CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.45


 Marylou Bradford
 7 months ago
 Views:
Transcription
1 1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.45 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa
2 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES Today s objectives: Students will be able to 1 Apply Newton s second law to determine forces and accelerations for particles in rectilinear motion. Inclass activities: Reading Quiz Applications Equations of Motion Using Rectangular (Cartesian) Coordinates Concept Quiz Group Problem Solving Attention Quiz
3 3 / 40 READING QUIZ 1 In dynamics, the friction force acting on a moving object is always (a) in the direction of its motion. (b) a kinetic friction. (c) a static friction. (d) zero. ANS: (b) 2 If a particle is connected to a spring, the elastic spring force is expressed by F = ks. The s in this equation is the (a) spring constant. (b) undeformed length of the spring. (c) difference between deformed and undeformed lengths. (d) deformed length of the spring. ANS: (c)
4 APPLICATIONS If a man is trying to move a 100 lb crate, how large a force F must he exert to start moving the crate? What factors influence how large this force must be to start moving the crate? If the crate starts moving, is there acceleration present? What would you have to know before you could find these answers? 4 / 40
5 5 / 40 APPLICATIONS(continued) Objects that move in air (or other fluid) have a drag force acting on them. This drag force is a function of velocity. If the dragster is traveling with a known velocity and the magnitude of the opposing drag force at any instant is given as a function of velocity, can we determine the time and distance required for dragster to come to a stop if its engine is shut off? How?
6 6 / 40 RECTANGULAR COORDINATES (Section 13.4) The equation of motion, F = ma, is best used when the problem requires finding forces (especially forces perpendicular to the path), accelerations, velocities, or mass. Remember, unbalanced forces cause acceleration! Three scalar equations can be written from this vector equation. The equation of motion, being a vector equation, may be expressed in terms of its three components in the Cartesian (rectangular) coordinate system as ΣF = ma or (ΣF x ) i + (ΣF y ) j + (ΣF z ) k = m(a x i + a y j + a z k) or, as scalar equations, ΣF x = ma x, ΣF y = ma y, ΣF z = ma z
7 7 / 40 PROCEDURE FOR ANALYSIS Free Body Diagram Establish your coordinate system and draw the particle s free body diagram showing only external forces. These external forces usually include the weight, normal forces, friction forces, and applied forces. Show the ma vector (sometimes called the inertial force) on a separate diagram. Make sure any friction forces act opposite to the direction of motion! If the particle is connected to an elastic linear spring, a spring force equal to ks should be included on the FBD.
8 8 / 40 PROCEDURE FOR ANALYSIS (continued) Equation of Motion If the forces can be resolved directly from the freebody diagram (often the case in 2D problems), use the scalar form of the equation of motion. In more complex cases (usually 3D), a Cartesian vector is written for every force and a vector analysis is often best. A Cartesian vector formulation of the second law is ΣF = ma or ΣF x i + ΣF y j + ΣF z k = m(a x i + a y j + a z k) Three scalar equations can be written from this vector equation. You may only need two equations if the motion is in 2D.
9 9 / 40 PROCEDURE FOR ANALYSIS (continued) Kinematics The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion. Any of the kinematics tools learned in Chapter 12 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem!
10 10 / 40 EXAMPLE Given: The 200lb mine car is hoisted up the incline. The motor M pulls in the cable with an acceleration of 4 ft/s 2. Find: The acceleration of the mine car and the tension in the cable. Plan: 1 Draw the freebody and kinetic diagrams of the car. 2 Using a dependent motion equation, determine an acceleration relationship between cable and mine car. 3 Apply the equation of motion to determine the cable tension.
11 11 / 40 Solution EXAMPLE (solution) 1. Draw the freebody and kinetic diagrams of the mine car: Since the motion is up the incline, rotate the x y axes. Motion occurs only in the xdirection. We are also neglecting any friction in the wheel bearings, etc., on the cart.
12 12 / The cable equation results in s p + 2s c = l t Taking the derivative twice yields a p + 2a c = 0 (1) The relative acceleration is a p = a c + a p/c. As the motor is mounted on the car, a p/c = 4 ft/s 2. So, a p = a c + 4 ft/s 2 (2) Solving equations (1) and (2), yields a C = ft/s 2
13 3. Apply the equation of motion in the xdirection: m = 200/32.2 = slug (+ ) F x = ma x 3T mg(sin 30 ) = ma x 3T (200)(sin 30 ) = 6.211(1.333) T = 36.1 lb 13 / 40
14 14 / 40 CHECK YOUR UNDERSTANDING QUIZ 1. If the cable has a tension of 3 N, determine the acceleration of block B. (a) 4.26 m/s 2 (b) 4.26 m/s 2 (c) 8.31 m/s 2 (d) 8.31 m/s 2 ANS: (d) 2. Determine the acceleration of the block. (a) 2.20 m/s 2 (b) 3.17 m/s 2 (c) 11.0 m/s 2 (d) 4.26 m/s 2 ANS: (a)
15 15 / 40 GROUP PROBLEM SOLVING Given: W A = 10 lb, W B = 20 lb, v oa = 2 ft/s( ), and µ k = 0.2 Find: v A when A has moved 4 feet to the right. Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics. First, draw free body diagrams of A and B. Apply the equation of motion to each. Using dependent motion equations, derive a relationship between a A and a B and use with the equation of motion formulas.
