rotating, but it is turned to the a What are the S.I. units of α ? left to follow the curve of the A. m/s2 highway. Are you accelerating? B. rad / s • ANSWER: YES! Any change in velocity, either C. N/m magnitude or speed, implies you are accelerating. D. rad • If so, in what direction? E. rad /s2 • ANSWER: West. If your speed is constant, acceleration is always perpendicular to the velocity, toward the centre of circular path. Circular Motion r = constant Angular Velocity s and θ both change as the particle moves s = “arc length” θ = “angular position” when θ is measured in radians when ω is measured in rad/s 1 Special case of circular motion: Uniform Circular Motion A carnival has a Ferris wheel where some seats are located halfway between the center Tangential velocity is and the outside rim.
    [Show full text]
  • Lecture 24 Angular Momentum
    LECTURE 24 ANGULAR MOMENTUM Instructor: Kazumi Tolich Lecture 24 2 ¨ Reading chapter 11-6 ¤ Angular momentum n Angular momentum about an axis n Newton’s 2nd law for rotational motion Angular momentum of an rotating object 3 ¨ An object with a moment of inertia of � about an axis rotates with an angular speed of � about the same axis has an angular momentum, �, given by � = �� ¤ This is analogous to linear momentum: � = �� Angular momentum in general 4 ¨ Angular momentum of a point particle about an axis is defined by � � = �� sin � = ��� sin � = �-� = ��. � �- ¤ �⃗: position vector for the particle from the axis. ¤ �: linear momentum of the particle: � = �� �⃗ ¤ � is moment arm, or sometimes called “perpendicular . Axis distance.” �. Quiz: 1 5 ¨ A particle is traveling in straight line path as shown in Case A and Case B. In which case(s) does the blue particle have non-zero angular momentum about the axis indicated by the red cross? A. Only Case A Case A B. Only Case B C. Neither D. Both Case B Quiz: 24-1 answer 6 ¨ Only Case A ¨ For a particle to have angular momentum about an axis, it does not have to be Case A moving in a circle. ¨ The particle can be moving in a straight path. Case B ¨ For it to have a non-zero angular momentum, its line of path is displaced from the axis about which the angular momentum is calculated. ¨ An object moving in a straight line that does not go through the axis of rotation has an angular position that changes with time. So, this object has an angular momentum.
    [Show full text]
  • Chapter 8: Rotational Motion
    TODAY: Start Chapter 8 on Rotation Chapter 8: Rotational Motion Linear speed: distance traveled per unit of time. In rotational motion we have linear speed: depends where we (or an object) is located in the circle. If you ride near the outside of a merry-go-round, do you go faster or slower than if you ride near the middle? It depends on whether “faster” means -a faster linear speed (= speed), ie more distance covered per second, Or - a faster rotational speed (=angular speed, ω), i.e. more rotations or revolutions per second. • Linear speed of a rotating object is greater on the outside, further from the axis (center) Perimeter of a circle=2r •Rotational speed is the same for any point on the object – all parts make the same # of rotations in the same time interval. More on rotational vs tangential speed For motion in a circle, linear speed is often called tangential speed – The faster the ω, the faster the v in the same way v ~ ω. directly proportional to − ω doesn’t depend on where you are on the circle, but v does: v ~ r He’s got twice the linear speed than this guy. Same RPM (ω) for all these people, but different tangential speeds. Clicker Question A carnival has a Ferris wheel where the seats are located halfway between the center and outside rim. Compared with a Ferris wheel with seats on the outside rim, your angular speed while riding on this Ferris wheel would be A) more and your tangential speed less. B) the same and your tangential speed less.
