DOCSLIB.ORG

Angular Momentum

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Angular Momentum Angular Momentum 8.01 W10D1 Today’s Reading Assignment: MIT 8.01 Course Notes: Chapter 19 Angular Momentum Sections 19.1-19.6 Announcements Problem Set 7 due Week 10 Tuesday at 9 pm in box outside 26-152 Math Review Week 10 Tuesday at 9 pm in 26-152 Exam 3 Tuesday Nov 26 7:30-9:30 pm Conflict Exam 3 Wednesday Nov 27 8-10 am, 10-12 noon Nov 27 Drop Date Overview: Torque and Angular Momentum Angular momentum is defined as LS = rS ×p Torque about a point S is equal to the time derivative of the angular momentum about S . dL τ = S S dt When torque about S is zero, angular momentum is constant L = L S , f S ,i Angular Momentum of a Point Particle Point particle of mass m moving with a velocity v Momentum p = mv Fix a point S Vector r S from the point S to the location of the object Angular momentum about the point S LS = rS ×p 2 -1 SI Unit [kg ⋅ m ⋅s ] Cross Product: Angular Momentum of a Point Particle L = r ×p Magnitude: S S L = r p sinθ S S a) moment arm rS ,⊥ = rS sinθ LS = rS ,⊥ p b) perpendicular momentum L r p pS ,⊥ = p sinθ S = S ⊥ Angular Momentum of a Point Particle: Direction Direction: Right Hand Rule Concept Q. Mag. of Angular Momentum In the above situation where a particle is moving in the x-y plane with a constant velocity, the magnitude of the angular momentum about the point S (the origin) 1) decreases then increases 2) increases then decreases 3) is constant 4) is zero because this is not circular motion Table Problem: Angular Momentum and Cross Product A particle of mass m moves with a uniform velocity v = v ˆi + v ˆj x y At time t, the position vector of the particle with respect to the point S is r x ˆi y ˆj S = + Find the direction and the magnitude of the angular momentum about the origin, (the point S) at time t. Angular Momentum and Circular Motion of a Point Particle: Fixed axis of rotation: z-axis Angular velocity ω ≡ ω kˆ z Velocity ˆ ˆ v = vθθ = Rω zθ Angular momentum about the point S 2 L = r × p = r × mv = Rmv kˆ = mR ω kˆ S S S θ z Table Prob.: Angular momentum Along Axis of Rotation for Circular Motion A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum L0 relative to the origin. b) Find the z component of L 0 . Hint: Use r = rrˆ + hkˆ 0 Concept Q.: Direction of Ang. Mom. A particle of mass m moves in a circle of radius R at an angular speed ω about the z axis in a plane parallel to but above the x-y plane. Relative to the origin 1. L is constant. 0 2. is constant but direction of is not. L0 L0 L 3. Direction of L 0 is constant but 0 is not. 4. L has no z-component. 0 Worked Example: Angular Momentum of Two Particles Two identical particles of mass m move in a circle of radius r, 180º out of phase, at an angular speed ω about the z axis in a plane parallel to but a distance h above the x-y plane. a) Find the magnitude and the direction of the angular momentum L 0 relative to the origin. b) Is this angular momentum relative to the origin constant? If yes, why? If no, why is it not constant? Worked Example: Angular Momentum of Two Particles r = xˆi + yˆj+ hkˆ = rrˆ + hkˆ 0,1 1 r = (x2 + y2 )1/2 ˆ ˆ v = vθ = rωθ L = r × mv = (rrˆ + hkˆ ) × mrωθˆ 0,1 0,1 1 2 ˆ L0,1 = mr ωk − hmrωrˆ1 ! 2 L = r × mv = (rrˆ + hkˆ ) × mrωθˆ = mr ωkˆ − hmrωrˆ 0,2 0,2 2 2 rˆ rˆ 1 = − 2 L + L = mr 2ωkˆ − hmrωrˆ + mr 2ωkˆ − hmrωrˆ = 2mr 2ωkˆ 0,1 0,2 1 2 Angular Momentum of a Ring A circular ring of radius R and mass M rotates at an angular speed ω about the z- axis in a plane parallel to but a distance h above the x-y plane. Find the magnitude and the direction of the angular momentum L 0 relative to the origin. Divide ring into pairs of small objects with mass 2 L + L = 2Δmr ωkˆ 0,1 0,2 L m r 2 kˆ Mr 2 kˆ 0 = ∑ pair ω = ω pairs Concept Q. Symmetric Body A rigid body with rotational symmetry body rotates at a constant angular speed ω about it symmetry (z) axis. In this case 1. is constant. L0 2. L is constant but is not. 0 L0 / L0 3. is constant but is not. L0 / L0 L0 4. L 0 has no z-component. 