Handout #1 by Hejie Lin MET 327 APPLIED ENGINEERING II (DYNAMICS) 1. Introduction to Statics and Dynamics 1.1 Statics vs. Dynamics 1 Ch 9 Moment of Inertia A dynamic system is characterized with mass (M), damping coefficient (C) and Stiffness (K). For our analysis, damping coefficient (C) is assumed to be zero. Dynamic motion of a mechanics system can be analyzed with Equation of Motion (EOM). EOM relates forces to the acceleration, velocity and displacement. 1-D Dynamic System Equation of Motion (EOM) Where, F: Force, M: Mass, C: Damping Coefficient, K: Stiffness, a: Acceleration, v: Velocity, x: Displacement. Statics analysis calculates the relation between forces and displacement under motionless static condition. Therefore both velocity (v=0) and acceleration (a=0) are zero. As a result, both mass (M) and the damping coefficient (C) can exist in a system but do not need to be known for solving the equation of motion (EOM). Under static condition, the equation of motion (EOM) can be simplified to the Hook s Law ( F = Kx ). The displacement (x) due to applied force ( F ) can be calculated from Hook s Law and vise verse. The stiffness, K, is a constant square matrix for a large linear system. On the other hand, a dynamic system does not have spring or damper. Therefore, the stiffness ( K = 0 ) and damping coefficient ( C = 0 ) are both zero. Under this condition the equation of motion (EOM) can be simplified to the Newton s 2 nd Law ( F = Ma ). The acceleration (a) due to applied force ( F ) can be calculated from Newton s 2 nd Law and vise verse. Statics (MET 227) Dynamics (MET 327) Static Conditions: v = 0, a = 0 Equation of Motion: F = Kx System Parameters: K = 0, C = 0 Equation of Motion: F = Ma
Handout #1 by Hejie Lin 2 Ch 9 Moment of Inertia 1.2 Statics Analysis (MET227): Statics studies the stress and strain distribution of motionless static structures. The standard procedures for static analysis can be described in the following two steps. Step 1: Calculate Reaction Forces and Moments Calculate reaction forces and moments from applied forces and moments using Free Body Diagram (FBD). All forces and moments should be balanced on any FBD. Only force exists on a truss member because of the zero-moment points if no force acting between the joints of the truss member. The self weight of the truss member is neglected or lumped to the end nodes (joints). Otherwise, the member should be analyzed as a beam as described as following. Both forces ( F x, Fy ) and moments ( M ) exist on beam member. Beam moment diagram and shear diagram can be calculated using Free Body Diagram (FBD). In a beam structure, forces are categorized as normal forces ( F x ) and shear forces ( F y ) as shown in the figure below. Truss Structure Beam Structure Reaction Forces ( A, A, B ) Only x y x Reaction Forces ( F, F ) and Moment ( M ) x y
Handout #1 by Hejie Lin 3 Ch 9 Moment of Inertia Step 2: Calculate Normal and Shear Stresses For each face of a finite volume, the resultant force can be decomposed into a normal force ( F x ) and two shear forces ( F y, Fz ). Normal forces are forces perpendicular to the surface and the shear forces are forces parallel to the surface. Stress is defined as force divided by area. The units for stress is pound per square inch (psi= 2 2 lb / in ) or Newton per square meter ( N / m ). Normal stresses ( σ xx ) are calculated by normal force divided by the surface area and shear stresses ( σ xx, σ yx ) are calculated by shear forces ( F y, Fz ) divided by the surface area respectively. As a result, the stress on a surface is a vector with 3 components ( σ, σ, σ ). The stress and strain relation is called the constitution law. The constitution law is studied in the field of Mechanics of Material. xx yx zx Normal and Shear Forces in a Finite Volume Normal and Shear Stress in a Finite Volume = F / A F / A F / A σ xx x x σ yx = y x σ zx = z x
Handout #1 by Hejie Lin 4 Ch 9 Moment of Inertia 1.3 Dynamic Analysis (MET 327) Dynamics analysis can be divided into the following two categories: Kinematics and Kinetics. Kinematics Kinematics studies the relationship between displacement, velocity and acceleration without considering the mass and force of the system. (Ch. 10,11) Force and mass are not considered in Kinematics analysis. If we neglect the deformation of a solid body, we can consider it as a rigid body. It takes six Degree of Freedom (DOF) to fully describe a motion of a rigid body. Three translational DOF corresponds to forces and three rotational DOF corresponds to the moments. For a system with many bodies, we also need to study the relationship between the motions of each rigid body. In the example below, the DOF of the cart is constrained from 6 ( x, y, z, θ x, θ y, θ z ) to 1 (x ) with 5 motionless DOF ( y = z = θ x = θ y = θ z = 0 ). The DOF of the wheel is also constrained from 6 ( x, y, z, θ x, θ y, θ z ) to 1 ( θ z ) with 5 motionless DOF ( x = y = z = θ x = θ y = 0). Together two DOF ( x, θ z ) are finally constrained into 1 DOF ( x or θ z ) with a non-elongation string ( x = rθ z ). 1-DOF Kinematics System Rectilinear Motion Rotational Motion x θ v ω a α Constrain Equations x = rθ v = rω a = rα
Handout #1 by Hejie Lin 5 Ch 9 Moment of Inertia Kinetics Kinetic studies the relation between force and mass of a system. The stiffness and Damping coefficient are zero in a dynamic analysis. Newton s 2 nd Law is extended into rectilinear motion and rotational motion. In any Free Body Diagram (FBD), the summation of force vectors (including applied forces, reaction forces and initial forces) will be zero. The mass moment of inertia is defined (based ont = Iα ) as a measurement of inertia torque to balance the applied torque similar to the mass is defined (based on F = Ma ) as a measurement of inertia force to balance the applied force. Rectilinear and Rotational Kinetics Force Balance for Torque Balance for Rectilinear Motion Rotational Motion 1 3 T = Iα F F = Ma Force Balance for Constrain Equations F = 2 F 3 Both mass (m) and mass moment of inertia ( I ) will be calculated for EOM. (Ch 9) Three methods will be used to calculate the dynamic motion: (1) Newton s 2 nd Law, (2) Conservation of energy and (3) Conservation of momentum.
