MET 327 APPLIED ENGINEERING II (DYNAMICS) 1-D Dynamic System Equation of Motion (EOM)

Size: px
Start display at page:

Download "MET 327 APPLIED ENGINEERING II (DYNAMICS) 1-D Dynamic System Equation of Motion (EOM)"

Transcription

1 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

2 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

3 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

4 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

5 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.

6 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

7 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 = m( r 2 + h 2 ) I zc = m( r 2 + h 2 ) Sphere V=(4/3) п r²

8 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

9 Handout #1 by Hejie Lin Calculation Procedures 9 Ch 9 Moment of Inertia Step One: Calculate Centroidal Mass Moment of Inertia Formulas From Table I xc = mr I yc = m( 3r + l ) 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:

10 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

11 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:

12 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.

13 Handout #1 by Hejie Lin Example 9-6 #1 (book problem 9-44 on page 332) 13 Ch 9 Moment of Inertia Sol:

14 Handout #1 by Hejie Lin 14 Ch 9 Moment of Inertia Example 9-6 #2 (book problem 9-45 on page 332) Sol:

15 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:

16 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

17 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:

18 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:

19 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 = slug ft², k y = ft) 19 Ch 9 Moment of Inertia Sol:

20 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:

21 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 = slug ft², k y =0.389 ft)

22 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.

23 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.

24 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

25 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:

26 Handout #2 by Hejie Lin Example #3 (based on Example 10-3) 26 Ch10 Kinematics: Rectilinear Motion Sol:

27 Handout #2 by Hejie Lin 27 Ch10 Kinematics: Rectilinear Motion

28 Handout #2 by Hejie Lin Example #4 (based on Example 10-3) 28 Ch10 Kinematics: Rectilinear Motion Sol:

29 Handout #2 by Hejie Lin 29 Ch10 Kinematics: Rectilinear Motion

30 Handout #2 by Hejie Lin Example #5 (based on Example 10-9) 30 Ch10 Kinematics: Rectilinear Motion Sol:

31 Handout #2 by Hejie Lin 31 Ch10 Kinematics: Rectilinear Motion

32 Handout #2 by Hejie Lin 10-5 Rectilinear Motion With Uniform Acceleration (Page ) 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.

33 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.

34 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.

35 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.

36 Handout #2 by Hejie Lin Example #6 (Book Example 10-4, page 345) 36 Ch10 Kinematics: Rectilinear Motion Sol:

37 Handout #2 by Hejie Lin Example #7 (based on Example 10-19) 37 Ch10 Kinematics: Rectilinear Motion Sol:

38 Handout #2 by Hejie Lin Example #8 (based on Example 10-25) 38 Ch10 Kinematics: Rectilinear Motion Sol:

39 Handout #2 by Hejie Lin Example #9 (based on Example 10-27) 39 Ch10 Kinematics: Rectilinear Motion Sol:

40 Handout #2 by Hejie Lin 40 Ch10 Kinematics: Rectilinear Motion 10-6 Projectiles (Page ) 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.

41 Handout #2 by Hejie Lin Example #10 (based on Example 10-7) 41 Ch10 Kinematics: Rectilinear Motion Sol:

42 Handout #2 by Hejie Lin Example #11 (based on Example 10-8) 42 Ch10 Kinematics: Rectilinear Motion Sol:

43 Handout #2 by Hejie Lin 43 Ch10 Kinematics: Rectilinear Motion

44 Handout #2 by Hejie Lin Example #12 (based on Example 10-8) 44 Ch10 Kinematics: Rectilinear Motion Sol:

45 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:

46 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 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 m, 7.33 s, 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 s, 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 m/s, º)

47 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

48 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.

49 Handout #3 by Hejie Lin Example #1 (based on Example 11-4) 49 Ch11 Kinematics: Angular Motion Sol:

50 Handout #3 by Hejie Lin Example #2 (based on Book Example 11-5) 50 Ch11 Kinematics: Angular Motion Sol:

51 Handout #3 by Hejie Lin Example #3 (based on Example 10-6) 51 Ch11 Kinematics: Angular Motion Sol:

52 Handout #3 by Hejie Lin 52 Ch11 Kinematics: Angular Motion

53 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

54 Handout #3 by Hejie Lin Example #4 (based on Example 11-6) 54 Ch11 Kinematics: Angular Motion Sol:

55 Handout #3 by Hejie Lin Example #5 (Problem 11-27) 55 Ch11 Kinematics: Angular Motion Sol:

56 Handout #3 by Hejie Lin Example #6 (Problem 11-38) 56 Ch11 Kinematics: Angular Motion Sol:

57 Handout #3 by Hejie Lin Example #7 (based on Problem 11-40) 57 Ch11 Kinematics: Angular Motion Sol:

58 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

59 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:

