Chapter 15 Kinematics of a Particle: Impulse and Momentum. Lecture Notes for Section 15-5~7
|
|
- Morgan Josephine Chapman
- 5 years ago
- Views:
Transcription
1 Chapter 15 Kinematics of a Particle: Impulse and Momentum Lecture Notes for Section 15-5~7
2 ANGULAR MOMENTUM, MOMENT OF A FORCE AND PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM Today s Objectives: Students will be able to: 1. Determine the angular momentum of a particle and apply the principle of angular impulse & momentum. 2. Use conservation of angular momentum to solve problems. In-Class Activities: Check Homework Reading Quiz Applications Angular Momentum Angular Impulse and Momentum Principle Conservation of Angular Momentum Concept Quiz Group Problem Solving Attention Quiz
3 READING QUIZ 1. Select the correct expression for the angular momentum of a particle about a point. A) r v B) r (m v) C) v r D) (m v) r 2. The sum of the moments of all external forces acting on a particle is equal to A) angular momentum of the particle. B) linear momentum of the particle. C) time rate of change of angular momentum. D) time rate of change of linear momentum.
4 APPLICATIONS Planets and most satellites move in elliptical orbits. This motion is caused by gravitational attraction forces. Since these forces act in pairs, the sum of the moments of the forces acting on the system will be zero. This means that angular momentum is conserved. If the angular momentum is constant, does it mean the linear momentum is also constant? Why or why not?
5 APPLICATIONS (continued) The passengers on the amusement-park ride experience conservation of angular momentum about the axis of rotation (the z-axis). As shown on the free body diagram, the line of action of the normal force, N, passes through the z-axis and the weight s line of action is parallel to it. Therefore, the sum of moments of these two forces about the z-axis is zero. If the passenger moves away from the z- axis, will his speed increase or decrease? Why?
6 ANGULAR MOMENTUM (Section 15.5) The angular momentum of a particle about point O is defined as the moment of the particle s linear momentum about O. i j k H o = r mv = r x r y r z mv x mv y mv z The magnitude of H o is (H o ) z = mv d
7 RELATIONSHIP BETWEEN MOMENT OF A FORCE AND ANGULAR MOMENTUM (Section 15.6) The resultant force acting on the particle is equal to the time rate of change of the particle s linear momentum. Showing the time derivative using the familiar dot notation results in the equation F = L = mv We can prove that the resultant moment acting on the particle about point O is equal to the time rate of change of the particle s angular momentum about point O or M o = r F = H o
8 PRINCIPLE OF ANGULAR IMPULSE AND MOMENTUM (Section 15.7) Considering the relationship between moment and time rate of change of angular momentum t t M o = H o = dh o /dt By integrating between the time interval t 1 to t ( H H M dt ( o ) 2 o ) or o ( H o ) 1 1 This equation is referred to as the principle of angular impulse and momentum. The second term on the left side, M o dt, is called the angular impulse. In cases of 2D motion, it can be applied as a scalar equation using components about the z-axis. t 2 + dt t 1 M o ( H o ) 2
9 CONSERVATION OF ANGULAR MOMENTUM When the sum of angular impulses acting on a particle or a system of particles is zero during the time t 1 to t 2, the angular momentum is conserved. Thus, (H O ) 1 = (H O ) 2 An example of this condition occurs when a particle is subjected only to a central force. In the figure, the force F is always directed toward point O. Thus, the angular impulse of F about O is always zero, and angular momentum of the particle about O is conserved.
10 EXAMPLE Given:A satellite has an elliptical orbit about earth. m satellite = 700 kg m earth = kg v A = 10 km/s r A = m A = 70 Find: The speed, v B, of the satellite at its closest distance, r B, from the center of the earth. Plan: Apply the principles of conservation of energy and conservation of angular momentum to the system.
11 Solution: EXAMPLE (continued) Conservation of energy: T A + V A = T B + V B becomes 1 m s v A2 G m s m e = 1 m s v B2 G m s m e 2 r A 2 r B where G = m 3 /(kg s 2 ). Dividing through by m s and substituting values yields: 0.5 (10,000 ) ( x ) or = 0.5 (v B ) 2 ( )/r B v 2 B (5.976 r B )
12 Solution: EXAMPLE (continued) Now use Conservation of Angular Momentum. (r A m s v A ) sin A = r B m s v B ( )(10,000) sin 70 = r B v B or r B = ( )/v B Solving the two equations for r B and v B yields r B = m v B = 10.2 km/s
13 CONCEPT QUIZ 1. If a particle moves in the x - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) x - y direction. 2. If there are no external impulses acting on a particle A) only linear momentum is conserved. B) only angular momentum is conserved. C) both linear momentum and angular momentum are conserved. D) neither linear momentum nor angular momentum are conserved.