16 GROUP PROBLEM SOLVING (Solution) Freebody and kinetic diagrams of B: Apply the equation of motion to B: + ΣF y = ma y W B 2T = m B a B 20 2T = a B (3) Freebody and kinetic diagrams of A: Apply the equations of motion to A: (+ ) ΣF y = ma y = 0 (+ ) ΣF x = ma x N = W A = 10 lb f = µ k N = 2 lb f T = m A a A 2 T = a A (4) 16 / 40
17 17 / 40 Now consider the kinematics. Constraint equation: Therefore s A + 2s B = constant v A + 2v B = 0 a A + 2a B = 0 a A = 2a B (5) (Notice a A is considered positive to the left and a B is positive downward.)
18 18 / 40 Now combine equations (3), (4), and (5) Constraint equation: T = 22 = 7.33 lb 3 a A = ft/s 2 = ft/s 2 ( ) Now use the kinematic equation: (v A ) 2 = (v 0A ) 2 + 2a A (s A s 0A ) (v A ) 2 = (2) 2 + 2( 17.16)( 4) v A = 11.9 ft/s( )
19 19 / 40 ATTENTION QUIZ 1. Determine the tension in the cable when the 400 kg box is moving upward with a 4 m/s 2 acceleration. (a) 2265 N (b) 3365 N (c) 5524 N (d) 6543 N ANS: (c) T 60 a = 4 m/s
20 20 / 40 ATTENTION QUIZ 2. A 10 lb particle has forces of F 1 = (3i + 5j) lb and F 2 = ( 7i + 9j) lb acting on it. Determine the acceleration of the particle. (a) ( 0.4i + 1.4j) ft/s 2 (b) ( 0.4i + 14j) ft/s 2 (c) ( 12.9i + 45j) ft/s 2 (d) (13i + 4j) ft/s 2 ANS: (c)
21 21 / 40 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today s objectives: Students will be able to 1 Apply the equation of motion using normal and tangential coordinates. Inclass activities: Reading Quiz Applications Equation of Motion in n t Coordinates Concept Quiz Group Problem Solving Attention Quiz
22 22 / 40 READING QUIZ 1 The normal component of the equation of motion is written as ΣF n = ma n, where ΣF n is referred to as the? (a) impulse (b) centripetal force (c) tangential force (d) inertia force ANS: (b) 2 The positive n direction of the normal and tangential coordinates is. (a) normal to the tangential component (b) always directed toward the center of curvature (c) normal to the binormal component (d) All of the above. ANS: (d)
23 23 / 40 APPLICATIONS Race tracks are often banked in the turns to reduce the frictional forces required to keep the cars from sliding up to the outer rail at high speeds. If the car s maximum velocity and a minimum coefficient of friction between the tires and track are specified, how can we determine the minimum banking angle (θ) required to prevent the car from sliding up the track?
24 APPLICATIONS(continued) The picture shows a ride at the amusement park. The hydraulicallypowered arms turn at a constant rate, which creates a centrifugal force on the riders. We need to determine the smallest angular velocity of the cars A and B so that the passengers do not loose contact with the seat. What parameters do we need for this calculation? 24 / 40
25 25 / 40 APPLICATIONS(continued) Satellites are held in orbit around the earth by using the earth s gravitational pull as the centripetal force  the force acting to change the direction of the satellite s velocity. Knowing the radius of orbit of the satellite, we need to determine the required speed of the satellite to maintain this orbit. What equation governs this situation?
26 26 / 40 NORMAL & TANGENTIAL COORDINATES (Sec. 13.5) When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. The normal direction (n) always points toward the path s center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle.
27 27 / 40 EQUATIONS OF MOTION Since the equation of motion is a vector equation, ΣF = ma, it may be written in terms of the n and t coordinates as ΣF t u t + ΣF n u n + ΣF b u b = ma t + ma n F t and F n are the sums of the force components acting in the t and n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations: Ft = ma t and Fn = ma n Since there is no motion in the binormal (b) direction, we can also write F b = 0.