60 Handout #3 by Hejie Lin Example #8 (based on Example 11-8) 60 Ch11 Kinematics: Angular Motion Sol:

61 Handout #3 by Hejie Lin Example #9 (based on Example 11-9) 61 Ch11 Kinematics: Angular Motion Sol:

62 Handout #3 by Hejie Lin Example #10 (based on Problem 11-61) 62 Ch11 Kinematics: Angular Motion Sol:

63 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 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, m/s, rad/s, 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 m/s, 84.8º) [40 points]

64 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 v = v + 2as 0 CHAPTER 11: Kinetics of Angular Motion: a = αr v = ωr s = θr a n v = r 2

65 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

66 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:

67 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

68 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

69 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:

70 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 = rad 2 s

71 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

72 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.

73 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

74 Handout #4 by Hejie Lin 74 Ch13 Kinetics

75 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

76 Handout #4 by Hejie Lin 76 Ch13 Kinetics

77 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 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 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)

78 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

79 Handout #5 by Hejie Lin 79 Chapter 14: Work, Energy, and Power

80 Handout #5 by Hejie Lin WORK DONE: TRANLATIONAL Energy Gain 80 Chapter 14: Work, Energy, and Power Energy Lose

81 Handout #5 by Hejie Lin Potential Energy Due to Gravity (PEg) 81 Chapter 14: Work, Energy, and Power

82 Handout #5 by Hejie Lin 82 Chapter 14: Work, Energy, and Power

83 Handout #5 by Hejie Lin Potential Energy Due to Spring Force (PEs) 83 Chapter 14: Work, Energy, and Power

84 Handout #5 by Hejie Lin 84 Chapter 14: Work, Energy, and Power

85 Handout #5 by Hejie Lin 85 Chapter 14: Work, Energy, and Power

86 Handout #5 by Hejie Lin 86 Chapter 14: Work, Energy, and Power

87 Handout #5 by Hejie Lin Kinetic Energy: Translational (KE) 87 Chapter 14: Work, Energy, and Power

88 Handout #5 by Hejie Lin 88 Chapter 14: Work, Energy, and Power

89 Handout #5 by Hejie Lin 89 Chapter 14: Work, Energy, and Power

90 Handout #5 by Hejie Lin WORK DONE: ANGULAR 90 Chapter 14: Work, Energy, and Power

91 Handout #5 by Hejie Lin Kinetic Energy: Angular (KE) 91 Chapter 14: Work, Energy, and Power

92 Handout #5 by Hejie Lin FORCE & WORK 92 Chapter 14: Work, Energy, and Power

93 Handout #5 by Hejie Lin POWER & EFFICIENCY 93 Chapter 14: Work, Energy, and Power

94 Handout #5 by Hejie Lin 94 Chapter 14: Work, Energy, and Power

95 Handout #5 by Hejie Lin 95 Chapter 14: Work, Energy, and Power

96 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:

97 Handout #5 by Hejie Lin Conservation of Energy Approach in Kinetics System: 97 Chapter 14: Work, Energy, and Power

98 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 rad/s²) Solution:

99 Handout #5 by Hejie Lin 99 Chapter 14: Work, Energy, and Power

100 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 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 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)

101 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

102 Handout #6 (8/12) by Hejie Lin The Conservation of Momentum 102 Chapter 15: Impulse and Momentum

103 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

104 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.