Handout #1 by Hejie Lin 6 Ch 9 Moment of Inertia Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes 9-5. Mass Moment of Inertia (page317) Definition of Mass Moment of Inertia (Chapter 9-5; page 317) Mass Moment of Inertia will create an angular inertial toque when the mass is rotating with angular acceleration. The definition for mass moment of inertia is as following. Mass Moment of Inertia about axis x Definition of Mass Moment of Inertia = r M I x 2 (Eq.9-5 on page 316) Where, M = mass; r= perpendicular distance from axis; I x = Mass Moment of Inertia about x-axis Physical Meaning of Mass Moment of Inertia (Chapter 9-5) Parallel to the Newton s 2 nd Law ( F = Ma ), the relation between Torque, mass moment of inertia and angular acceleration can be formulated ast = Iα. Note that torque is defined as force multiply the distance to the center of rotationt = Fr. The following example will demonstrate the meaning and show the definition of mass moment of inertia. A mass m connected to a mass-less rigid bar of length equal to l. When the mass and the massless bar is under a angular 2 accelerationα, we will show the torque required is T = Iα where I = mr. A Mass with Mass-less Rigid Bar Derivation of Mass Moment of Inertia (1) give angular acceleration α (2) translational acceleration a = αr (3) force required F = ma = m( αr) (4) equivalent torque required 2 T = Fr = m αr r = mr α = I ( ) ( ) α
Handout #1 by Hejie Lin Table of Centroidal Mass Moment of Inertia (Table9-2; page 319) 7 Ch 9 Moment of Inertia Volume (V) 1. Circular Cylinder V=п r² l 4. Circular Cone V=(1/3) п r² h x = (3/4)h c I yc = 3 1 1 m( r 2 + h 2 ) 5 4 16 I zc = 3 1 1 m( r 2 + h 2 ) 5 4 16 5. Sphere V=(4/3) п r²
Handout #1 by Hejie Lin 8 Ch 9 Moment of Inertia 6. Hemisphere V=(2/3) п r² x = (-3/8) r I c yc = I zc =(83/320) m r²
Handout #1 by Hejie Lin Calculation Procedures 9 Ch 9 Moment of Inertia Step One: Calculate Centroidal Mass Moment of Inertia Formulas From Table 9-2 1 2 I xc = mr 2 1 2 2 I yc = m( 3r + l ) 12 1 2 2 I zc = m( 3r + l ) 12 Where, m= mass; l is the length, r is the radius; I xc, I yc, I zc = mass moment of inertia about centroidal axis. Example 9-5 #1 Determine the mass moment of inertia about the centroidal longitudinal axis of a shaft that has a mass of 100 kg and diameter of 120mm. Sol:
Handout #1 by Hejie Lin Units of Mass Moment of Inertia (Chapter 9-5) 10 Ch 9 Moment of Inertia The most confusion part of dealing with S.I. units and British units in the mass. Mass is a very important quantity in the dynamic analysis. Both S.I. and British units are used in Canada and the U.S. S.I. Units British Units M=mass Kg (= N s²/m) slug (= lb s²/ ft) r =perpendicular distance from axis m ft I x = mass moment of inertia Kg m² (= N m s²) slug ft² (= lb ft s²) T=torque N m lb ft F=Force N (= Kg m/s²) lb (= slug ft/s²) a=acceleration m/s² ft/s² α =angular acceleration rad/s² rad/s² In S.I. system, mass (Kg) is the basic units and force Newton (N) is a derived units. In British system, force or more precisely weight (lb) is the basic units and the mass (slug) is the derived units. Note that, g = gravity acceleration = 9.81 m/s² = 32.2 ft/s². Therefore, the gravity force acting on 1 Kg mass is 9.81 N, and the gravity force acting on 1 slug mass is 32.2 lb. W=M g 9.81 Kg m/s² = (1 Kg) (9.81 m/ s²) 32.2 lb= (1 lb s²/ ft) (32.2 ft/s²) However, it is conventionally to replace Kg m/s² with N and replace lb s²/ ft with slug. That is Kg m/s² = N lb s²/ ft = slug (Force Units) (Mass Units) The same equations is therefore more commonly presented as 9.81 N = (1 Kg) (9.81 m/ s²) 32.2 lb= (1 slug) (32.2 ft/s²) For example, something weighted 10 lb will be equal to M=W/g= (10 lb)/ (32.2 ft/s²)= 0.31 lb s²/ ft = 0.31 slug
Handout #1 by Hejie Lin 11 Ch 9 Moment of Inertia Example 9-5 #2 Determine the mass moment of inertia about the centroidal axis of a sphere 2 ft in diameter and weighting 64.4.lb. Sol:
Handout #1 by Hejie Lin 9-6 Mass Moment of Inertia of Composite Bodies 12 Ch 9 Moment of Inertia Procedures for Calculating the Mass Moment of Inertia The formula for calculating the mass moment of inertia about any axis is simple. However, the calculation could be very longsome and tedious. The parallel axis theorem gives us a very power solution to calculate the mass moment of inertia about some important axes. Note that one of two parallel axes about which the mass moment of inertia rotates must pass the center of gravity (C.G.) when the parallel axis theorem is used. Step Two: Move the Centroidal mass moment of inertia to a parallel axis by Parallel Axis Theorem Formula: Parallel Axis Theorem I I c + md 2 = (Eq.9-6 on page 321) I I I x y z = I = I = I xc yc zc + md + md + md 2 x 2 y 2 z Where, I, I, I = mass moment of inertia x y z about x-, y- z- axis respectively. d, d, d is the perpendicular distance x y z from C.G. to the x-, y- z- axis respectively. Step Three: Apply Parallel Axis Theorem to other masses and add together. Formula: Parallel Axis Theorem I = Ii = I 1 + I 2 i I I I x y z = I = I = I 1x 1y 1z + I + I + I 2x 2 y 2z Where, I1 x, I1y, I1z = mass moment of inertia of cylinder about x-, y- z- axis respectively. I 2 x, I 2 y, I 2z = mass moment of inertia of prism about x-, y- z- axis respectively.