5. Two of the above are true. Angular Momentum of Cylindrically Symmetric Body A cylindrically symmetric rigid body rotating about its symmetry axis at a constant angular velocity ω = ω kˆ with ω > 0 has angular z z momentum about any point on its axis L = dL = dm r 2 ω kˆ z ∫ z ∫ dm z body body ⎛ ⎞ = dm r 2 ω kˆ ⎜ ∫ dm ⎟ z ⎝ body ⎠ = I ω kˆ z z Kinetic Energy of Cylindrically Symmetric Body A cylindrically symmetric rigid body with moment of inertia Iz rotating about its symmetry axis at a constant angular velocity ω = ω kˆ z ω > 0 z Angular Momentum L = I ω z z z 2 Kinetic energy 1 2 L K = I ω = z rot 2 z z 2I z Concept Q.: Angular Momentum of Disk A disk with mass M and radius R is spinning with angular speed ω about an axis that passes through the rim of the disk perpendicular to its plane. The magnitude of its angular momentum is: 1 1 1. MR2ω 2 4. MR2ω 4 4 1 1 2. MR2ω 2 5. MR2ω 2 2 3 3 3. MR2ω 2 6. MR2ω 2 2 Concept Q.: Non-Symmetric Body A body rotates with constant angular speed ω about the z axis which is not a symmetry axis of the body. Relative to the origin 1. L 0 is constant. 2. L 0 is constant but L 0 / L 0 is not. 3. L 0 / L 0 is constant but L 0 is not. 4. L 0 has no z-component. Table Problem: Angular Momentum of a Two Particles About Different Points Two point like particles 1 and 2, each of mass m, are rotating at a constant angular speed about point A. How does the angular momentum about the point B compare to the angular momentum about point A? What about at a later time when the particles have rotated by 90 degrees? Conservation of Angular Momentum Time Derivative of Angular Momentum for a Point Particle Angular momentum of a point particle about S: L = r × p S S Time derivative of the angular momentum about S: dLS d = (rS ×p) dt dt dLS d drS d Product rule = (rS ×p)= ×p + rS × p dt dt dt dt dr dr Key Fact: v = S ⇒ S × mv = v× mv = 0 dt dt Result: dL d S = r × p = r ×F = τ dt S dt S S Torque and the Time Derivative of Angular Momentum: Point Particle Torque about a point S is equal to the time derivative of the angular momentum about S . dLS τ = S dt Concept Q.: Change in Angular Mom. A person spins a tennis ball on a string in a horizontal circle with velocity v (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow (force F ) in the forward direction. This causes a change in angular momentum Δ L in the 1. rˆ direction ˆ 2. θ direction 3. kˆ direction Conservation of Angular Momentum dL Rotational dynamics S τ S = dt No torques dL 0 = τ = S S dt Change in Angular momentum is zero ΔLS ≡ LS , f − LS ,i = 0 Angular Momentum is conserved L = L S , f S ,i Demo: Rotating on a Chair A person holding dumbbells in his/her arms spins in a rotating stool. When he/she pulls the dumbbells inward, the moment of inertia changes and he/she spins faster. Concept Question: Figure Skater A figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational moment of inertia and her angular speed increases. Assume that her angular momentum is constant. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be 1. the same. 2. larger. 3. smaller. 4. not enough information is given to decide. Table Problem: Meteor Encounter A meteor of mass m is approaching earth as shown on the sketch. The distance h on the sketch below is called the impact parameter. The radius of the earth is Re. The mass of the earth is me. Suppose the meteor has an initial speed of v0. Assume that the meteor started very far away from the earth. Suppose the meteor just grazes the earth. You may ignore all other gravitational forces except the earth.