28 28 / 40 NORMAL AND TANGENTIAL ACCERLERATIONS The tangential acceleration, a t = dv/dt, represents the time rate of change in the magnitude of the velocity. Depending on the direction of ΣF t, the particle s speed will either be increasing or decreasing. The normal acceleration, a n = v 2 /ρ, represents the time rate of change in the direction of the velocity vector. Remember, a n always acts toward the path s center of curvature. Thus, ΣF n will always be directed toward the center of the path. Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from ρ = dy [1 + ( d2 y dx 2 dx )2 ] 3/2
29 29 / 40 SOLVING PROBLEMS WITH n t COORDINATES Use n t coordinates when a particle is moving along a known, curved path. Establish the n t coordinate system on the particle. Draw freebody and kinetic diagrams of the particle. The normal acceleration (a n ) always acts inward (the positive ndirection). The tangential acceleration (a t ) may act in either the positive or negative t direction. Apply the equations of motion in scalar form and solve. It may be necessary to employ the kinematic relations: a t = dv dt = v dv ds a n = v2 ρ (6) (7)
30 EXAMPLE Given: At the instant θ = 45, the boy with a mass of 75 kg, moves a speed of 6 m/s, which is increasing at 0.5 m/s 2. Neglect his size and the mass of the seat and cords. The seat is pin connected to the frame BC. Find: Horizontal and vertical reactions of the seat on the boy. Plan: 1 Since the problem involves a curved path and requires finding the force perpendicular to the path, use n t coordinates. 2 Draw the boy s freebody and kinetic diagrams. 3 Apply the equation of motion in the n t directions. 30 / 40
31 31 / 40 EXAMPLE (Solution) 1. The n t coordinate system can be established on the boy at angle θ = 45. Approximating the boy and seat together as a particle, the freebody and kinetic diagrams can be drawn.
32 EXAMPLE (continued) .//01234#35,6.,+# 75"/89#35,6.,+#! '#! +,! # $ % # ()*# +, " # 2. Apply the equations of motion in the n t directions. a. ma n = R x cos θ R y sin θ + W sin θ Using a n = v 2 /ρ = 6 2 /10, W = 75(9.81) N, and m = 75 kg, we get: R x cos 45 R y sin = (75)(6 2 /10) b. ma t = R x sin θ + R y cos θ W cos θ we get: R x sin 45 + R y cos = (75)(0.5) Using above two equations, solve for R x, R y. R x = 217 N and R y = 572 N. "# $ & # "# 32 / 40
33 CONCEPT QUIZ &"!"#"$"%" 1. A 10 kg sack slides down a smooth surface. If the normal force on the surface at the flat spot, A, is 98.1 N ( ), the radius of curvature is (a) 0.2 m (b) 0.4 m (c) 1.0 m (d) None of the above ANS: (d) 2. A 20 lb block is moving along a smooth surface. If the normal force on the surface at A is 10 lb, the velocity is. (a) 7.6 ft/s (b) 9.6 ft/s (c) 10.6 ft/s (d) 12.6 ft/s ANS: (c) 33 / 40
34 GROUP PROBLEM SOLVING Given: A 800 kg car is traveling over the hill having the shape of a parabola. When it is at point A, it is traveling at 9 m/s and increasing its speed at 3 m/s 2. Find: The resultant normal force and resultant frictional force exerted on the road at point A. Plan: 1 Treat the car as a particle. 2 Draw the freebody and kinetic diagrams. 3 Apply the equations of motion in the n t directions. 4 Use calculus to determine the slope and radius of curvature of the path at point A. 34 / 40
35 1. The n t coordinate system can be established on the car at point A. Treat the car as a particle and draw the freebody and kinetic diagrams: W = mg = weight of car N = resultant normal force on road F = resultant friction force on road 35 / 40
36 36 / Apply the equations of motion in the n t directions: W cos θ N = ma n Using W = mg and a n = v 2 /ρ = (9) 2 /ρ (800)(9.81) cos θ N = (800)(81/ρ) N = 7848 cos θ 64800/ρ (8) W sin θ F = ma t Using W = mg and a t = 3m/s 2 (given) (800)(9.81) sin θ F = (800)(3) F = 7848 sin θ 2400 (9)
37 37 / Determine ρ by differentiating y = f(x) at x = 80 m: y = 20(1 x 2 /6400) dy/dx = ( 40)x/6400 d 2 y/dx 2 = ( 40)/640 (10) dy [1 + ( dx ρ x=80 = )2 ] 3/2 [1 + ( 0.5) 2 ] 3/2 d2 y = = meter dx 2 Determine θ from the slope of the curve at A: tan θ = dy/dx x=80 = 40x 6400 = 0.5 x=80 θ = tan 1 ( dy dx ) = tan 1 ( 0.5) = 26.6
38 38 / 40 From Eq. (8): N = 7848 cos θ 64800/ρ = 7848 cos(26.6 ) 64800/223.6 = 6728 N (11) From Eq.(9) : F = 7848 sin θ 2400 = 7848 sin(26.6 o ) 2400 = 1114 N (12)
39 39 / 40 ATTENTION QUIZ 1 The tangential acceleration of an object (a) represents the rate of change of the velocity vector s direction. (b) represents the rate of change in the magnitude of the velocity. (c) is a function of the radius of curvature. (d) Both (b) and (c). ANS: (b)
40 40 / 40 ATTENTION QUIZ 1 The block has a mass of 20 kg and a speed of v = 30 m/s at the instant it is at its lowest point. Determine the tension in the cord at this instant. (a) 1596 N (b) 1796 N (c) 1996 N (d) 2196 N #$"%"!" &"'"($"%)*" ANS: (c)
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today s Objectives: Students will be able to: 1. Apply the equation of motion using normal and tangential coordinates. InClass Activities: Check
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5)
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5) Today s Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. APPLICATIONS Race
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 28: Ch.17, Sec.2 3
1 / 20 CEE 271: Applied Mechanics II, Dynamics Lecture 28: Ch.17, Sec.2 3 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Monday, November 1, 2011 2 / 20 PLANAR KINETIC
More informationTHE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES
THE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES Today s Objectives: Students will be able to: 1. Calculate the work of a force. 2. Apply the principle of work and energy to
More informationNEWTON S LAWS OF MOTION (EQUATION OF MOTION) (Sections )