105 Handout #6 (8/12) by Hejie Lin 105 Chapter 15: Impulse and Momentum

106 Handout #6 (8/12) by Hejie Lin Example #2 (Conservation of Momentum-Linear) 106 Chapter 15: Impulse and Momentum

107 Handout #6 (8/12) by Hejie Lin 107 Chapter 15: Impulse and Momentum

108 Handout #6 (8/12) by Hejie Lin 108 Chapter 15: Impulse and Momentum

109 Handout #6 (8/12) by Hejie Lin Example #3 (Impulse to Momentum) 109 Chapter 15: Impulse and Momentum

110 Handout #6 (8/12) by Hejie Lin Example #4 (Conservation of Momentum-Angular) 110 Chapter 15: Impulse and Momentum

111 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

112 Handout #6 (8/12) by Hejie Lin 112 Chapter 15: Impulse and Momentum (b) conservation of energy method

113 Handout #6 (8/12) by Hejie Lin 113 Chapter 15: Impulse and Momentum

114 Handout #6 (8/12) by Hejie Lin (c) conservation of momentum method. 114 Chapter 15: Impulse and Momentum

115 Handout #6 (8/12) by Hejie Lin 115 Chapter 15: Impulse and Momentum

116 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

117 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):

118 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 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 m/s², 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)

E 490 FE Exam Prep. Engineering Mechanics

E 490 FE Exam Prep. Engineering Mechanics E 490 FE Exam Prep Engineering Mechanics 2008 E 490 Course Topics Statics Newton s Laws of Motion Resultant Force Systems Moment of Forces and Couples Equilibrium Pulley Systems Trusses Centroid of an

More information

CEE 271: Applied Mechanics II, Dynamics Lecture 25: Ch.17, Sec.4-5

CEE 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 information

PLANAR KINETIC EQUATIONS OF MOTION (Section 17.2)

PLANAR 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 information

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work

Translational vs Rotational. m x. Connection Δ = = = = = = Δ = = = = = = Δ =Δ = = = = = 2 / 1/2. Work Translational vs Rotational / / 1/ Δ m x v dx dt a dv dt F ma p mv KE mv Work Fd / / 1/ θ ω θ α ω τ α ω ω τθ Δ I d dt d dt I L I KE I Work / θ ω α τ Δ Δ c t s r v r a v r a r Fr L pr Connection Translational

More information

. d. v A v B. e. none of these.

. d. v A v B. e. none of these. General Physics I Exam 3 - Chs. 7,8,9 - Momentum, Rotation, Equilibrium Oct. 28, 2009 Name Rec. Instr. Rec. Time For full credit, make your work clear to the grader. Show the formulas you use, the essential

More information

Plane Motion of Rigid Bodies: Forces and Accelerations

Plane Motion of Rigid Bodies: Forces and Accelerations Plane Motion of Rigid Bodies: Forces and Accelerations Reference: Beer, Ferdinand P. et al, Vector Mechanics for Engineers : Dynamics, 8 th Edition, Mc GrawHill Hibbeler R.C., Engineering Mechanics: Dynamics,

More information

Chapter 8 Lecture Notes

Chapter 8 Lecture Notes Chapter 8 Lecture Notes Physics 2414 - Strauss Formulas: v = l / t = r θ / t = rω a T = v / t = r ω / t =rα a C = v 2 /r = ω 2 r ω = ω 0 + αt θ = ω 0 t +(1/2)αt 2 θ = (1/2)(ω 0 +ω)t ω 2 = ω 0 2 +2αθ τ

More information

Chapter 10. Rotation

Chapter 10. Rotation Chapter 10 Rotation Rotation Rotational Kinematics: Angular velocity and Angular Acceleration Rotational Kinetic Energy Moment of Inertia Newton s nd Law for Rotation Applications MFMcGraw-PHY 45 Chap_10Ha-Rotation-Revised

More information

= o + t = ot + ½ t 2 = o + 2

= 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 information

Your Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon

Your Name: PHYSICS 101 MIDTERM. Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon 1 Your Name: PHYSICS 101 MIDTERM October 26, 2006 2 hours Please circle your section 1 9 am Galbiati 2 10 am Kwon 3 11 am McDonald 4 12:30 pm McDonald 5 12:30 pm Kwon Problem Score 1 /13 2 /20 3 /20 4

More information

VALLIAMMAI ENGINEERING COLLEGE SRM NAGAR, KATTANKULATHUR DEPARTMENT OF MECHANICAL ENGINEERING

VALLIAMMAI ENGINEERING COLLEGE SRM NAGAR, KATTANKULATHUR DEPARTMENT OF MECHANICAL ENGINEERING VALLIAMMAI ENGINEERING COLLEGE SRM NAGAR, KATTANKULATHUR 603203 DEPARTMENT OF MECHANICAL ENGINEERING BRANCH: MECHANICAL YEAR / SEMESTER: I / II UNIT 1 PART- A 1. State Newton's three laws of motion? 2.