Load more

Recommended publications
  • Glossary Physics (I-Introduction)
    1 Glossary Physics (I-introduction) - Efficiency: The percent of the work put into a machine that is converted into useful work output; = work done / energy used [-]. = eta In machines: The work output of any machine cannot exceed the work input (<=100%); in an ideal machine, where no energy is transformed into heat: work(input) = work(output), =100%. Energy: The property of a system that enables it to do work. Conservation o. E.: Energy cannot be created or destroyed; it may be transformed from one form into another, but the total amount of energy never changes. Equilibrium: The state of an object when not acted upon by a net force or net torque; an object in equilibrium may be at rest or moving at uniform velocity - not accelerating. Mechanical E.: The state of an object or system of objects for which any impressed forces cancels to zero and no acceleration occurs. Dynamic E.: Object is moving without experiencing acceleration. Static E.: Object is at rest.F Force: The influence that can cause an object to be accelerated or retarded; is always in the direction of the net force, hence a vector quantity; the four elementary forces are: Electromagnetic F.: Is an attraction or repulsion G, gravit. const.6.672E-11[Nm2/kg2] between electric charges: d, distance [m] 2 2 2 2 F = 1/(40) (q1q2/d ) [(CC/m )(Nm /C )] = [N] m,M, mass [kg] Gravitational F.: Is a mutual attraction between all masses: q, charge [As] [C] 2 2 2 2 F = GmM/d [Nm /kg kg 1/m ] = [N] 0, dielectric constant Strong F.: (nuclear force) Acts within the nuclei of atoms: 8.854E-12 [C2/Nm2] [F/m] 2 2 2 2 2 F = 1/(40) (e /d ) [(CC/m )(Nm /C )] = [N] , 3.14 [-] Weak F.: Manifests itself in special reactions among elementary e, 1.60210 E-19 [As] [C] particles, such as the reaction that occur in radioactive decay.
    [Show full text]
  • The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected]
    The University of Akron IdeaExchange@UAkron The Dr. Gary B. and Pamela S. Williams Honors Honors Research Projects College Fall 2017 The Experimental Determination of the Moment of Inertia of a Model Airplane Michael Koken [email protected] Please take a moment to share how this work helps you through this survey. Your feedback will be important as we plan further development of our repository. Follow this and additional works at: http://ideaexchange.uakron.edu/honors_research_projects Part of the Aerospace Engineering Commons, Aviation Commons, Civil and Environmental Engineering Commons, Mechanical Engineering Commons, and the Physics Commons Recommended Citation Koken, Michael, "The Experimental Determination of the Moment of Inertia of a Model Airplane" (2017). Honors Research Projects. 585. http://ideaexchange.uakron.edu/honors_research_projects/585 This Honors Research Project is brought to you for free and open access by The Dr. Gary B. and Pamela S. Williams Honors College at IdeaExchange@UAkron, the institutional repository of The nivU ersity of Akron in Akron, Ohio, USA. It has been accepted for inclusion in Honors Research Projects by an authorized administrator of IdeaExchange@UAkron. For more information, please contact [email protected], [email protected]. 2017 THE EXPERIMENTAL DETERMINATION OF A MODEL AIRPLANE KOKEN, MICHAEL THE UNIVERSITY OF AKRON Honors Project TABLE OF CONTENTS List of Tables ................................................................................................................................................
    [Show full text]
  • Rotational Motion (The Dynamics of a Rigid Body)
    University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat­ ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de­ scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates.