NEWTON S LAWS OF MOTION (EQUATION OF MOTION) (Sections 13.113.3) Today s Objectives: Students will be able to: a) Write the equation of motion for an accelerating body. b) Draw the freebody and kinetic
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 1: Ch.12, Sec.13h
1 / 30 CEE 271: Applied Mechanics II, Dynamics Lecture 1: Ch.12, Sec.13h Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, August 21, 2012 2 / 30 INTRODUCTION
More informationME 230 Kinematics and Dynamics
ME 230 Kinematics and Dynamics WeiChih Wang Department of Mechanical Engineering University of Washington Lecture 8 Kinetics of a particle: Work and Energy (Chapter 14)  14.114.3 W. Wang 2 Kinetics
More informationEQUATIONS OF MOTION: CYLINDRICAL COORDINATES
Today s Objectives: Students will be able to: 1. Analyze the kinetics of a particle using cylindrical coordinates. EQUATIONS OF MOTION: CYLINDRICAL COORDINATES InClass Activities: Check Homework Reading
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 33: Ch.19, Sec.1 2
1 / 36 CEE 271: Applied Mechanics II, Dynamics Lecture 33: Ch.19, Sec.1 2 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Thursday, December 6, 2012 2 / 36 LINEAR
More informationUNIT07. Newton s Three Laws of Motion
1. Learning Objectives: UNIT07 Newton s Three Laws of Motion 1. Understand the three laws of motion, their proper areas of applicability and especially the difference between the statements of the first
More informationAP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force).
AP Physics C: Mechanics Practice (Newton s Laws including friction, resistive forces, and centripetal force). 1981M1. A block of mass m, acted on by a force of magnitude F directed horizontally to the
More information66 Chapter 6: FORCE AND MOTION II
Chapter 6: FORCE AND MOTION II 1 A brick slides on a horizontal surface Which of the following will increase the magnitude of the frictional force on it? A Putting a second brick on top B Decreasing the
More informationHATZIC SECONDARY SCHOOL
HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT CIRCULAR MOTION MULTIPLE CHOICE / 30 OPEN ENDED / 65 TOTAL / 95 NAME: 1. An object travels along a path at constant speed. There is a constant
More informationChapter 5 Newton s Laws of Motion. Copyright 2010 Pearson Education, Inc.
Chapter 5 Newton s Laws of Motion Force and Mass Units of Chapter 5 Newton s First Law of Motion Newton s Second Law of Motion Newton s Third Law of Motion The Vector Nature of Forces: Forces in Two Dimensions
More informationChapter 4: Newton s Second Law F = m a. F = m a (4.2)
Lecture 7: Newton s Laws and Their Applications 1 Chapter 4: Newton s Second Law F = m a First Law: The Law of Inertia An object at rest will remain at rest unless, until acted upon by an external force.
More informationAnnouncements. Equilibrium of a Particle in 2D
nnouncements Equilibrium of a Particle in 2D Today s Objectives Draw a free body diagram (FBD) pply equations of equilibrium to solve a 2D problem Class ctivities pplications What, why, and how of a
More informationAlgebra Based Physics Uniform Circular Motion
1 Algebra Based Physics Uniform Circular Motion 2016 07 20 www.njctl.org 2 Uniform Circular Motion (UCM) Click on the topic to go to that section Period, Frequency and Rotational Velocity Kinematics of
More informationChapter Four Holt Physics. Forces and the Laws of Motion
Chapter Four Holt Physics Forces and the Laws of Motion Physics Force and the study of dynamics 1.Forces  a. Force  a push or a pull. It can change the motion of an object; start or stop movement; and,
More informationPHYS 101 Previous Exam Problems. Force & Motion I
PHYS 101 Previous Exam Problems CHAPTER 5 Force & Motion I Newton s Laws Vertical motion Horizontal motion Mixed forces Contact forces Inclines General problems 1. A 5.0kg block is lowered with a downward
More informationCircular Motion & Gravitation FR Practice Problems
1) A mass m is attached to a length L of string and hung straight strainght down from a pivot. Small vibrations at the pivot set the mass into circular motion, with the string making an angle θ with the
More informationCIRCULAR MOTION AND GRAVITATION
CIRCULAR MOTION AND GRAVITATION An object moves in a straight line if the net force on it acts in the direction of motion, or is zero. If the net force acts at an angle to the direction of motion at any
More informationPreparing for Six Flags Physics Concepts
Preparing for Six Flags Physics Concepts uniform means constant, unchanging At a uniform speed, the distance traveled is given by Distance = speed x time At uniform velocity, the displacement is given
More informationContents. Objectives Circular Motion Velocity and Acceleration Examples Accelerating Frames Polar Coordinates Recap. Contents
Physics 121 for Majors Today s Class You will see how motion in a circle is mathematically similar to motion in a straight line. You will learn that there is a centripetal acceleration (and force) and
More informationVersion PREVIEW Semester 1 Review Slade (22222) 1
Version PREVIEW Semester 1 Review Slade () 1 This printout should have 48 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Holt SF 0Rev 10A
More informationIsaac Newton ( )
Isaac Newton (16421727) In the beginning of 1665 I found the rule for reducing any degree of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions
More informationProficient. a. The gravitational field caused by a. The student is able to approximate a numerical value of the
Unit 6. Circular Motion and Gravitation Name: I have not failed. I've just found 10,000 ways that won't work. Thomas Edison Big Idea 1: Objects and systems have properties such as mass and charge. Systems
More informationPLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work.
PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY Today s Objectives: Students will be able to: 1. Define the various ways a force and couple do work. InClass Activities: 2. Apply the principle of work
More informationPhysics 207 Lecture 10. Lecture 10. Employ Newton s Laws in 2D problems with circular motion
Lecture 10 Goals: Employ Newton s Laws in 2D problems with circular motion Assignment: HW5, (Chapters 8 & 9, due 3/4, Wednesday) For Tuesday: Finish reading Chapter 8, start Chapter 9. Physics 207: Lecture
More informationEQUATIONS OF MOTION: CYLINDRICAL COORDINATES (Section 13.6)
EQUATIONS OF MOTION: CYLINDRICAL COORDINATES (Section 13.6) Today s Objectives: Students will be able to analyze the kinetics of a particle using cylindrical coordinates. APPLICATIONS The forces acting
More information= o + t = ot + ½ t 2 = o + 2
Chapters 89 Rotational Kinematics and Dynamics Rotational motion Rotational motion refers to the motion of an object or system that spins about an axis. The axis of rotation is the line about which the
More informationChapter 4 Dynamics: Newton s Laws of Motion
Chapter 4 Dynamics: Newton s Laws of Motion Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal Force Applications
More informationPLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION
PLANAR KINETICS OF A RIGID BODY FORCE AND ACCELERATION I. Moment of Inertia: Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate.
More informationCentripetal force keeps an Rotation and Revolution
Centripetal force keeps an object in circular motion. Which moves faster on a merrygoround, a horse near the outside rail or one near the inside rail? While a hamster rotates its cage about an axis,
More informationPhys101 Lecture 5 Dynamics: Newton s Laws of Motion
Phys101 Lecture 5 Dynamics: Newton s Laws of Motion Key points: Newton s second law is a vector equation Action and reaction are acting on different objects FreeBody Diagrams Ref: 41,2,3,4,5,6,7. Page
More informationSECOND MIDTERM  REVIEW PROBLEMS
Physics 10 Spring 009 George A. WIllaims SECOND MIDTERM  REVIEW PROBLEMS A solution set is available on the course web page in pdf format. A data sheet is provided. No solutions for the following problems:
More informationLecture Presentation. Chapter 6 Preview Looking Ahead. Chapter 6 Circular Motion, Orbits, and Gravity
Chapter 6 Preview Looking Ahead Lecture Presentation Chapter 6 Circular Motion, Orbits, and Gravity Text: p. 160 Slide 62 Chapter 6 Preview Looking Back: Centripetal Acceleration In Section 3.8, you learned
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) You are standing in a moving bus, facing forward, and you suddenly fall forward as the
More informationUse the following to answer question 1:
Use the following to answer question 1: On an amusement park ride, passengers are seated in a horizontal circle of radius 7.5 m. The seats begin from rest and are uniformly accelerated for 21 seconds to
More informationDynamics II Motion in a Plane. Review Problems
Dynamics II Motion in a Plane Review Problems Problem 1 A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exerts an 8.0 N thrust
More informationCutnell/Johnson Physics
Cutnell/Johnson Physics Classroom Response System Questions Chapter 5 Dynamics of Uniform Circular Motion Interactive Lecture Questions 5.1.1. An airplane flying at 115 m/s due east makes a gradual turn
More informationPHYS 1303 Final Exam Example Questions
PHYS 1303 Final Exam Example Questions (In summer 2014 we have not covered questions 3035,40,41) 1.Which quantity can be converted from the English system to the metric system by the conversion factor
More informationPRACTICE TEST for Midterm Exam
South Pasadena AP Physics PRACTICE TEST for Midterm Exam FORMULAS Name Period Date / / d = vt d = v o t + ½ at 2 d = v o + v 2 t v = v o + at v 2 = v 2 o + 2ad v = v x 2 + v y 2 = tan 1 v y v v x = v cos
More informationForces I. Newtons Laws
Forces I Newtons Laws Kinematics The study of how objects move Dynamics The study of why objects move Newton s Laws and Forces What is force? What are they? Force A push or a pull Symbol is F Unit is N
More informationAdvanced Higher Physics. Rotational motion
Wallace Hall Academy Physics Department Advanced Higher Physics Rotational motion Problems AH Physics: Rotational Motion 1 2013 Data Common Physical Quantities QUANTITY SYMBOL VALUE Gravitational acceleration
More informationPLANAR KINETIC EQUATIONS OF MOTION: TRANSLATION
PLANAR KINETIC EQUATIONS OF MOTION: TRANSLATION Today s Objectives: Students will be able to: 1. Apply the three equations of motion for a rigid body in planar motion. 2. Analyze problems involving translational