More information

JNTU World. Subject Code: R13110/R13

JNTU World. Subject Code: R13110/R13 Set No - 1 I B. Tech I Semester Regular Examinations Feb./Mar. - 2014 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question

More information

General Physics (PHY 2130)

General Physics (PHY 2130) General Physics (PHY 130) Lecture 0 Rotational dynamics equilibrium nd Newton s Law for rotational motion rolling Exam II review http://www.physics.wayne.edu/~apetrov/phy130/ Lightning Review Last lecture:

More information

KINGS COLLEGE OF ENGINEERING ENGINEERING MECHANICS QUESTION BANK UNIT I - PART-A

KINGS COLLEGE OF ENGINEERING ENGINEERING MECHANICS QUESTION BANK UNIT I - PART-A KINGS COLLEGE OF ENGINEERING ENGINEERING MECHANICS QUESTION BANK Sub. Code: CE1151 Sub. Name: Engg. Mechanics UNIT I - PART-A Sem / Year II / I 1.Distinguish the following system of forces with a suitable

More information

Concept Question: Normal Force

Concept 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 information

6. Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm.

6. Find the net torque on the wheel in Figure about the axle through O if a = 10.0 cm and b = 25.0 cm. 1. During a certain period of time, the angular position of a swinging door is described by θ = 5.00 + 10.0t + 2.00t 2, where θ is in radians and t is in seconds. Determine the angular position, angular

More information

Uniform Circular Motion

Uniform Circular Motion Uniform Circular Motion Motion in a circle at constant angular speed. ω: angular velocity (rad/s) Rotation Angle The rotation angle is the ratio of arc length to radius of curvature. For a given angle,

More information

We define angular displacement, θ, and angular velocity, ω. What's a radian?

We define angular displacement, θ, and angular velocity, ω. What's a radian? We define angular displacement, θ, and angular velocity, ω Units: θ = rad ω = rad/s What's a radian? Radian is the ratio between the length of an arc and its radius note: counterclockwise is + clockwise

More information

Final Exam April 30, 2013

Final Exam April 30, 2013 Final Exam Instructions: You have 120 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use a calculator during the exam. Usage of mobile phones and other electronic

More information

PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011

PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011 PHYSICS 1, FALL 011 EXAM SOLUTIONS WEDNESDAY, NOVEMBER, 011 Note: The unit vectors in the +x, +y, and +z directions of a right-handed Cartesian coordinate system are î, ĵ, and ˆk, respectively. In this

More information

Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 2

Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 2 Pleeeeeeeeeeeeeease mark your UFID, exam number, and name correctly. 20 problems 3 problems from exam 1 3 problems from exam 2 6 problems 13.1 14.6 (including 14.5) 8 problems 1.1---9.6 Go through the

More information

Mechanics Topic D (Rotation) - 1 David Apsley

Mechanics Topic D (Rotation) - 1 David Apsley TOPIC D: ROTATION SPRING 2019 1. Angular kinematics 1.1 Angular velocity and angular acceleration 1.2 Constant-angular-acceleration formulae 1.3 Displacement, velocity and acceleration in circular motion

More information

Rotational Dynamics Smart Pulley

Rotational Dynamics Smart Pulley Rotational Dynamics Smart Pulley The motion of the flywheel of a steam engine, an airplane propeller, and any rotating wheel are examples of a very important type of motion called rotational motion. If

More information

Physics 201 Midterm Exam 3

Physics 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 information

PHYSICS I RESOURCE SHEET

PHYSICS I RESOURCE SHEET PHYSICS I RESOURCE SHEET Cautions and Notes Kinematic Equations These are to be used in regions with constant acceleration only You must keep regions with different accelerations separate (for example,

More information

CEE 271: Applied Mechanics II, Dynamics Lecture 24: Ch.17, Sec.1-3

CEE 271: Applied Mechanics II, Dynamics Lecture 24: Ch.17, Sec.1-3 1 / 38 CEE 271: Applied Mechanics II, Dynamics Lecture 24: Ch.17, Sec.1-3 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, Nov. 13, 2012 2 / 38 MOMENT OF

More information

Exam 3 Practice Solutions

Exam 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 information

Rotational Motion and Torque

Rotational Motion and Torque Rotational Motion and Torque Introduction to Angular Quantities Sections 8- to 8-2 Introduction Rotational motion deals with spinning objects, or objects rotating around some point. Rotational motion is

More information

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym

FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym FALL TERM EXAM, PHYS 1211, INTRODUCTORY PHYSICS I Thursday, 11 December 2014, 6 PM to 9 PM, Field House Gym NAME: STUDENT ID: INSTRUCTION 1. This exam booklet has 13 pages. Make sure none are missing 2.