14 GROUP PROBLEM SOLVING Given: The four 5 lb spheres are rigidly attached to the crossbar frame, which has a negligible weight. A moment acts on the shaft as shown, M = 0.5t lb ft). Find: The velocity of the spheres after 4 seconds, starting from rest. Plan: Apply the principle of angular impulse and momentum about the axis of rotation (z-axis).
15 GROUP PROBLEM SOLVING (continued) Solution: Angular momentum: H Z = r mv reduces to a scalar equation. (H Z ) 1 = 0 and (H Z ) 2 = 4 {(5/32.2) (0.6) v 2 } = v 2 Angular impulse: t 2 t 2 M dt = (0.5t + 0.8) dt = [(0.5/2) t t] = 7.2 lb ft s t t Apply the principle of angular impulse and momentum = v 2 v 2 = 19.4 ft/s 4
16 ATTENTION QUIZ 1. A ball is traveling on a smooth surface in a 3 ft radius circle with a speed of 6 ft/s. If the attached cord is pulled down with a constant speed of 2 ft/s, which of the following principles can be applied to solve for the velocity of the ball when r = 2 ft? A) Conservation of energy B) Conservation of angular momentum C) Conservation of linear momentum D) Conservation of mass 2. If a particle moves in the z - y plane, its angular momentum vector is in the A) x direction. B) y direction. C) z direction. D) z - y direction.
17 Example The box has a mass m and is traveling down the smooth circular ramp such that when it is at the angle θ it is a speed v. Determine its angular momentum about point O at this instant and the rate of increase in its speed, i.e., a t.
18 Example Solution Since v is tangent to the path, the angular momentum is HO rmv From the FBD, the weight W = mg contributes a moment about O ; ( sin ) d M ( ) O H O mg r rmv dt Since r and m are constant, dv dv mgr sin rm g sin dt dt
19 Example The 0.4 kg ball B is attached to a cord which passes through a hole at A in a smooth table. When the ball is r 1 = 0.5 m from the hole, it is rotating around in a circle such that its speed is v 1 = 1.2 m/s. By applying a force F the cord is pulled downward through the hole with a constant speed v c = 2 m/s.
20 Example Determine (a) the speed of the ball at the instant it is r 2 = 0.2 m from the hole, and (b) the amount of work done by F in shortening the radial distance from r 1 to r 2.
21 Example Solution Part (a) Free-Body Diagram As the ball the cord force F on the ball passes through the z axis. Weight and N B are parallel and the conservation of angular momentum applies about the z axis. Conservation of Angular Momentum H1 H2 rm 1 Bv1 rm 2 Bv 2 (0.5)(0.4)(1.2) (0.2)(0.4) v v 3m/s 2 2
22 Example Solution Conservation of Angular Momentum The speed of the ball is thus 2 2 v2 (3.0) (2) 3.606m/s Part (b) The only force that does work on the ball is F. T U T (0.4)(1.2) UF (0.4)(3.606) UF 2.313J 2 2
23
24 Chapter 15 Kinematics of a Particle: Impulse and Momentum Lecture Notes for Section 15-8~9
25 15.8 Steady Flow of a Fluid Stream Consider the diversion of a steady stream of fluid by a fixed pipe The impulse and momentum diagrams for the fluid are as shown below
26 15.8 Steady Flow of a Fluid Stream The force F represents the resultant of all the external forces acting on the fluid stream Since flow is steady, F will be constant during the time interval dt Applying the principle of linear impulse and momentum to the fluid stream, A B dm v mv Fdt dm v mv
27 15.8 Steady Flow of a Fluid Stream Principle of Impulse and Momentum Solving for the resultant force yields dm F ( vb va) dt It is convenient to express vector equation in the form of two scalar component equations dm Fx ( vbx vax) dt dm Fy ( vby vay) dt
28 15.8 Steady Flow of a Fluid Stream Principle of Impulse and Momentum Since flow is steady in the x-y plane, hence we have dm + M O ( dobvb doava) dt Once the velocity of the fluid flowing onto the device is determined, the mass flow is calculated using dm AvA A A BvA B B AQA BQB dt
29 15.8 Steady Flow of a Fluid Stream Procedure of Analysis Free-Body Diagram Draw a FBD of the device which is directing the fluid Equations of Steady Flow Apply the equations of steady flow, dm M O ( dobvb doava) dt
30 Example 15.6 Determine the components of reaction which the fixed pipe joint at A exerts on the elbow. If water flowing through the pipe is subjected to a static gauge pressure of 100 kpa at A. The discharge at B is QB = 0.2 m 3 /s. Water has a density ρw = 1000kg/m 3, and the water-filled elbow has a mass of 20 kg and center of mass at G.