Handout #1 by Hejie Lin Example 9-6 #1 (book problem 9-44 on page 332) 13 Ch 9 Moment of Inertia Sol:
Handout #1 by Hejie Lin 14 Ch 9 Moment of Inertia Example 9-6 #2 (book problem 9-45 on page 332) Sol:
Handout #1 by Hejie Lin Example 9-6 #3 15 Ch 9 Moment of Inertia The right circular cone shown has a mass of 90 Kg. Determine (a) the mass moment of inertia about y-axis and (b) the mass moment of inertia about x-axis Sol:
Handout #1 by Hejie Lin 16 Ch 9 Moment of Inertia 9-7 Radius of Gyration ( 9-7; page324) Mass moment of inertia about any axis can be condensed into mass multiply a radius of gyration about that particular axis as shown in the condensed equation below. Note that each radius of gyration is axis dependent. Which mean the selection of axis of which the mass moment rotates will change the value of radius of gyration. The radius of gyration is a fictitious value. No physical meaning is related to this number. However, it provides a simple and clean formula to present mass moment of inertia about any axis. Step Four: Condensed Mass Moment of Inertia: Condense the mass moment of inertia to a mass multiply the radius of gyration. Using Radius of Gyration ( k ): 2 I = mk (Eq.9-8 on, page 324) I x I y I z = = = I xc I yc I zc + md 2 2 x = mk x + md 2 mk 2 y = y + md 2 mk 2 z = z Where, k x, k y, k z = radius of gyration about x-, y- z- axis respectively. kx = k y = k = z i i i i i i I xi m I I i yi m i zi m i
Handout #1 by Hejie Lin 17 Ch 9 Moment of Inertia Example 9-7 #1 Determine radius of gyration about the centroidal longitudinal axis of a shaft that has a mass of 100 kg and diameter of 120mm. Sol:
Handout #1 by Hejie Lin Example 9-7 #2 18 Ch 9 Moment of Inertia (1) The slender rods have a mass per unit length of 3 Kg/m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the plane of the page and pass through point A. Also, calculate the radius of gyration about A. (Ans. I A =8 Kg m², k=1.155 m) Sol:
Handout #1 by Hejie Lin Example 9-7 #3 The shape in figure 1 shown weight 0.2 lb/in.³. Determine (a) mass moment of inertia about y-axis, (b) radius of gyration about the y-axis. (Ans: I y = 0.0763 slug ft², k y =0.2019 ft) 19 Ch 9 Moment of Inertia Sol:
Handout #1 by Hejie Lin 20 Ch 9 Moment of Inertia Example 9-7 #4 A fly wheel can be considered as being composed of a thin disc and a rim. The rim weights 322 lb and has diameter of 24 in. and 30 in. The disc weight 64.4 lb. Determine the mass moment of inertia about the centroidal axis about which the flywheel rotates (b) the radius of gyration about the same centroidal axis. Sol:
Handout #1 by Hejie Lin UNSOLVED PROBLEM 21 Ch 9 Moment of Inertia (1) The slender rods have a mass per unit length of 3 Kg/m. Determine the mass moment of inertia of the assembly about an axis perpendicular to the plane of the page and pass through point A. Also, calculate the radius of gyration about A. (Ans. I =72 Kg m², k= 2 m) A (2) The shape as shown weight 0.2 lb/in.³ Determine (a) mass moment of inertia about y-axis, (b) radius of gyration about the y-axis. (Ans: I y = 0.284 slug ft², k y =0.389 ft)
Handout #2 by Hejie Lin 22 Ch10 Kinematics: Rectilinear Motion Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes Chapter 10. Kinematics: Rectilinear Motion (page337) Introduction A dynamic system is characterized with mass (M), damping coefficient (C) and Stiffness (K). For our analysis, damping coefficient (C) is assumed to be zero. Dynamic motion of a mechanics system can be analyzed with Equation of Motion (EOM). EOM relates forces to the acceleration, velocity and displacement. Statics analysis calculates the relation between forces and displacement under motionless static condition. Therefore both velocity (v=0) and acceleration (a=0) are zero. As a result, both mass (M) and the damping coefficient (C) can exist in a system but do not need to be known for solving the equation of motion (EOM). Under static condition, the equation of motion (EOM) can be simplified to the Hook s Law ( F = Kx ). The displacement (x) due to applied force ( F ) can be calculated from Hook s Law and vise verse. The stiffness, K, is a constant square matrix for a large linear system. On the other hand, a dynamic system does not have spring or damper. Therefore, the stiffness ( K = 0 ) and damping coefficient ( C = 0) are both zero. Under this condition the equation of motion (EOM) can be simplified to the Newton s 2 nd Law ( F = Ma ). The acceleration (a) due to applied force ( F ) can be calculated from Newton s 2 nd Law and vise verse. Vibration System Statics System(MET 227) Kinetics System Conditions: Conditions: v = 0, a = 0 System Parameters: K = 0, C = 0 EOM: F= Ma+Cv+Kx EOM: F= Kx EOM: F= Ma Where, F: Force, M: Mass, C: Damping Coefficient, K: Stiffness, a: Acceleration, v: Velocity, x: Displacement.