    [Show full text]
  • Motion Projectile Motion,And Straight Linemotion, Differences Between the Similaritiesand 2 CHAPTER 2 Acceleration Make Thefollowing As Youreadthe ●
    026_039_Ch02_RE_896315.qxd 3/23/10 5:08 PM Page 36 User-040 113:GO00492:GPS_Reading_Essentials_SE%0:XXXXXXXXXXXXX_SE:Application_File chapter 2 Motion section ●3 Acceleration What You’ll Learn Before You Read ■ how acceleration, time, and velocity are related Describe what happens to the speed of a bicycle as it goes ■ the different ways an uphill and downhill. object can accelerate ■ how to calculate acceleration ■ the similarities and differences between straight line motion, projectile motion, and circular motion Read to Learn Study Coach Outlining As you read the Velocity and Acceleration section, make an outline of the important information in A car sitting at a stoplight is not moving. When the light each paragraph. turns green, the driver presses the gas pedal and the car starts moving. The car moves faster and faster. Speed is the rate of change of position. Acceleration is the rate of change of velocity. When the velocity of an object changes, the object is accelerating. Remember that velocity is a measure that includes both speed and direction. Because of this, a change in velocity can be either a change in how fast something is moving or a change in the direction it is moving. Acceleration means that an object changes it speed, its direction, or both. How are speeding up and slowing down described? ●D Construct a Venn When you think of something accelerating, you probably Diagram Make the following trifold Foldable to compare and think of it as speeding up. But an object that is slowing down contrast the characteristics of is also accelerating. Remember that acceleration is a change in acceleration, speed, and velocity.
    [Show full text]
  • Frames of Reference
    Galilean Relativity 1 m/s 3 m/s Q. What is the women velocity? A. With respect to whom? Frames of Reference: A frame of reference is a set of coordinates (for example x, y & z axes) with respect to whom any physical quantity can be determined. Inertial Frames of Reference: - The inertia of a body is the resistance of changing its state of motion. - Uniformly moving reference frames (e.g. those considered at 'rest' or moving with constant velocity in a straight line) are called inertial reference frames. - Special relativity deals only with physics viewed from inertial reference frames. - If we can neglect the effect of the earth’s rotations, a frame of reference fixed in the earth is an inertial reference frame. Galilean Coordinate Transformations: For simplicity: - Let coordinates in both references equal at (t = 0 ). - Use Cartesian coordinate systems. t1 = t2 = 0 t1 = t2 At ( t1 = t2 ) Galilean Coordinate Transformations are: x2= x 1 − vt 1 x1= x 2+ vt 2 or y2= y 1 y1= y 2 z2= z 1 z1= z 2 Recall v is constant, differentiation of above equations gives Galilean velocity Transformations: dx dx dx dx 2 =1 − v 1 =2 − v dt 2 dt 1 dt 1 dt 2 dy dy dy dy 2 = 1 1 = 2 dt dt dt dt 2 1 1 2 dz dz dz dz 2 = 1 1 = 2 and dt2 dt 1 dt 1 dt 2 or v x1= v x 2 + v v x2 =v x1 − v and Similarly, Galilean acceleration Transformations: a2= a 1 Physics before Relativity Classical physics was developed between about 1650 and 1900 based on: * Idealized mechanical models that can be subjected to mathematical analysis and tested against observation.