31 Example Solution Since the density of water is constant, Q B = Q A = Q, dm wq 200 kg/s dt Q 0.2 Q vb m/s ; v 6.37 m/s 2 A A (0.05) A B Free-Body Diagram Since 1 kpa = 1000 N/m 2, F p A A A A 3 2 [100(10 )][ (0.1) ] N A
32 Example Solution Equations of Steady Flow dm Fx ( vbx vax); dt Fx (0 6.37) Fx 4.41kN dm Fy ( vby vay); dt Fy 20(9.81) 200( ) F 4.90kN y
33 Example Solution Equations of Steady Flow If moments are summed about point O, then F x, F y, and static pressure F A are eliminated, as well as moment of momentum of water entering at A, dm MO ( dobvb doava) dt M O 20(9.81)(0.125) 200[(0.3)(25.46) 0] M 1.50kN m O
34 15.9 Propulsion with Variable Mass A Control Volume That Loses Mass Consider a device such as a rocket which at an instant of time has a mass m and is moving forward with a velocity v The closed system included both the mass m of the device and the expelled mass m e
35 15.9 Propulsion with Variable Mass A Control Volume That Loses Mass Applying the principle of impulse and momentum to the system, mv m v F dt ( m dm )( v dv) ( m dm ) v e e CV e e e e F dt vdm m dv dm dv v dm CV e e e e The relative velocity of the device as seen by the observer moving with the particles of the ejected mass is v D/e = (v + v e ), dv FCV m v dt D/ e dm dt e
36 15.9 Propulsion with Variable Mass A Control Volume That Gain Mass A device such as a scoop or a shovel may gain mass as it moves forward The device shown has a mass m and is moving forward with a velocity v
37 15.9 Propulsion with Variable Mass A Control Volume That Gain Mass Applying the principle of impulse and momentum to the system, mv m v F dt ( m dm )( v dv) ( m dm ) v i i CV i i i i We may write this equation as dv dmi FCV m ( vvi ) dt dt
38 Example The initial combined mass of a rocket and its fuel is m 0. A total mass mf of fuel is consumed as a constant rate of dm e /dt = c and expelled at a constant speed of u relative to the rocket. Determine the max velocity of the rocket i.e., at the instant the fuel runs out.
39 Example Solution The rocket is losing mass as it moves upward. The only external force acting on the system consisting of the rocket and a portion of the expelled mass is the weight W dv dme dv Fs m vd/ e ; W m uc dt dt dt Since W = mg, ( m0 ct) g ( m0 ct) dv uc dt
40 Example Solution Separating the variables and integrating, realizing that v = 0 at t = 0, we have uc m v t t 0 dv ln( 0 ) ln 0 g dt v u m ct gt u gt 0 0 m0 ct m0 ct The time t needed to consume all the fuel is given by By sub, we get dme mf t ct t mf / c dt v max uln m gm 0 m0 m f c f
41
42 End section 15-1 _01: 15-2, 15-8, 15-13, End section 15-2 _02: 15-31, 15-48, End section 15-3 _03: 15-82, CH015 HW
ME 012 Engineering Dynamics
ME 012 Engineering Dynamics Lecture 17 Angular Momentum, Relation Between Moment of a Force and Angular Momentum, and Principle of Angular Impulse and Momentum (Chapter 15, Section 5, 6, and 7) Thursday,
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5
1 / 40 CEE 271: Applied Mechanics II, Dynamics Lecture 9: Ch.13, Sec.4-5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa 2 / 40 EQUATIONS OF MOTION:RECTANGULAR COORDINATES
More informationAPPLICATIONS. CEE 271: Applied Mechanics II, Dynamics Lecture 17: Ch.15, Sec.4 7. IMPACT (Section 15.4) APPLICATIONS (continued) IMPACT READING QUIZ
APPLICATIONS CEE 271: Applied Mechanics II, Dynamics Lecture 17: Ch.15, Sec.4 7 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Date: The quality of a tennis ball
More informationEQUATIONS 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. In-Class Activities: Check
More informationPRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (Section 15.1)
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM (Section 15.1) Linear momentum: L = mv vector mv is called the linear momentum denoted as L (P in 1120) vector has the same direction as v. units of (kg m)/s or
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 informationPhysics 2211 ABC Quiz #3 Solutions Spring 2017
Physics 2211 ABC Quiz #3 Solutions Spring 2017 I. (16 points) A block of mass m b is suspended vertically on a ideal cord that then passes through a frictionless hole and is attached to a sphere of mass
More informationME 230 Kinematics and Dynamics
ME 230 Kinematics and Dynamics Wei-Chih Wang Department of Mechanical Engineering University of Washington Lecture 6: Particle Kinetics Kinetics of a particle (Chapter 13) - 13.4-13.6 Chapter 13: Objectives