Handout #2 by Hejie Lin 23 Ch10 Kinematics: Rectilinear Motion Kinematics studies the relationship between displacement, velocity and acceleration without considering the mass and force of the system. (Ch. 10,11) Force and mass are not considered in Kinematics analysis. Position vector is a vector used to represent a position (location) of any point using a vector with its tail at the origin of the coordinate and its head at the point. Displacement is the difference between original position vector and some later position vector. When we select the origin of the coordinate at the original position, the later position vector becomes the displacement. Both displacement and position vector are vector quantities. Velocity is the rate of change of displacement with respect to time. Velocity is a vector quantity. Acceleration is the rate of change of velocity with respect to time. Acceleration is a vector quantity.
Handout #2 by Hejie Lin 24 Ch10 Kinematics: Rectilinear Motion Based on the definition of displacement, velocity and acceleration, the velocity is the rate of change (differentiation) of displacement with respect to time and the acceleration is the rate of change (differentiation) of the velocity with respect to time. Also based on the definition of displacement, velocity and acceleration, the velocity is the summation (integration) of the acceleration with respect to time and the displacement is the summation (integration) of velocity with respect to time. From Top Down: SLOPE From Bottom up: AREA
Handout #2 by Hejie Lin Example #1 (based on Example 10-1) 25 Ch10 Kinematics: Rectilinear Motion A car is driven 8 km north, 9 km east, and then another 4 km north. Calculate the displacement of the car traveled. Sol: Example #2 (based on Book Problem 10-2, page 353) A forklift truck lifts a pallet 2 m off the floor, moves 7 m ahead, and sets the pallet on a stack 1.5 m high. Determine the displacement of the pallet. Sol:
Handout #2 by Hejie Lin Example #3 (based on Example 10-3) 26 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin 27 Ch10 Kinematics: Rectilinear Motion
Handout #2 by Hejie Lin Example #4 (based on Example 10-3) 28 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin 29 Ch10 Kinematics: Rectilinear Motion
Handout #2 by Hejie Lin Example #5 (based on Example 10-9) 30 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin 31 Ch10 Kinematics: Rectilinear Motion
Handout #2 by Hejie Lin 10-5 Rectilinear Motion With Uniform Acceleration (Page 342-348) 32 Ch10 Kinematics: Rectilinear Motion Given a uniform acceleration, the velocity can be calculated from acceleration with initial velocity. Given a uniform acceleration, the displacement can be calculated from velocity with a initial velocity. Note that the initial time and initial displacement are forced to be zero to simplify the solution. Instead of integration, we can be derived the same equation by using the summation.
Handout #2 by Hejie Lin 33 Ch10 Kinematics: Rectilinear Motion Here we will use integration to derive the relation between acceleration, velocity and displacement. The integration technique gives an easier derivation but sometime difficult to understand the physical meaning behind the mathematics. It is important to remember that the integration is the same as summation.
Handout #2 by Hejie Lin 34 Ch10 Kinematics: Rectilinear Motion We can have an equation to relate velocity and acceleration without time. This equation can be easily derived from the following two equations.
Handout #2 by Hejie Lin 35 Ch10 Kinematics: Rectilinear Motion These three equations will be used to solve the uniform acceleration problem. The initial time and displacement are zero The acceleration are uniform (constant) at all time. Each equation relates four unknown variables. We can use three known variable to find the missing variable. We can arbitrary select the positive direction. Once the positive direction is determined, the positive velocity and acceleration are determined.
Handout #2 by Hejie Lin Example #6 (Book Example 10-4, page 345) 36 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin Example #7 (based on Example 10-19) 37 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin Example #8 (based on Example 10-25) 38 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin Example #9 (based on Example 10-27) 39 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin 40 Ch10 Kinematics: Rectilinear Motion 10-6 Projectiles (Page 348-352) The projectile motion is like a baseball hit for a home run. The main assumption is that air has no resistant. Projectile motion consists two rectilinear (translational) motions connected by the common time factor. The first translational motion is vertical motion with a uniform acceleration of gravity (9.81 m/s²). The second translational motion is horizontal motion with zero acceleration. Both vertical displacement, x(t), and horizontal displacement, y(t), are functions of the same time variable. The displacement, velocity and acceleration follow the same positive sign convention. We can arbitrary assign a positive direction for the displacement, velocity and acceleration.