More informationUnit 4 Work, Power & Conservation of Energy Workbook
Name: Per: AP Physics C Semester 1  Mechanics Unit 4 Work, Power & Conservation of Energy Workbook Unit 4  Work, Power, & Conservation of Energy Supplements to Text Readings from Fundamentals of Physics
More informationChapter 10: Friction A gem cannot be polished without friction, nor an individual perfected without
Chapter 10: Friction 101 Chapter 10 Friction A gem cannot be polished without friction, nor an individual perfected without trials. Lucius Annaeus Seneca (4 BC  65 AD) 10.1 Overview When two bodies are
More informationP211 Spring 2004 Form A
1. A 2 kg block A traveling with a speed of 5 m/s as shown collides with a stationary 4 kg block B. After the collision, A is observed to travel at right angles with respect to the initial direction with
More informationREVISING MECHANICS (LIVE) 30 JUNE 2015 Exam Questions
REVISING MECHANICS (LIVE) 30 JUNE 2015 Exam Questions Question 1 (Adapted from DBE November 2014, Question 2) Two blocks of masses 20 kg and 5 kg respectively are connected by a light inextensible string,
More informationPLANAR RIGID BODY MOTION: TRANSLATION & ROTATION
PLANAR RIGID BODY MOTION: TRANSLATION & ROTATION Today s Objectives : Students will be able to: 1. Analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis. InClass
More information9/20/11. Physics 101 Tuesday 9/20/11 Class 8" Chapter " Weight and Normal forces" Frictional Forces"
Reading Quiz Physics 101 Tuesday 9/20/11 Class 8" Chapter 5.6 6.1" Weight and Normal forces" Frictional Forces" The force due to kinetic friction is usually larger than the force due to static friction.
More informationPractice Problems from Chapters 1113, for Midterm 2. Physics 11a Fall 2010
Practice Problems from Chapters 1113, for Midterm 2. Physics 11a Fall 2010 Chapter 11 1. The Ferris wheel shown below is turning at constant speed. Draw and label freebody diagrams showing the forces
More informationINTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION
INTRODUCTION & RECTILINEAR KINEMATICS: CONTINUOUS MOTION (Sections 12.112.2) Today s Objectives: Students will be able to find the kinematic quantities (position, displacement, velocity, and acceleration)
More informationCircular Velocity and Centripetal Acceleration
1. An object is spun around in circular motion such that it completes 100 cycles in 25 s. a. What is the period of its rotation? [0.25 s] b. If the radius is 0.3 m what is the velocity? [7.54 m/s] c. Draw
More informationApplying Newton s Laws
Chapter 5 Applying Newton s Laws PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Goals for Chapter 5 To use and apply Newton s Laws
More informationDynamics; Newton s Laws of Motion
Dynamics; Newton s Laws of Motion Force A force is any kind of push or pull on an object. An object at rest needs a force to get it moving; a moving object needs a force to change its velocity. The magnitude
More informationAngular Speed and Angular Acceleration Relations between Angular and Linear Quantities
Angular Speed and Angular Acceleration Relations between Angular and Linear Quantities 1. The tires on a new compact car have a diameter of 2.0 ft and are warranted for 60 000 miles. (a) Determine the
More informationPES Physics 1 Practice Questions Exam 2. Name: Score: /...
Practice Questions Exam /page PES 0 003  Physics Practice Questions Exam Name: Score: /... Instructions Time allowed for this is exam is hour 5 minutes... multiple choice (... points)... written problems
More informationHint 1. The direction of acceleration can be determined from Newton's second law
Chapter 5 [ Edit ] Overview Summary View Diagnostics View Print View with Answers Chapter 5 Due: 11:59pm on Sunday, October 2, 2016 To understand how points are awarded, read the Grading Policy for this
More informationConstant Acceleration. Physics General Physics Lecture 7 Uniform Circular Motion 9/13/2016. Fall 2016 Semester Prof.
Physics 22000 General Physics Lecture 7 Uniform Circular Motion Fall 2016 Semester Prof. Matthew Jones 1 2 Constant Acceleration So far we have considered motion when the acceleration is constant in both
More informationTue Sept 15. Dynamics  Newton s Laws of Motion. Forces: Identifying Forces Freebody diagram Affect on Motion
Tue Sept 15 Assignment 4 Friday Preclass Thursday Lab  Print, do prelab Closed toed shoes Exam Monday Oct 5 7:159:15 PM email me if class conflict or extended time Dynamics  Newton s Laws of Motion
More informationName Period Date A) B) C) D)
Example Problems 9.2 E1. A car rounds a curve of constant radius at a constant speed. Which diagram best represents the directions of both the car s velocity and acceleration? Explain: A) B) C) D) E2.