More information

TOPIC D: ROTATION EXAMPLES SPRING 2018

TOPIC D: ROTATION EXAMPLES SPRING 2018 TOPIC D: ROTATION EXAMPLES SPRING 018 Q1. A car accelerates uniformly from rest to 80 km hr 1 in 6 s. The wheels have a radius of 30 cm. What is the angular acceleration of the wheels? Q. The University

More information

AP Physics 1: Rotational Motion & Dynamics: Problem Set

AP Physics 1: Rotational Motion & Dynamics: Problem Set AP Physics 1: Rotational Motion & Dynamics: Problem Set I. Axis of Rotation and Angular Properties 1. How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? 2. How many degrees are

More information

3. Kinetics of Particles

3. 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 information

Big Idea 4: Interactions between systems can result in changes in those systems. Essential Knowledge 4.D.1: Torque, angular velocity, angular

Big Idea 4: Interactions between systems can result in changes in those systems. Essential Knowledge 4.D.1: Torque, angular velocity, angular Unit 7: Rotational Motion (angular kinematics, dynamics, momentum & energy) Name: Big Idea 3: The interactions of an object with other objects can be described by forces. Essential Knowledge 3.F.1: Only

More information

Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum

Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion. Torque and angular momentum Handout 7: Torque, angular momentum, rotational kinetic energy and rolling motion Torque and angular momentum In Figure, in order to turn a rod about a fixed hinge at one end, a force F is applied at a

More information

SECTION A. 8 kn/m. C 3 m 3m

SECTION A. 8 kn/m. C 3 m 3m SECTION Question 1 150 m 40 kn 5 kn 8 kn/m C 3 m 3m D 50 ll dimensions in mm 15 15 Figure Q1(a) Figure Q1(b) The horizontal beam CD shown in Figure Q1(a) has a uniform cross-section as shown in Figure

More information

Rotation. PHYS 101 Previous Exam Problems CHAPTER

Rotation. PHYS 101 Previous Exam Problems CHAPTER PHYS 101 Previous Exam Problems CHAPTER 10 Rotation Rotational kinematics Rotational inertia (moment of inertia) Kinetic energy Torque Newton s 2 nd law Work, power & energy conservation 1. Assume that

More information

Chapter 8 Lecture. Pearson Physics. Rotational Motion and Equilibrium. Prepared by Chris Chiaverina Pearson Education, Inc.

Chapter 8 Lecture. Pearson Physics. Rotational Motion and Equilibrium. Prepared by Chris Chiaverina Pearson Education, Inc. Chapter 8 Lecture Pearson Physics Rotational Motion and Equilibrium Prepared by Chris Chiaverina Chapter Contents Describing Angular Motion Rolling Motion and the Moment of Inertia Torque Static Equilibrium

More information

Rotational Kinematics and Dynamics. UCVTS AIT Physics

Rotational Kinematics and Dynamics. UCVTS AIT Physics Rotational Kinematics and Dynamics UCVTS AIT Physics Angular Position Axis of rotation is the center of the disc Choose a fixed reference line Point P is at a fixed distance r from the origin Angular Position,

More information

Dynamics of Rotation

Dynamics of Rotation Dynamics of Rotation 1 Dynamic of Rotation Angular velocity and acceleration are denoted ω and α respectively and have units of rad/s and rad/s. Relationship between Linear and Angular Motions We can show

More information

PHYSICS 149: Lecture 21

PHYSICS 149: Lecture 21 PHYSICS 149: Lecture 21 Chapter 8: Torque and Angular Momentum 8.2 Torque 8.4 Equilibrium Revisited 8.8 Angular Momentum Lecture 21 Purdue University, Physics 149 1 Midterm Exam 2 Wednesday, April 6, 6:30

More information

Chapter 8: Momentum, Impulse, & Collisions. Newton s second law in terms of momentum:

Chapter 8: Momentum, Impulse, & Collisions. Newton s second law in terms of momentum: linear momentum: Chapter 8: Momentum, Impulse, & Collisions Newton s second law in terms of momentum: impulse: Under what SPECIFIC condition is linear momentum conserved? (The answer does not involve collisions.)