Handout #2 by Hejie Lin Example #10 (based on Example 10-7) 41 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin Example #11 (based on Example 10-8) 42 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin 43 Ch10 Kinematics: Rectilinear Motion
Handout #2 by Hejie Lin Example #12 (based on Example 10-8) 44 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin Example #13 (Example 10-8) The projectile reaches its maximum elevation at point A. Determine the initial velocity and angle Ө. 45 Ch10 Kinematics: Rectilinear Motion Sol:
Handout #2 by Hejie Lin UNSOLVED PROBLEM 46 Ch10 Kinematics: Rectilinear Motion (1) The object moves from point (1) to (2) in a circular path at a constant speed of 30 m/s in a time of 3 second. Determine the magnitude and direction of average acceleration. (Ans. 5.18 m/s², -15º) (2) An object with an initial velocity of 25m/s upward lands 80 m below its starting point. Determine (a) its maximum height, (b) its total time in the air, and (c) the velocity at the time of its landing. (Ans. 31.86 m, 7.33 s, 46.85 m/s) (3) A projectile that is fired with an initial velocity of 250 m/s is inclined upward at an angle of 40 º. It lands at a point 100 m higher than the initial point. Determine (a) the time of flight, (b) the horizontal displacement, and (c) the final velocity at the point of landing. (Ans. 32.13 s, 6152.73 m/s, 248 m/s, -38.9º) (4) A golfer hits a ball, giving it a maximum height of 20 m above his tee-off point. The tee-off point is 10 m above the landing spot of the ball. If the ball lands a horizontal distance of 130 m away from which he hit it, what was its initial velocity? (Ans. 35.06 m/s, 34.38 º)
Handout #3 by Hejie Lin 47 Ch11 Kinematics: Angular Motion Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes Chapter 11. Kinematics: Angular Motion (page365) Definition The angular displacement usually considers about one rotational axis direction. Angular acceleration is the rate of change of angular velocity with respect to time. Angular displacement is the difference between original angle and some later angle. Radian is used in Both S.I. and British system. The angular displacement in the formula must use radian for S.I. and British system. 1 revolution= 360 degree = 2п radians Angular velocity is the rate of change of angular displacement with respect to time. The units for angular velocity are rpm (revolution per minute) or rad/s. Rpm is the conventional units but can not be used in the Mathematics formulas. revolution minute or rad second The relationship between rpm and rad/s is as following: revolution 1minute 2π radians 1 rpm = 1 = minute 60 seconds 1 revolution 2π 60 rad s
Handout #3 by Hejie Lin 48 Ch11 Kinematics: Angular Motion Relationship Between Rectilinear and Angular Motion (Page 371) The following equations are under the condition of uniform angular acceleration. The initial time and initial angular displacement are defined as zero.
Handout #3 by Hejie Lin Example #1 (based on Example 11-4) 49 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #2 (based on Book Example 11-5) 50 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #3 (based on Example 10-6) 51 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin 52 Ch11 Kinematics: Angular Motion
Handout #3 by Hejie Lin 53 Ch11 Kinematics: Angular Motion 10-6 Relationship Between Rectilinear and Angular Motion (Page 368) The relationship between linear displacement and angular displacement is the fundamental constrain equation. The relationship between rectilinear and angular motion for velocity and acceleration can be derived from the relationship between rectilinear and angular motion of displacement. In the figure below, s can replaced with linear velocity and acceleration. Where r is constant. θ 2π = s 2πr s θ = r t t v ω = r t t s = rθ v = rω a = rα
Handout #3 by Hejie Lin Example #4 (based on Example 11-6) 54 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #5 (Problem 11-27) 55 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #6 (Problem 11-38) 56 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #7 (based on Problem 11-40) 57 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin 10-7 Normal and Tangential Acceleration (Page 374) 58 Ch11 Kinematics: Angular Motion Tangential acceleration a t is due to the change of magnitude of it linear velocity v and angular velocity ω values. The direction of tangential acceleration is tangential to the arc of the travel and is therefore perpendicular to the radius. Normal acceleration a n is due to the change in direction of velocityv. The direction of normal acceleration is normal to the arc of the travel and is therefore parallel to the radius. a v n t A wheel with radius r turns at constant speed v will have a normal acceleration a n as 2 v a n r
Handout #3 by Hejie Lin 59 Ch11 Kinematics: Angular Motion The normal acceleration of a wheel with radius r turns at constant speed v can be calculated as following:
Handout #3 by Hejie Lin Example #8 (based on Example 11-8) 60 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #9 (based on Example 11-9) 61 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin Example #10 (based on Problem 11-61) 62 Ch11 Kinematics: Angular Motion Sol:
Handout #3 by Hejie Lin UNSOLVED PROBLEM 63 Ch11 Kinematics: Angular Motion (1) A pulley turns initially at 40 rpm clockwise, then reverse its direction of rotation to 30 counterclockwise in 14 seconds, at a constant deceleration and acceleration rate. Determine (a) the deceleration rate, (b) the total number of revolutions of the pulley during the 14-second interval, and (c) the angular displacement of the pulley at t=14 seconds. (Ans. 0.523 rad/s², 4.17 rev, 1.17 rev) (2) The combined pulley has two cables wounded around it at different diameters and fastened to point C and block E, respectively. If the velocity of block E is 200 mm/s downward, determine (a) the angular velocity of D, (b) the velocity of point C, (c) the angular velocity of lever AC, and (d) the velocity of point A, at the instance shown. (Ans. 2.5 rad/s, 0.125 m/s, 0.568 rad/s, 0.148 m/s) (3) The object rotates about A and decelerates from an initial speed of 40 rpm to 10 rpm in 5 seconds. At t=2.5 seconds, determine the total acceleration of point B for the position shown. (Ans. 27.6 m/s, 84.8º) [40 points]