More informationPotential Energy. Serway 7.6, 7.7;
Potential Energy Conservative and nonconservative forces Gravitational and elastic potential energy Mechanical Energy Serway 7.6, 7.7; 8.1 8.2 Practice problems: Serway chapter 7, problems 41, 43 chapter
More informationFALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003
FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Saturday, 14 December 2013, 1PM to 4 PM, AT 1003 NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 14 pages. Make sure none are missing 2. There is
More informationPhysics General Physics. Lecture 7 Uniform Circular Motion. Fall 2016 Semester Prof. Matthew Jones
Physics 22000 General Physics Lecture 7 Uniform Circular Motion Fall 2016 Semester Prof. Matthew Jones 1 2 Constant Acceleration So far we have considered motion when the acceleration is constant in both
More informationCircular Motion and Gravitation Notes 1 Centripetal Acceleration and Force
Circular Motion and Gravitation Notes 1 Centripetal Acceleration and Force This unit we will investigate the special case of kinematics and dynamics of objects in uniform circular motion. First let s consider
More informationAddis Ababa University Addis Ababa Institute of Technology School Of Mechanical and Industrial Engineering Extension Division Assignment 2
Addis Ababa University Addis Ababa Institute of Technology School Of Mechanical and Industrial Engineering Extension Division Assignment 2 1. The 50kg crate is projected along the floor with an initial
More informationExam 3 Practice Solutions
Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at
More informationLECTURE 9 FRICTION & SPRINGS. Instructor: Kazumi Tolich
LECTURE 9 FRICTION & SPRINGS Instructor: Kazumi Tolich Lecture 9 2 Reading chapter 61 to 62 Friction n Static friction n Kinetic friction Springs Static friction 3 Static friction is the frictional force
More informationPhysics 207 Lecture 12. Lecture 12
Lecture 12 Goals: Chapter 8: Solve 2D motion problems with friction Chapter 9: Momentum & Impulse v Solve problems with 1D and 2D Collisions v Solve problems having an impulse (Force vs. time) Assignment:
More informationPhysics 1401V October 28, 2016 Prof. James Kakalios Quiz No. 2
This is a closed book, closed notes, quiz. Only simple (nonprogrammable, nongraphing) calculators are permitted. Define all symbols and justify all mathematical expressions used. Make sure to state all
More informationKINETIC ENERGY AND WORK
Chapter 7: KINETIC ENERGY AND WORK 1 Which of the following is NOT a correct unit for work? A erg B ft lb C watt D newton meter E joule 2 Which of the following groups does NOT contain a scalar quantity?
More informationPhysics, Chapter 3: The Equilibrium of a Particle
University of Nebraska  Lincoln DigitalCommons@University of Nebraska  Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 11958 Physics, Chapter 3: The Equilibrium of a Particle
More informationThere are two main types of friction:
Section 4.15: Friction Friction is needed to move. Without friction, a car would sit in one spot spinning its tires, and a person would not be able to step forward. However, the motion of an object along
More informationCircular Motion Dynamics
Circular Motion Dynamics 8.01 W04D2 Today s Reading Assignment: MIT 8.01 Course Notes Chapter 9 Circular Motion Dynamics Sections 9.19.2 Announcements Problem Set 3 due Week 5 Tuesday at 9 pm in box outside
More informationdt 2 x = r cos(θ) y = r sin(θ) r = x 2 + y 2 tan(θ) = y x A circle = πr 2
v = v i + at a dv dt = d2 x dt 2 A sphere = 4πr 2 x = x i + v i t + 1 2 at2 x = r cos(θ) V sphere = 4 3 πr3 v 2 = v 2 i + 2a x F = ma R = v2 sin(2θ) g y = r sin(θ) r = x 2 + y 2 tan(θ) = y x a c = v2 r
More informationFriction is always opposite to the direction of motion.
6. Forces and MotionII Friction: The resistance between two surfaces when attempting to slide one object across the other. Friction is due to interactions at molecular level where rough edges bond together:
More informationPhysics for Scientists and Engineers. Chapter 6 Dynamics I: Motion Along a Line
Physics for Scientists and Engineers Chapter 6 Dynamics I: Motion Along a Line Spring, 008 Ho Jung Paik Applications of Newton s Law Objects can be modeled as particles Masses of strings or ropes are negligible
More informationSteps to Solving Newtons Laws Problems.