More information

Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!

Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm! Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm! Final Exam Review December 9, 009 3 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector

More information

5. Plane Kinetics of Rigid Bodies

5. Plane Kinetics of Rigid Bodies 5. Plane Kinetics of Rigid Bodies 5.1 Mass moments of inertia 5.2 General equations of motion 5.3 Translation 5.4 Fixed axis rotation 5.5 General plane motion 5.6 Work and energy relations 5.7 Impulse

More information

A) 4.0 m/s B) 5.0 m/s C) 0 m/s D) 3.0 m/s E) 2.0 m/s. Ans: Q2.

A) 4.0 m/s B) 5.0 m/s C) 0 m/s D) 3.0 m/s E) 2.0 m/s. Ans: Q2. Coordinator: Dr. W. Al-Basheer Thursday, July 30, 2015 Page: 1 Q1. A constant force F ( 7.0ˆ i 2.0 ˆj ) N acts on a 2.0 kg block, initially at rest, on a frictionless horizontal surface. If the force causes

More information

NAME. (2) Choose the graph below that represents the velocity vs. time for constant, nonzero acceleration in one dimension.

NAME. (2) Choose the graph below that represents the velocity vs. time for constant, nonzero acceleration in one dimension. (1) The figure shows a lever (which is a uniform bar, length d and mass M), hinged at the bottom and supported steadily by a rope. The rope is attached a distance d/4 from the hinge. The two angles are

More information

1. Replace the given system of forces acting on a body as shown in figure 1 by a single force and couple acting at the point A.

1. Replace the given system of forces acting on a body as shown in figure 1 by a single force and couple acting at the point A. Code No: Z0321 / R07 Set No. 1 I B.Tech - Regular Examinations, June 2009 CLASSICAL MECHANICS ( Common to Mechanical Engineering, Chemical Engineering, Mechatronics, Production Engineering and Automobile

More information

Mechanics. Time (s) Distance (m) Velocity (m/s) Acceleration (m/s 2 ) = + displacement/time.

Mechanics. Time (s) Distance (m) Velocity (m/s) Acceleration (m/s 2 ) = + displacement/time. Mechanics Symbols: Equations: Kinematics The Study of Motion s = distance or displacement v = final speed or velocity u = initial speed or velocity a = average acceleration s u+ v v v u v= also v= a =

More information

5/2/2015 7:42 AM. Chapter 17. Plane Motion of Rigid Bodies: Energy and Momentum Methods. Mohammad Suliman Abuhaiba, Ph.D., PE

5/2/2015 7:42 AM. Chapter 17. Plane Motion of Rigid Bodies: Energy and Momentum Methods. Mohammad Suliman Abuhaiba, Ph.D., PE 5//05 7:4 AM Chapter 7 Plane Motion of Rigid Bodies: Energy and Momentum Methods 5//05 7:4 AM Chapter Outline Principle of Work and Energy for a Rigid Body Work of Forces Acting on a Rigid Body Kinetic

More information

Write your name legibly on the top right hand corner of this paper

Write your name legibly on the top right hand corner of this paper NAME Phys 631 Summer 2007 Quiz 2 Tuesday July 24, 2007 Instructor R. A. Lindgren 9:00 am 12:00 am Write your name legibly on the top right hand corner of this paper No Books or Notes allowed Calculator

More information

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved:

Solution Only gravity is doing work. Since gravity is a conservative force mechanical energy is conserved: 8) roller coaster starts with a speed of 8.0 m/s at a point 45 m above the bottom of a dip (see figure). Neglecting friction, what will be the speed of the roller coaster at the top of the next slope,

More information

Set No - 1 I B. Tech I Semester Regular Examinations Jan./Feb ENGINEERING MECHANICS

Set No - 1 I B. Tech I Semester Regular Examinations Jan./Feb ENGINEERING MECHANICS 3 Set No - 1 I B. Tech I Semester Regular Examinations Jan./Feb. 2015 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Question Paper Consists

More information

JNTU World. Subject Code: R13110/R13 '' '' '' ''' '

JNTU World. Subject Code: R13110/R13 '' '' '' ''' ' Set No - 1 I B. Tech I Semester Supplementary Examinations Sept. - 2014 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question