More informationAP Physics QUIZ Gravitation
AP Physics QUIZ Gravitation Name: 1. If F1 is the magnitude of the force exerted by the Earth on a satellite in orbit about the Earth and F2 is the magnitude of the force exerted by the satellite on the
More informationFigure 1 Answer: = m
Q1. Figure 1 shows a solid cylindrical steel rod of length =.0 m and diameter D =.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel
More informationDynamics Kinetics of a particle Section 4: TJW Force-mass-acceleration: Example 1
Section 4: TJW Force-mass-acceleration: Example 1 The beam and attached hoisting mechanism have a combined mass of 1200 kg with center of mass at G. If the inertial acceleration a of a point P on the hoisting
More informationEQUATIONS OF MOTION: RECTANGULAR COORDINATES
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. In-Class
More informationME 230 Kinematics and Dynamics
ME 230 Kinematics and Dynamics Wei-Chih Wang Department of Mechanical Engineering University of Washington Lecture 8 Kinetics of a particle: Work and Energy (Chapter 14) - 14.1-14.3 W. Wang 2 Kinetics
More informationEQUATIONS OF MOTION: GENERAL PLANE MOTION (Section 17.5) Today s Objectives: Students will be able to analyze the planar kinetics of a rigid body
EQUATIONS OF MOTION: GENERAL PLANE MOTION (Section 17.5) Today s Objectives: Students will be able to analyze the planar kinetics of a rigid body undergoing general plane motion. APPLICATIONS As the soil
More informationNEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES
NEWTON S LAWS OF MOTION, EQUATIONS OF MOTION, & EQUATIONS OF MOTION FOR A SYSTEM OF PARTICLES Objectives: Students will be able to: 1. Write the equation of motion for an accelerating body. 2. Draw the
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 14: Ch.15, Sec.1-3
1 / 20 CEE 271: Applied Mechanics II, Dynamics Lecture 14: Ch.15, Sec.1-3 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Thursday, October 4, 2012 PRINCIPLE OF LINEAR
More informationChapter 4 Kinetics of Particle: Impulse and Momentum
Chapter 4 Kinetics of Particle: Impulse and Momentum Dr. Khairul Salleh Basaruddin Applied Mechanics Division School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) khsalleh@unimap.edu.my
More informationNEWTON S LAWS OF MOTION (EQUATION OF MOTION) (Sections )
NEWTON S LAWS OF MOTION (EQUATION OF MOTION) (Sections 13.1-13.3) Today s Objectives: Students will be able to: a) Write the equation of motion for an accelerating body. b) Draw the free-body and kinetic
More information1 Motion of a single particle - Linear momentum, work and energy principle
1 Motion of a single particle - Linear momentum, work and energy principle 1.1 In-class problem A block of mass m slides down a frictionless incline (see Fig.). The block is released at height h above
More information= o + t = ot + ½ t 2 = o + 2
Chapters 8-9 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 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 In-Class Activities: Check Homework Reading
More informationTherefore, the control volume in this case can be treated as a solid body, with a net force or thrust of. bm # V
When the mass m of the control volume remains nearly constant, the first term of the Eq. 6 8 simply becomes mass times acceleration since 39 CHAPTER 6 d(mv ) CV m dv CV CV (ma ) CV Therefore, the control
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term Exam 2 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 801 Physics Fall Term 013 Problem 1 of 4 (5 points) Exam Solutions Answers without work shown will not be given any credit A block of mass m
More informationAnnouncements. Principle of Work and Energy - Sections Engr222 Spring 2004 Chapter Test Wednesday
Announcements Test Wednesday Closed book 3 page sheet sheet (on web) Calculator Chap 12.6-10, 13.1-6 Principle of Work and Energy - Sections 14.1-3 Today s Objectives: Students will be able to: a) Calculate
More informationMOMENT OF A FORCE SCALAR FORMULATION, CROSS PRODUCT, MOMENT OF A FORCE VECTOR FORMULATION, & PRINCIPLE OF MOMENTS
MOMENT OF A FORCE SCALAR FORMULATION, CROSS PRODUCT, MOMENT OF A FORCE VECTOR FORMULATION, & PRINCIPLE OF MOMENTS Today s Objectives : Students will be able to: a) understand and define moment, and, b)
More informationCEE 271: Applied Mechanics II, Dynamics Lecture 25: Ch.17, Sec.4-5