    [Show full text]
  • L-9 Conservation of Energy, Friction and Circular Motion Kinetic Energy Potential Energy Conservation of Energy Amusement Pa
    L-9 Conservation of Energy, Friction and Circular Motion Kinetic energy • If something moves in • Kinetic energy, potential energy and any way, it has conservation of energy kinetic energy • kinetic energy (KE) • What is friction and what determines how is energy of motion m v big it is? • If I drive my car into a • Friction is what keeps our cars moving tree, the kinetic energy of the car can • What keeps us moving in a circular path? do work on the tree – KE = ½ m v2 • centripetal vs. centrifugal force it can knock it over KE does not depend on which direction the object moves Potential energy conservation of energy • If I raise an object to some height (h) it also has • if something has energy W stored as energy – potential energy it doesn’t loose it GPE = mgh • If I let the object fall it can do work • It may change from one • We call this Gravitational Potential Energy form to another (potential to kinetic and F GPE= m x g x h = m g h back) h • KE + PE = constant mg mg m in kg, g= 10m/s2, h in m, GPE in Joules (J) • example – roller coaster • when we do work in W=mgh PE regained • the higher I lift the object the more potential lifting the object, the as KE energy it gas work is stored as • example: pile driver, spring launcher potential energy. Amusement park physics Up and down the track • the roller coaster is an excellent example of the conversion of energy from one form into another • work must first be done in lifting the cars to the top of the first hill.
    [Show full text]
  • 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]
  • Multidisciplinary Design Project Engineering Dictionary Version 0.0.2
    Multidisciplinary Design Project Engineering Dictionary Version 0.0.2 February 15, 2006 . DRAFT Cambridge-MIT Institute Multidisciplinary Design Project This Dictionary/Glossary of Engineering terms has been compiled to compliment the work developed as part of the Multi-disciplinary Design Project (MDP), which is a programme to develop teaching material and kits to aid the running of mechtronics projects in Universities and Schools. The project is being carried out with support from the Cambridge-MIT Institute undergraduate teaching programe. For more information about the project please visit the MDP website at http://www-mdp.eng.cam.ac.uk or contact Dr. Peter Long Prof. Alex Slocum Cambridge University Engineering Department Massachusetts Institute of Technology Trumpington Street, 77 Massachusetts Ave. Cambridge. Cambridge MA 02139-4307 CB2 1PZ. USA e-mail: [email protected] e-mail: [email protected] tel: +44 (0) 1223 332779 tel: +1 617 253 0012 For information about the CMI initiative please see Cambridge-MIT Institute website :- http://www.cambridge-mit.org CMI CMI, University of Cambridge Massachusetts Institute of Technology 10 Miller’s Yard, 77 Massachusetts Ave. Mill Lane, Cambridge MA 02139-4307 Cambridge. CB2 1RQ. USA tel: +44 (0) 1223 327207 tel. +1 617 253 7732 fax: +44 (0) 1223 765891 fax. +1 617 258 8539 . DRAFT 2 CMI-MDP Programme 1 Introduction This dictionary/glossary has not been developed as a definative work but as a useful reference book for engi- neering students to search when looking for the meaning of a word/phrase. It has been compiled from a number of existing glossaries together with a number of local additions.
    [Show full text]
  • Exploring Robotics Joel Kammet Supplemental Notes on Gear Ratios
    CORC 3303 – Exploring Robotics Joel Kammet Supplemental notes on gear ratios, torque and speed Vocabulary SI (Système International d'Unités) – the metric system force torque axis moment arm acceleration gear ratio newton – Si unit of force meter – SI unit of distance newton-meter – SI unit of torque Torque Torque is a measure of the tendency of a force to rotate an object about some axis. A torque is meaningful only in relation to a particular axis, so we speak of the torque about the motor shaft, the torque about the axle, and so on. In order to produce torque, the force must act at some distance from the axis or pivot point. For example, a force applied at the end of a wrench handle to turn a nut around a screw located in the jaw at the other end of the wrench produces a torque about the screw. Similarly, a force applied at the circumference of a gear attached to an axle produces a torque about the axle. The perpendicular distance d from the line of force to the axis is called the moment arm. In the following diagram, the circle represents a gear of radius d. The dot in the center represents the axle (A). A force F is applied at the edge of the gear, tangentially. F d A Diagram 1 In this example, the radius of the gear is the moment arm. The force is acting along a tangent to the gear, so it is perpendicular to the radius. The amount of torque at A about the gear axle is defined as = F×d 1 We use the Greek letter Tau ( ) to represent torque.
    [Show full text]