More information

Rotational & Rigid-Body Mechanics. Lectures 3+4

Rotational & Rigid-Body Mechanics. Lectures 3+4 Rotational & Rigid-Body Mechanics Lectures 3+4 Rotational Motion So far: point objects moving through a trajectory. Next: moving actual dimensional objects and rotating them. 2 Circular Motion - Definitions

More information

CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5

CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5 1 / 42 CEE 271: Applied Mechanics II, Dynamics Lecture 27: Ch.18, Sec.1 5 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, November 27, 2012 2 / 42 KINETIC

More information

Force, Energy & Periodic Motion. Preparation for unit test

Force, Energy & Periodic Motion. Preparation for unit test Force, Energy & Periodic Motion Preparation for unit test Summary of assessment standards (Unit assessment standard only) In the unit test you can expect to be asked at least one question on each sub-skill.

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. Physics 8.01 Fall Problem Set 2: Applications of Newton s Second Law Solutions

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics. Physics 8.01 Fall Problem Set 2: Applications of Newton s Second Law Solutions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall 2012 Problem 1 Problem Set 2: Applications of Newton s Second Law Solutions (a) The static friction force f s can have a magnitude

More information

Centripetal acceleration ac = to2r Kinetic energy of rotation KE, = \lto2. Moment of inertia. / = mr2 Newton's second law for rotational motion t = la

Centripetal acceleration ac = to2r Kinetic energy of rotation KE, = \lto2. Moment of inertia. / = mr2 Newton's second law for rotational motion t = la The Language of Physics Angular displacement The angle that a body rotates through while in rotational motion (p. 241). Angular velocity The change in the angular displacement of a rotating body about

More information

PHYS 1114, Lecture 33, April 10 Contents:

PHYS 1114, Lecture 33, April 10 Contents: PHYS 1114, Lecture 33, April 10 Contents: 1 This class is o cially cancelled, and has been replaced by the common exam Tuesday, April 11, 5:30 PM. A review and Q&A session is scheduled instead during class

More information

Rotation. Rotational Variables

Rotation. Rotational Variables Rotation Rigid Bodies Rotation variables Constant angular acceleration Rotational KE Rotational Inertia Rotational Variables Rotation of a rigid body About a fixed rotation axis. Rigid Body an object that

More information

AP Physics QUIZ Chapters 10

AP Physics QUIZ Chapters 10 Name: 1. Torque is the rotational analogue of (A) Kinetic Energy (B) Linear Momentum (C) Acceleration (D) Force (E) Mass A 5-kilogram sphere is connected to a 10-kilogram sphere by a rigid rod of negligible

More information

TOPIC E: OSCILLATIONS EXAMPLES SPRING Q1. Find general solutions for the following differential equations:

TOPIC E: OSCILLATIONS EXAMPLES SPRING Q1. Find general solutions for the following differential equations: TOPIC E: OSCILLATIONS EXAMPLES SPRING 2019 Mathematics of Oscillating Systems Q1. Find general solutions for the following differential equations: Undamped Free Vibration Q2. A 4 g mass is suspended by

More information

PROBLEM 16.4 SOLUTION

PROBLEM 16.4 SOLUTION PROBLEM 16.4 The motion of the.5-kg rod AB is guided b two small wheels which roll freel in horizontal slots. If a force P of magnitude 8 N is applied at B, determine (a) the acceleration of the rod, (b)

More information

Chapter 10. Rotation of a Rigid Object about a Fixed Axis

Chapter 10. Rotation of a Rigid Object about a Fixed Axis Chapter 10 Rotation of a Rigid Object about a Fixed Axis Angular Position Axis of rotation is the center of the disc Choose a fixed reference line. Point P is at a fixed distance r from the origin. A small

More information

Review questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right.