1 / 36 CEE 271: Applied Mechanics II, Dynamics Lecture 25: Ch.17, Sec.4-5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Date: 2 / 36 EQUATIONS OF MOTION: ROTATION
More informationPhysics 201 Midterm Exam 3
Name: Date: _ Physics 201 Midterm Exam 3 Information and Instructions Student ID Number: Section Number: TA Name: Please fill in all the information above Please write and bubble your Name and Student
More informationChapter 8: Newton s Laws Applied to Circular Motion
Chapter 8: Newton s Laws Applied to Circular Motion Centrifugal Force is Fictitious? F actual = Centripetal Force F fictitious = Centrifugal Force Center FLEEing Centrifugal Force is Fictitious? Center
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 informationω = k/m x = A cos (ωt + ϕ 0 ) L = I ω a x = ω 2 x P = F v P = de sys J = F dt = p w = m g F G = Gm 1m 2 D = 1 2 CρAv2 a r = v2
PHYS 2211 A, B, & C Final Exam Formulæ & Constants Spring 2017 Unless otherwise directed, drag is to be neglected and all problems take place on Earth, use the gravitational definition of weight, and all
More informationPhysics 201 Midterm Exam 3
Physics 201 Midterm Exam 3 Information and Instructions Student ID Number: Section Number: TA Name: Please fill in all the information above. Please write and bubble your Name and Student Id number on
More informationPhysics 2211 A & B Quiz #3 Solutions Fall 2016
Physics 2211 A & B Quiz #3 Solutions Fall 2016 I. (16 points) A block of mass m 1 is connected by an ideal rope passing over an ideal pulley to a block of mass m 2. The block of mass m 1 slides up a plane
More informationPhysics 5A Final Review Solutions
Physics A Final Review Solutions Eric Reichwein Department of Physics University of California, Santa Cruz November 6, 0. A stone is dropped into the water from a tower 44.m above the ground. Another stone
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 information31 ROTATIONAL KINEMATICS
31 ROTATIONAL KINEMATICS 1. Compare and contrast circular motion and rotation? Address the following Which involves an object and which involves a system? Does an object/system in circular motion have
More informationN - W = 0. + F = m a ; N = W. Fs = 0.7W r. Ans. r = 9.32 m
91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 479 R1 1. The ball is thrown horizontally with a speed of 8 m>s. Find the equation of the path, y = f(x), and then determine the ball s velocity and the normal
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 informationVARIABLE MASS PROBLEMS
VARIABLE MASS PROBLEMS Question 1 (**) A rocket is moving vertically upwards relative to the surface of the earth. The motion takes place close to the surface of the earth and it is assumed that g is the
More informationAP Physics II Summer Packet
Name: AP Physics II Summer Packet Date: Period: Complete this packet over the summer, it is to be turned it within the first week of school. Show all work were needed. Feel free to use additional scratch
More informationGyroscopes and statics
Gyroscopes and statics Announcements: Welcome back from Spring Break! CAPA due Friday at 10pm We will finish Chapter 11 in H+R on angular momentum and start Chapter 12 on stability. Friday we will begin
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 informationDistance travelled time taken and if the particle is a distance s(t) along the x-axis, then its instantaneous speed is:
Chapter 1 Kinematics 1.1 Basic ideas r(t) is the position of a particle; r = r is the distance to the origin. If r = x i + y j + z k = (x, y, z), then r = r = x 2 + y 2 + z 2. v(t) is the velocity; v =
More informationPhysics 2211 M Quiz #2 Solutions Summer 2017
Physics 2211 M Quiz #2 Solutions Summer 2017 I. (16 points) A block with mass m = 10.0 kg is on a plane inclined θ = 30.0 to the horizontal, as shown. A balloon is attached to the block to exert a constant
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 informationAP Physics 1 Lesson 9 Homework Outcomes. Name
AP Physics 1 Lesson 9 Homework Outcomes Name Date 1. Define uniform circular motion. 2. Determine the tangential velocity of an object moving with uniform circular motion. 3. Determine the centripetal
More informationSt. Joseph s Anglo-Chinese School
Time allowed:.5 hours Take g = 0 ms - if necessary. St. Joseph s Anglo-Chinese School 008 009 First Term Examination Form 6 ASL Physics Section A (40%) Answer ALL questions in this section. Write your