Handout #4 by Hejie Lin Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes 64 Ch13 Kinetics Chapter 13. Kinetics (page437) Introduction A dynamic system is characterized with mass (M), damping coefficient (C) and Stiffness (K). For our analysis, damping coefficient (C) is assumed to be zero. Vibration System Statics (MET 227) Kinetics (MET 327) Conditions: Conditions: v = 0, a = 0 System Parameters: K = 0, C = 0 EOM: F= Ma+Cv+Kx EOM: F= Kx EOM: F= Ma CHAPTER 9: Mass Moment of Inertia: I = ( ) m r 2 CHAPTER 10: Kinetics of Rectilinear Motion: v a t s v t s r a = constant v = v 0 + at 1 2 s = v0t + at 2 2 2 v = v + 2as 0 CHAPTER 11: Kinetics of Angular Motion: a = αr v = ωr s = θr a n v = r 2
Handout #4 by Hejie Lin Newton s Three Laws of Motion 65 Ch13 Kinetics First Law: Every object remains at rest or maintains a constant velocity in a straight line unless an unbalanced force acts upon it. F = 0 F a = = 0 M v = constant Second Law: A body that has a resultant force acting upon it behaves as follows: a a r F = F 1 M r Ma F = Ma, Third Law: For every action force there is an equal and opposite reaction force. F = ACTION F REACTION The opposing reaction force is the inertia force. F REACTION = F INERTIA Whether an object is accelerating form zero or from some existing velocity, the inertia force will act opposite to the acceleration. Based on Newton s 2nd Law of motion, the inertia force can be represented as Therefore, we can conclude that F INERTIA = -Ma F REACTION = -Ma
Handout #4 by Hejie Lin Three Methods for Kinetics Analysis Kinetics, the study of unbalance forces causes motion, can be analyzed by the following three methods. CHAPTER 13: Inertia Force and Torque Method Dynamic Equilibrium Approach Using Free - Body Diagram (FBD) CHAPTER 14: Work and Energy Method 66 Ch13 Kinetics Conservati on of Energy CHAPTER 15: Impulse and Momentum Method Conservati on of Momentum Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 2. Select Degree of Freedom (DOF): 3. Assign Positive Direction for Each DOF: 4. Draw All Isolated Masses / Components: 5. Assume Required Positive Acceleration: 6. Show All Action Forces / Reaction Forces: 7. Show All Inertia Force / Friction Forces: 8. Balance all Forces / Moments:
Handout #4 by Hejie Lin Example #1 67 Ch13 Kinetics Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): x 3. Assign Positive Direction for Each DOF: x + 4. Draw All Isolated Masses / Components: M 5. Assume Required Positive Acceleration: a > 0, v > 0, x > 0 6. Show All Action Forces / Reaction Forces: P 7. Show All Inertia Force / Friction Forces: Ma, Mgµ, Mg K 8. Balance all Forces / Moments F = 0 x P Ma Mgµ K = 0
Handout #4 by Hejie Lin Example #2 68 Ch13 Kinetics Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): y 3. Assign Positive Direction for Each DOF: y upward positive 4. Draw All Isolated Masses / Components: 5. Assume Required Positive Acceleration: a>0 6. Show All Action Forces / Reaction Forces: M*g, P 7. Show All Inertia Force / Friction Forces: M*a 8. Balance all Forces / Moments: P-M*g-M*a=0
Handout #4 by Hejie Lin Example #3 (modified from Example 13-5) 69 Ch13 Kinetics Neglect the inertia of the pulleys and rolling resistance of the mass B and determine the acceleration of mass A and B when the system is release from the rest. Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): x 3. Assign Positive Direction for Each DOF: x downward positive 4. Draw All Isolated Masses / Components: 5. Assume Required Positive Acceleration: a>0 6. Show All Action Forces / Reaction Forces: Ma*g, Mb*g, T, 2T 7. Show All Inertia Force / Friction Forces: Ma*a, Mb*(2a), Mb*g*uk 8. Balance all Forces / Moments:
Handout #4 by Hejie Lin 13-5 Angular Dynamic Equilibrium Angular Inertia is used as a mass in Newton s 2 nd law of motion as: T = Iα 70 Ch13 Kinetics Where T is the toque force, α is the angular acceleration is related to linear motion as v α = r and I is mass moment of inertia and is defined as I = m r Or, is the mass moment of inertia could be calculated from radius of gyration k I = M k 2 2 Example #4 (modified from Example 13-7) A rotor with a mass moment of inertia ( I C ) of 6 kg m² about its center of mass has a torque T of 90 N m applied to it. Determine the angular acceleration of the rotor. Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): θ 3. Assign Positive Direction for Each DOF: θ ccw + 4. Draw All Isolated Masses / Components: I 5. Assume Required Positive Acceleration: α > 0 6. Show All Action Forces / Reaction Forces: T 7. Show All Inertia Force / Friction Forces: α I 8. Balance all Forces / Moments: M = 0 T Iα = 0 α = T I 90 N = 6 kg m m kg s kg m 2 = 15 = 15 2 2 rad 2 s
Handout #4 by Hejie Lin Example #5 (from Example 13-10) 71 Ch13 Kinetics A 6-ft slender rod weighting 64.4 lb is initially at rest when the force of 10 lb is applied as shown. Determine the angular acceleration of the rod using the summation of moment of the rod about (a) point A and (b) point C. Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): θ 3. Assign Positive Direction for Each DOF: θ ccw + 4. Draw All Isolated Masses / Components: rod 5. Assume Required Positive Acceleration: α > 0, ω > 0 6. Show All Action Forces / Reaction Forces: P, 7. Show All Inertia Force / Friction Forces: I, Ma α A X, A Y 8. Balance all Forces / Moments M 0 or M = 0, F = 0, F = 0 A = C y x
Handout #4 by Hejie Lin 72 Ch13 Kinetics Note that it is much easier to use summation of moment about point A because the reactions have zero moment about point A. The force balance of vertical and horizontal forces can be used to find the reaction forces at point A.