Mathematical Analysis With Newtons Laws similar to projectiles (x y) isolation Steps to Solving Newtons Laws Problems. 1) FBD 2) Axis 3) Components 4) Fnet (x) (y) 5) Subs 1 Visual Samples F 4 1) F 3 F
More informationPhysics 201, Lecture 10
Physics 201, Lecture 10 Today s Topics n Circular Motion and Newton s Law (Sect. 6.1,6.2) n Centripetal Force in Uniform Circular Motion n Examples n n Motion in Accelerated Frame (sec. 6.3, conceptual
More informationMarch 10, P12 Inclined Planes.notebook. Physics 12. Inclined Planes. Push it Up Song
Physics 12 Inclined Planes Push it Up Song 1 Bell Work A box is pushed up a ramp at constant velocity. Draw a neatly labeled FBD showing all of the forces acting on the box. direction of motion θ F p F
More informationConcept Question: Normal Force
Concept Question: Normal Force Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is 1. larger than 2. identical
More information4) Vector = and vector = What is vector = +? A) B) C) D) E)
1) Suppose that an object is moving with constant nonzero acceleration. Which of the following is an accurate statement concerning its motion? A) In equal times its speed changes by equal amounts. B) In
More informationChapter 5 Newton s Laws of Motion. Copyright 2010 Pearson Education, Inc.
Chapter 5 Newton s Laws of Motion Copyright 2010 Pearson Education, Inc. Force and Mass Copyright 2010 Pearson Education, Inc. Units of Chapter 5 Newton s First Law of Motion Newton s Second Law of Motion
More informationLecture Outline Chapter 5. Physics, 4 th Edition James S. Walker. Copyright 2010 Pearson Education, Inc.
Lecture Outline Chapter 5 Physics, 4 th Edition James S. Walker Chapter 5 Newton s Laws of Motion Force and Mass Units of Chapter 5 Newton s First Law of Motion Newton s Second Law of Motion Newton s Third
More informationNewton s 3 Laws of Motion
Newton s 3 Laws of Motion 1. If F = 0 No change in motion 2. = ma Change in motion Fnet 3. F = F 1 on 2 2 on 1 Newton s First Law (Law of Inertia) An object will remain at rest or in a constant state of
More informationCHAPTER 4 TEST REVIEW  Answer Key
AP PHYSICS Name: Period: Date: DEVIL PHYSICS BADDEST CLASS ON CAMPUS 50 Multiple Choice 45 Single Response 5 MultiResponse Free Response 3 Short Free Response 2 Long Free Response AP EXAM CHAPTER TEST
More informationBase your answers to questions 5 and 6 on the information below.
1. A car travels 90. meters due north in 15 seconds. Then the car turns around and travels 40. meters due south in 5.0 seconds. What is the magnitude of the average velocity of the car during this 20.second
More information第 1 頁, 共 7 頁 Chap10 1. Test Bank, Question 3 One revolution per minute is about: 0.0524 rad/s 0.105 rad/s 0.95 rad/s 1.57 rad/s 6.28 rad/s 2. *Chapter 10, Problem 8 The angular acceleration of a wheel
More informationChapter 4 Dynamics: Newton s Laws of Motion
Chapter 4 Dynamics: Newton s Laws of Motion Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal Force Applications
More information112 A General Method, and Rolling without Slipping
112 A General Method, and Rolling without Slipping Let s begin by summarizing a general method for analyzing situations involving Newton s Second Law for Rotation, such as the situation in Exploration
More informationSpeed and Velocity. Circular and Satellite Motion
Name: Speed and Velocity Read from Lesson 1 of the Circular and Satellite Motion chapter at The Physics Classroom: http://www.physicsclassroom.com/class/circles/u6l1a.html MOP Connection: Circular Motion
More informationUnit 2: Vector Dynamics
Multiple Choice Portion Unit 2: Vector Dynamics 1. Which one of the following best describes the motion of a projectile close to the surface of the Earth? (Assume no friction) Vertical Acceleration Horizontal
More information7. Two forces are applied to a 2.0kilogram block on a frictionless horizontal surface, as shown in the diagram below.
1. Which statement about the movement of an object with zero acceleration is true? The object must be at rest. The object must be slowing down. The object may be speeding up. The object may be in motion.
More information(a) On the dots below that represent the students, draw and label freebody diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
More informationMultiple Choice (A) (B) (C) (D)
Multiple Choice 1. A ball is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle its velocity and acceleration directions are: (A) (B) (C) (D) 2.
More informationChapter 4 Dynamics: Newton s Laws of Motion
Chapter 4 Dynamics: Newton s Laws of Motion Force Newton s First Law of Motion Mass Newton s Second Law of Motion Newton s Third Law of Motion Weight the Force of Gravity; and the Normal Force Applications
More informationPhysics 207 Lecture 17
Physics 207, Lecture 17, Oct. 31 Agenda: Review for exam Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM Example Gravity, Normal Forces etc. Consider a women on a swing: Assignment: MP Homework
More informationWritten Homework problems. Spring (taken from Giancoli, 4 th edition)
Written Homework problems. Spring 014. (taken from Giancoli, 4 th edition) HW1. Ch1. 19, 47 19. Determine the conversion factor between (a) km / h and mi / h, (b) m / s and ft / s, and (c) km / h and m
More informationReview  Chapter 1. Ans: 2.12m
Review  Chapter 1 The distance d that a certain particle moves may be calculated from the expression d = at + bt 2 where a and b are constants; and t is the elapsed time. The dimensions of the quantities
More information