More informationTYPICAL NUMERIC QUESTIONS FOR PHYSICS I REGULAR QUESTIONS TAKEN FROM CUTNELL AND JOHNSON CIRCULAR MOTION CONTENT STANDARD IB
TYPICAL NUMERIC QUESTIONS FOR PHYSICS I REGULAR QUESTIONS TAKEN FROM CUTNELL AND JOHNSON CIRCULAR MOTION CONTENT STANDARD IB 1. A car traveling at 20 m/s rounds a curve so that its centripetal acceleration
More informationPRINCIPLE OF LINEAR IMPULSE AND MOMENTUM
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM Today s Objectives: Students will be able to: 1. Calculate the linear momentum of a particle and linear impulse of a force. 2. Apply the principle of linear impulse
More informationConservation of Energy Challenge Problems Problem 1
Conservation of Energy Challenge Problems Problem 1 An object of mass m is released from rest at a height h above the surface of a table. The object slides along the inside of the loop-the-loop track consisting
More informationwhere s is the horizontal range, in this case 50 yards. Note that as long as the initial speed of the bullet is great enough to let it hit a target at
1 PHYS 31 Homework Assignment Due Friday, 13 September 00 1. A monkey is hanging from a tree limbat height h above the ground. A hunter 50 yards away from the base of the tree sees him, raises his gun
More informationWebreview Torque and Rotation Practice Test
Please do not write on test. ID A Webreview - 8.2 Torque and Rotation Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A 0.30-m-radius automobile
More informationEQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS (Section 17.4) Today s Objectives: Students will be able to analyze the planar kinetics of a rigid
EQUATIONS OF MOTION: ROTATION ABOUT A FIXED AXIS (Section 17.4) Today s Objectives: Students will be able to analyze the planar kinetics of a rigid body undergoing rotational motion. APPLICATIONS The crank
More information3. Kinetics of Particles
3. Kinetics of Particles 3.1 Force, Mass and Acceleration 3.3 Impulse and Momentum 3.4 Impact 1 3.1 Force, Mass and Acceleration We draw two important conclusions from the results of the experiments. First,
More informationChapter 15 Kinematics of a Particle: Impulse and Momentum. Lecture Notes for Section 15-2~3
Chapter 15 Kinematics of a Particle: Impulse and Momentum Lecture Notes for Section 15-2~3 PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM AND CONSERVATION OF LINEAR MOMENTUM FOR SYSTEMS OF PARTICLES Today s
More informationEngineering Mechanics: Statics in SI Units, 12e
Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body
More informationName (please print): UW ID# score last first
Name (please print): UW ID# score last first Question I. (20 pts) Projectile motion A ball of mass 0.3 kg is thrown at an angle of 30 o above the horizontal. Ignore air resistance. It hits the ground 100
More informationLecture 16. Gravitation
Lecture 16 Gravitation Today s Topics: The Gravitational Force Satellites in Circular Orbits Apparent Weightlessness lliptical Orbits and angular momentum Kepler s Laws of Orbital Motion Gravitational
More informationl1, l2, l3, ln l1 + l2 + l3 + ln
Work done by a constant force: Consider an object undergoes a displacement S along a straight line while acted on a force F that makes an angle θ with S as shown The work done W by the agent is the product
More informationPLANAR RIGID BODY MOTION: TRANSLATION &
PLANAR RIGID BODY MOTION: TRANSLATION & Today s Objectives : ROTATION Students will be able to: 1. Analyze the kinematics of a rigid body undergoing planar translation or rotation about a fixed axis. In-Class
More informationExam 2: Equation Summary
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.01 Physics Fall Term 2012 Exam 2: Equation Summary Newton s Second Law: Force, Mass, Acceleration: Newton s Third Law: Center of Mass: Velocity
More information7.6 Journal Bearings
7.6 Journal Bearings 7.6 Journal Bearings Procedures and Strategies, page 1 of 2 Procedures and Strategies for Solving Problems Involving Frictional Forces on Journal Bearings For problems involving a
More informationChapter Units and Measurement
2 Chapter Units and Measurement 1. Identify the pair whose dimensions are equal [2002] torque and work stress and energy force and stress force and work 2. [2003] [L -1 T] ] [L -2 T 2 ] [L 2 T -2 ] [LT
More informationName Student ID Score Last First. I = 2mR 2 /5 around the sphere s center of mass?