Handout #4 by Hejie Lin Example #6 (modified from Example 13-8) 73 Ch13 Kinetics Neglect the pulley inertia. A power-driven winch is used to raise a mass of 300 kg. The winch drum is 0.5 m in diameter and has a mass moment of inertia about its center Ic equal to 8 kg m². When a torque of 629 N m is applied to the winch drum, determine the angular acceleration of drum. Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): θ A 3. Assign Positive Direction for Each DOF: θ ccw + 4. Draw All Isolated Masses/ Components: A, B, C 5. Assume Required Positive Acceleration: > 0 6. Show All Action Forces / Reaction Forces: T M g / A A, C, C α, B X, 7. Show All Inertia Force / Friction Forces: M BaB, I Aα A 8. Balance all Forces / Moments M = 0, F = 0 A A A Y X y Y
Handout #4 by Hejie Lin 74 Ch13 Kinetics
Handout #4 by Hejie Lin Example #7 (modified from Example 13-8) 75 Ch13 Kinetics Mass A and B are fastened together by a belt over pulley D. The mass moment of inertia of pulley D is 15 kg m². Determine the angular acceleration of pulley D. Sol: Procedures for FBD Method 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): y A 3. Assign Positive Direction for Each DOF: y + 4. Draw All Isolated Masses/ Components: A, B, D 5. Assume Required Positive Acceleration: 6. Show All Action Forces / Reaction Forces: M A y A g, M 7. Show All Inertia Force / Friction Forces: A B A, M Ba A, I D D A > A 0, v > 0 g / D X B, D M a α / M gµ 8. Balance all Forces / Moments A B D F = 0, F = 0, M = 0 y x c Y K
Handout #4 by Hejie Lin 76 Ch13 Kinetics
Handout #4 by Hejie Lin 77 Ch13 Kinetics SOLVED PROBLEMS (1) Neglect the inertia of pulley and rolling resistant of mass B. The coefficient of kinetic friction is 0.2. Determine (a) the acceleration of mass B, (b) the tension in the rope. (c) the distance mass A will fall in 2 seconds. (Ans. 5.88 m/s², 157 N, 11.8 m) (2) A 22 kg wheel B, 240 mm in diameter, is accelerated from rest by a 10 kg mass A moving downward. The wheel has a uniform distributed thickness and its motion is retarded by a braking force due to a spring force of 10 N. Determine (a) the radius of gyration of the wheel (b) the angular acceleration of the wheel A, (c) the acceleration of the mass A, (d) the tension in the rope and (e) the 1 2 distance mass A will fall in 6 seconds. (Assume that I = mr ) 2 (Ans. 84.85 mm, 38.1 rad/s², 4.58 m/s², 52.3 N, 82.4 m) (3) The spring is at its free length, and the system is at rest when weight B is dropped. Determine (a) the Equation of Motions (EOM) and (b) the distance mass B will fall in 2 seconds using the dynamic equilibrium method using freebody diagram (FBD). (Ans.? m)
Handout #5 by Hejie Lin 78 Chapter 14: Work, Energy, and Power Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes Chapter 14. Work, Energy, and Power (page473) Introduction A vibration system is characterized with mass (M), damping coefficient (C) and Stiffness (K). The vibration analysis becomes Statics analysis when the system is at rest (v=0 and a=0). The vibration analysis becomes Kinetics analysis when the vibration system has no spring. Vibration System Statics (MET 227) Kinetics System Conditions: Conditions: v = 0, a = 0 System Parameters: K = 0 F= Ma+Kx F= Kx F= Ma
Handout #5 by Hejie Lin 79 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin WORK DONE: TRANLATIONAL Energy Gain 80 Chapter 14: Work, Energy, and Power Energy Lose
Handout #5 by Hejie Lin Potential Energy Due to Gravity (PEg) 81 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 82 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin Potential Energy Due to Spring Force (PEs) 83 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 84 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 85 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 86 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin Kinetic Energy: Translational (KE) 87 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 88 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 89 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin WORK DONE: ANGULAR 90 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin Kinetic Energy: Angular (KE) 91 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin FORCE & WORK 92 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin POWER & EFFICIENCY 93 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 94 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 95 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin Applications of Conservation of Energy in Kinetics System 96 Chapter 14: Work, Energy, and Power Dynamic Equilibrium Approach in Kinetics System:
Handout #5 by Hejie Lin Conservation of Energy Approach in Kinetics System: 97 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin Examples (Conservation of Energy Approach for Kinetic System 98 Chapter 14: Work, Energy, and Power Cylinder A has of gyration of 85 mm. Determine the angular acceleration of A by conservation of energy method. [You may assume the system initially at rest to simplify the calculation.] (Ans. 38.8 rad/s²) Solution:
Handout #5 by Hejie Lin 99 Chapter 14: Work, Energy, and Power