    [Show full text]
  • Physics 3550, Fall 2012 Two Body, Central-Force Problem Relevant Sections in Text: §8.1 – 8.7
    Two Body, Central-Force Problem. Physics 3550, Fall 2012 Two Body, Central-Force Problem Relevant Sections in Text: x8.1 { 8.7 Two Body, Central-Force Problem { Introduction. I have already mentioned the two body central force problem several times. This is, of course, an important dynamical system since it represents in many ways the most fundamental kind of interaction between two bodies. For example, this interaction could be gravitational { relevant in astrophysics, or the interaction could be electromagnetic { relevant in atomic physics. There are other possibilities, too. For example, a simple model of strong interactions involves two-body central forces. Here we shall begin a systematic study of this dynamical system. As we shall see, the conservation laws admitted by this system allow for a complete determination of the motion. Many of the topics we have been discussing in previous lectures come into play here. While this problem is very instructive and physically quite important, it is worth keeping in mind that the complete solvability of this system makes it an exceptional type of dynamical system. We cannot solve for the motion of a generic system as we do for the two body problem. The two body problem involves a pair of particles with masses m1 and m2 described by a Lagrangian of the form: 1 2 1 2 L = m ~r_ + m ~r_ − V (j~r − ~r j): 2 1 1 2 2 2 1 2 Reflecting the fact that it describes a closed, Newtonian system, this Lagrangian is in- variant under spatial translations, time translations, rotations, and boosts.* Thus we will have conservation of total energy, total momentum and total angular momentum for this system.
    [Show full text]
  • 4. Central Forces
    4. Central Forces In this section we will study the three-dimensional motion of a particle in a central force potential. Such a system obeys the equation of motion mx¨ = V (r)(4.1) r where the potential depends only on r = x .Sincebothgravitationalandelectrostatic | | forces are of this form, solutions to this equation contain some of the most important results in classical physics. Our first line of attack in solving (4.1)istouseangularmomentum.Recallthatthis is defined as L = mx x˙ ⇥ We already saw in Section 2.2.2 that angular momentum is conserved in a central potential. The proof is straightforward: dL = mx x¨ = x V =0 dt ⇥ − ⇥r where the final equality follows because V is parallel to x. r The conservation of angular momentum has an important consequence: all motion takes place in a plane. This follows because L is a fixed, unchanging vector which, by construction, obeys L x =0 · So the position of the particle always lies in a plane perpendicular to L.Bythesame argument, L x˙ =0sothevelocityoftheparticlealsoliesinthesameplane.Inthis · way the three-dimensional dynamics is reduced to dynamics on a plane. 4.1 Polar Coordinates in the Plane We’ve learned that the motion lies in a plane. It will turn out to be much easier if we work with polar coordinates on the plane rather than Cartesian coordinates. For this reason, we take a brief detour to explain some relevant aspects of polar coordinates. To start, we rotate our coordinate system so that the angular momentum points in the z-direction and all motion takes place in the (x, y)plane.Wethendefinetheusual polar coordinates x = r cos ✓, y= r sin ✓ –48– Our goal is to express both the velocity and acceleration y ^ ^ θ r in polar coordinates.
    [Show full text]
  • PHYS 101 Lecture 1: Speed and Velocity 1 - 1
    PHYS 101 Lecture 1: Speed and velocity 1 - 1 Lecture 1: Speed and velocity What's important: • definitions of distance, displacement, speed and velocity Demonstrations: keratocyte video Logistics Textbook: Physics by Giancoli Further reading in biology (just for interest, not needed in this course, don't buy these!) Molecular Biology of the Cell by Alberts et al. (Garland) Random Walks in Biology by Howard Berg (Princeton) Grading: 10% 8 assignments Dates on website 40% 3 midterm exams (10 - 15 - 15) 50% Final exam Useful local websites: on-line lecture notes etc. for this course can be found through http://physics.sfu.ca or http://www.sfu.ca/~boal Cells See Supplement #1 on my PHYS 101 website for details on cell structure and classification. All we want to introduce at this point are the cell dimensions: Prokaryotes – cells with no internal structure dating back at least 3.5 billion years on Earth. Modern-day bacteria are prokaryotes. Typical dimensions are a few microns (1 µm = 10-6 m); for example, E. coli (one cause of gastrointestinal problems) is about 4 µm long and 1 µm across. Eukaryotes – thought to have appeared about 2 billion years ago, around the time of the rise of atmospheric oxygen. Lots of internal structure, including filaments, networks and internal compartments such as nuclei. Internal compartments are bacterial in size, and the overall cell size is in the 10 µm range. The cells of our bodies are eukaryotes. prokaryote eukaryote © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.
    [Show full text]