NOTE: ignore air resistance in all Questions. In all Questions choose the answer that is the closest!! Question I. (15 pts) Rotation 1. (5 pts) A bowling ball that has an 11 cm radius and a 7.2 kg mass
More informationPLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)
PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2) We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane. As discussed in Chapter 16, when
More informationLesson 8. Luis Anchordoqui. Physics 168. Thursday, October 11, 18
Lesson 8 Physics 168 1 Rolling 2 Intuitive Question Why is it that when a body is rolling on a plane without slipping the point of contact with the plane does not move? A simple answer to this question
More informationFriction is always opposite to the direction of motion.
6. Forces and Motion-II 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 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 informationPhys 111 Exam 2 October 18, Name Section University ID
Phys 111 Exam October 18, 016 Name Section University ID Please fill in your computer answer sheet as follows: 1) Use your previous answer sheet and start with 1. Note problem number of the second exam
More informationPhysics H7A, Fall 2011 Homework 4 Solutions
Physics H7A, Fall 20 Homework 4 Solutions. (K&K Problem 2.) A mass m is connected to a vertical revolving axle by two strings of length l, each making an angle of 45 with the axle, as shown. Both the axle
More informationChapter 8 continued. Rotational Dynamics
Chapter 8 continued Rotational Dynamics 8.6 The Action of Forces and Torques on Rigid Objects Chapter 8 developed the concepts of angular motion. θ : angles and radian measure for angular variables ω :
More informationPHYSICS. Chapter 8 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.
PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 8 Lecture RANDALL D. KNIGHT Chapter 8. Dynamics II: Motion in a Plane IN THIS CHAPTER, you will learn to solve problems about motion
More information= W Q H. ɛ = T H T C T H = = 0.20 = T C = T H (1 0.20) = = 320 K = 47 C
1. Four identical 0.18 kg masses are placed at the corners of a 4.0 x 3.0 m rectangle, and are held there by very light connecting rods which form the sides of the rectangle. What is the moment of inertia
More informationPhysics 351, Spring 2017, Homework #2. Due at start of class, Friday, January 27, 2017
Physics 351, Spring 2017, Homework #2. Due at start of class, Friday, January 27, 2017 Course info is at positron.hep.upenn.edu/p351 When you finish this homework, remember to visit the feedback page at
More informationCheck Homework. Reading Quiz Applications Equations of Equilibrium Example Problems Concept Questions Group Problem Solving Attention Quiz
THREE-DIMENSIONAL FORCE SYSTEMS Today s Objectives: Students will be able to solve 3-D particle equilibrium problems by a) Drawing a 3-D free body diagram, and, b) Applying the three scalar equations (based
More information@K302. Yasuyuki Matsuda
Introductory Physics (week 3) @K302 Yasuyuki Matsuda Today s Contents Velocity and Acceleration Newton s Laws of Motion Position, Velocity, Acceleration Particle Particle : An point-like object with its
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 informationPhysics 207 Lecture 11. Lecture 11. Chapter 8: Employ rotational motion models with friction or in free fall
Goals: Lecture 11 Chapter 8: Employ rotational motion models with friction or in free fall Chapter 9: Momentum & Impulse Understand what momentum is and how it relates to forces Employ momentum conservation
More informationA B = AB cos θ = 100. = 6t. a(t) = d2 r(t) a(t = 2) = 12 ĵ
1. A ball is thrown vertically upward from the Earth s surface and falls back to Earth. Which of the graphs below best symbolizes its speed v(t) as a function of time, neglecting air resistance: The answer
More informationPC 1141 : AY 2012 /13
NUS Physics Society Past Year Paper Solutions PC 1141 : AY 2012 /13 Compiled by: NUS Physics Society Past Year Solution Team Yeo Zhen Yuan Ryan Goh Published on: November 17, 2015 1. An egg of mass 0.050
More informationNewton s Laws of Motion
Chapter 4 Newton s Second Law: in vector form Newton s Laws of Motion σ റF = m റa in component form σ F x = ma x σ F y = ma y in equilibrium and static situations a x = 0; a y = 0 Strategy for Solving
More informationChapter 5 Gravitation Chapter 6 Work and Energy
Chapter 5 Gravitation Chapter 6 Work and Energy Chapter 5 (5.6) Newton s Law of Universal Gravitation (5.7) Gravity Near the Earth s Surface Chapter 6 (today) Work Done by a Constant Force Kinetic Energy,
More informationPhysics for Scientists and Engineers 4th Edition, 2017
A Correlation of Physics for Scientists and Engineers 4th Edition, 2017 To the AP Physics C: Mechanics Course Descriptions AP is a trademark registered and/or owned by the College Board, which was not
More information24 m / s. 4. The units N / kg are used for A. net force. B. gravitational force. C. electric field strength. D. gravitational field strength.