Handout #5 by Hejie Lin 100 Chapter 14: Work, Energy, and Power SOLVED PROBLEMS (1) Neglect the inertia of pulley and rolling resistant of mass B. The coefficient of kinetic friction is 0.2. Determine (a) the acceleration of mass B (b) and the tension in the rope by the conservation of energy method [you may assume the system is initially at rest to simplify the calculation.]. (Ans. 5.88 m/4s², 157 N) (2) A 22 kg wheel B, 240 mm in diameter, is accelerated from rest by a 10 kg mass A moving downward. The wheel has a uniform distributed thickness and its motion is retarded by a braking force due to a spring force of 10 N. Determine (a) the acceleration of the mass A, (b) the tension in the rope using the 1 2 conservation of energy method. (Assume that I = mr ) 2 (Ans. 4.58 m/s², 52.3 N) (3) The spring is at its free length, and the system is at rest when weight B is dropped. Determine the maximum distance that B will drop. (Ans. 1 m)
Handout #6 (8/12) by Hejie Lin 101 Chapter 15: Impulse and Momentum Reference to: (1) Applied Mechanics for Engineering Technology by Keith M. Walker; Pearson Prentice Hall 2008; 8 th Edition (2) Professor Raj Basu s Lecture Notes Chapter 15. Impulse and Momentum (page523) Review of Work and Kinetic Energy r r Work is a type energy and is defined as Force times distance ( W = F S ). Kinetic Energy (K.E.) of mass m at velocity ( V F ) is equal to the work required to push the mass a distance x to reach final velocity ( V F ). Definition of Impulse and Momentum
Handout #6 (8/12) by Hejie Lin The Conservation of Momentum 102 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin The Conservation of Momentum can be decomposed into 6 DOF as: 103 Chapter 15: Impulse and Momentum The Units of Impulse and Momentum
Handout #6 (8/12) by Hejie Lin Example #1(Conservation of Momentum-Linear) 104 Chapter 15: Impulse and Momentum When object A and B collide and remain together. Determine their resulting velocity and direction.
Handout #6 (8/12) by Hejie Lin 105 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin Example #2 (Conservation of Momentum-Linear) 106 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin 107 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin 108 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin Example #3 (Impulse to Momentum) 109 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin Example #4 (Conservation of Momentum-Angular) 110 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin Example #5 (Application of Conservation of Momentum) Cylinder A has a gyration of 85 mm. Determine the angular acceleration of A by (a) dynamic equilibrium method using free-body diagram (FBD), (b) conservation of energy method and (c) conservation of momentum method.[you may assume the system initially at rest to simplify the calculation.] 111 Chapter 15: Impulse and Momentum (a) Dynamic equilibrium method using free-body diagram (FBD) Procedures for FBD Technique 1. Determine Number of Degree of Freedom (DOF): 1 2. Select Degree of Freedom (DOF): θ A 3. Assign Positive Direction for Each DOF: θ A clockwise + 4. Draw All Isolated Masses / Components: A, B / 5. Assume Required Positive Acceleration / velocity: α A > 0 / ω A > 0 6. Show All Action Forces / Reaction Forces: M Ag, M B g / Ax, Ay 7. Show All Inertia Force / Friction Forces: M B α A r A, I A α A / 8. Balance all Forces / Moments
Handout #6 (8/12) by Hejie Lin 112 Chapter 15: Impulse and Momentum (b) conservation of energy method
Handout #6 (8/12) by Hejie Lin 113 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin (c) conservation of momentum method. 114 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin 115 Chapter 15: Impulse and Momentum
Handout #6 (8/12) by Hejie Lin 116 Chapter 15: Impulse and Momentum Example #6 (Conservation of Momentum vs. Conservation of Energy) A 30 gram bullet is fired horizontally with a speed of 502 m/s into a 12 kg block that is suspended on a long cord. Determine their resulting speed of block and bullet. Based on Conservation of Momentum Based on Conservation of Energy Without Considering the Energy Dissipation
Handout #6 (8/12) by Hejie Lin 117 Chapter 15: Impulse and Momentum The conservation of energy is not valid between the bullet and the block because of the energy dissipation due to the friction force between the bullet and the block. However, the conservation of momentum is valid because the momentum is related to the force multiples the time which are always equal between the bullet the block. APPENDIX A: The following show how energy is dissipated in a spring (elastic or plastic):
Handout #6 (8/12) by Hejie Lin 118 Chapter 15: Impulse and Momentum UNSOLVED PROBLEMS (1) Neglect the inertia of pulley and rolling resistant of mass B. The coefficient of kinetic friction is 0.2. Determine (a) the acceleration of mass B (b) and the tension in the rope by the conservation of momentum method. [you may assume the system is initially at rest to simplify the calculation.]. (Ans. 5.88 m/4s², 157 N) (2) A 22 kg wheel B, 240 mm in diameter, is accelerated from rest by a 10 kg mass A moving downward. The wheel has a uniform distributed thickness and its motion is retarded by a braking force due to a spring force of 10 N. Determine (a) the acceleration of the mass A, (b) the tension in the rope using the conservation of 1 2 momentum method. (Assume that I = mr ) (Ans. 4.58 m/s², 2 52.3 N) (3) When object A and B collide and remain in contact, the direction of A is diverted 10º to the left. Determine (a) the mass of B and (b) the resulting velocity of A and B. (Ans. 70.5kg, 1.64 m/s)