PHYSICS 12 JUNE 2004 PROVINCIAL EXAMINATION PART A: MULTIPLE CHOICE 1. Which of the following is a scalar quantity? A. work B. force C. velocity D. momentum 2. An astronaut on the moon throws a 5.0 kg
More informationForm #231 Page 1 of 6
Version Quiz #3 Form #231 Name: A Physics 2211 A & B Fall 2016 Recitation Section: Print your name, quiz form number (3 digits at the top of this form), and student number (9 digit Georgia Tech ID number)
More informationSolution to phys101-t112-final Exam
Solution to phys101-t112-final Exam Q1. An 800-N man stands halfway up a 5.0-m long ladder of negligible weight. The base of the ladder is.0m from the wall as shown in Figure 1. Assuming that the wall-ladder
More informationChapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5
Chapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5 CH 4: Uniform Circular Motion The velocity vector is tangent to the path The change in velocity vector is due to the change in direction.
More informationFaculty of Engineering and Department of Physics Engineering Physics 131 Final Examination Saturday April 21, 2018; 14:00 pm 16:30 pm
Faculty of Engineering and Department of Physics Engineering Physics 131 Final Examination Saturday April 21, 2018; 14:00 pm 16:30 pm 1. Closed book exam. No notes or textbooks allowed. 2. Formula sheets
More informationChapter 13. Gravitation
Chapter 13 Gravitation 13.2 Newton s Law of Gravitation Here m 1 and m 2 are the masses of the particles, r is the distance between them, and G is the gravitational constant. G =6.67 x10 11 Nm 2 /kg 2
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 informationNewton s Law of motion
5-A 11028 / 9, WEA, Sat Nagar, Karol Bagh New Delhi-110005 M : 9910915514, 9953150192 P : 011-45660510 E : keshawclasses@gmail.com W: www.keshawclasses.com Newton s Law of motion Q. 1. Two sphere A and
More informationAP Physics Multiple Choice Practice Gravitation
AP Physics Multiple Choice Practice Gravitation 1. Each of five satellites makes a circular orbit about an object that is much more massive than any of the satellites. The mass and orbital radius of each
More informationForces Part 1: Newton s Laws
Forces Part 1: Newton s Laws Last modified: 13/12/2017 Forces Introduction Inertia & Newton s First Law Mass & Momentum Change in Momentum & Force Newton s Second Law Example 1 Newton s Third Law Common
More informationPhysics 23 Notes Chapter 5
Physics 23 Notes Chapter 5 Dr. Alward Circular Motion Example A: An object is moving in a circular path of radius r = 1.2 m, and is completing 0.40 revolutions per second. What is the object s centripetal
More informationGet Discount Coupons for your Coaching institute and FREE Study Material at Force System
Get Discount Coupons for your Coaching institute and FEE Study Material at www.pickmycoaching.com Mechanics Force System When a member of forces simultaneously acting on the body, it is known as force
More information9.3 Worked Examples Circular Motion
9.3 Worked Examples Circular Motion Example 9.1 Geosynchronous Orbit A geostationary satellite goes around the earth once every 3 hours 56 minutes and 4 seconds, (a sidereal day, shorter than the noon-to-noon
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 informationPRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES AND CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES
PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM FOR A SYSTEM OF PARTICLES AND CONSERVATION OF LINEAR MOMENTUM FOR A SYSTEM OF PARTICLES Today s Objectives: Students will be able to: 1. Apply the principle of
More informationChapter 9-10 Test Review
Chapter 9-10 Test Review Chapter Summary 9.2. The Second Condition for Equilibrium Explain torque and the factors on which it depends. Describe the role of torque in rotational mechanics. 10.1. Angular
More informationPHY 141 Midterm 1 Oct 2, 2014 Version A
PHY 141 Midterm 1 Oct 2, 2014 Version A Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET! To avoid problems in grading: do all problems in order, write legibly, and show all
More informationChapter 9 Circular Motion Dynamics
Chapter 9 Circular Motion Dynamics Chapter 9 Circular Motion Dynamics... 9. Introduction Newton s Second Law and Circular Motion... 9. Universal Law of Gravitation and the Circular Orbit of the Moon...
More information