Review questions. Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. Review questions Before the collision, 70 kg ball is stationary. Afterward, the 30 kg ball is stationary and 70 kg ball is moving to the right. 30 kg 70 kg v (a) Is this collision elastic? (b) Find the

More information

Phys 270 Final Exam. Figure 1: Question 1

Phys 270 Final Exam. Figure 1: Question 1 Phys 270 Final Exam Time limit: 120 minutes Each question worths 10 points. Constants: g = 9.8m/s 2, G = 6.67 10 11 Nm 2 kg 2. 1. (a) Figure 1 shows an object with moment of inertia I and mass m oscillating

More information

Rotational Kinetic Energy

Rotational Kinetic Energy Lecture 17, Chapter 10: Rotational Energy and Angular Momentum 1 Rotational Kinetic Energy Consider a rigid body rotating with an angular velocity ω about an axis. Clearly every point in the rigid body

More information

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

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 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 information

Translational Motion Rotational Motion Equations Sheet

Translational Motion Rotational Motion Equations Sheet PHYSICS 01 Translational Motion Rotational Motion Equations Sheet LINEAR ANGULAR Time t t Displacement x; (x = rθ) θ Velocity v = Δx/Δt; (v = rω) ω = Δθ/Δt Acceleration a = Δv/Δt; (a = rα) α = Δω/Δt (

More information

Quantitative Skills in AP Physics 1

Quantitative Skills in AP Physics 1 This chapter focuses on some of the quantitative skills that are important in your AP Physics 1 course. These are not all of the skills that you will learn, practice, and apply during the year, but these

More information

Physics B Newton s Laws AP Review Packet

Physics B Newton s Laws AP Review Packet Force A force is a push or pull on an object. Forces cause an object to accelerate To speed up To slow down To change direction Unit: Newton (SI system) Newton s First Law The Law of Inertia. A body in

More information

Physics 4A Solutions to Chapter 10 Homework

Physics 4A Solutions to Chapter 10 Homework Physics 4A Solutions to Chapter 0 Homework Chapter 0 Questions: 4, 6, 8 Exercises & Problems 6, 3, 6, 4, 45, 5, 5, 7, 8 Answers to Questions: Q 0-4 (a) positive (b) zero (c) negative (d) negative Q 0-6

More information

Name (please print): UW ID# score last first

Name (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 information

Kinematics, Dynamics, and Vibrations FE Review Session. Dr. David Herrin March 27, 2012

Kinematics, Dynamics, and Vibrations FE Review Session. Dr. David Herrin March 27, 2012 Kinematics, Dynamics, and Vibrations FE Review Session Dr. David Herrin March 7, 0 Example A 0 g ball is released vertically from a height of 0 m. The ball strikes a horizontal surface and bounces back.

More information

Slide 1 / 133. Slide 2 / 133. Slide 3 / How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m?

Slide 1 / 133. Slide 2 / 133. Slide 3 / How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? 1 How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? Slide 1 / 133 2 How many degrees are subtended by a 0.10 m arc of a circle of radius of 0.40 m? Slide 2 / 133 3 A ball rotates

More information

Slide 2 / 133. Slide 1 / 133. Slide 3 / 133. Slide 4 / 133. Slide 5 / 133. Slide 6 / 133

Slide 2 / 133. Slide 1 / 133. Slide 3 / 133. Slide 4 / 133. Slide 5 / 133. Slide 6 / 133 Slide 1 / 133 1 How many radians are subtended by a 0.10 m arc of a circle of radius 0.40 m? Slide 2 / 133 2 How many degrees are subtended by a 0.10 m arc of a circle of radius of 0.40 m? Slide 3 / 133

More information

PHYS 1303 Final Exam Example Questions

PHYS 1303 Final Exam Example Questions PHYS 1303 Final Exam Example Questions 1.Which quantity can be converted from the English system to the metric system by the conversion factor 5280 mi f 12 f in 2.54 cm 1 in 1 m 100 cm 1 3600 h? s a. feet

More information

2015 ENGINEERING MECHANICS

2015 ENGINEERING MECHANICS Set No - 1 I B. Tech I Semester Supplementary Examinations Aug. 2015 ENGINEERING MECHANICS (Common to CE, ME, CSE, PCE, IT, Chem E, Aero E, AME, Min E, PE, Metal E) Time: 3 hours Max. Marks: 70 Question

More information

Physics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium

Physics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium Physics 101: Lecture 15 Torque, F=ma for rotation, and Equilibrium Strike (Day 10) Prelectures, checkpoints, lectures continue with no change. Take-home quizzes this week. See Elaine Schulte s email. HW

More information

Suggested Problems. Chapter 1

Suggested Problems. Chapter 1 Suggested Problems Ch1: 49, 51, 86, 89, 93, 95, 96, 102. Ch2: 9, 18, 20, 44, 51, 74, 75, 93. Ch3: 4, 14, 46, 54, 56, 75, 91, 80, 82, 83. Ch4: 15, 59, 60, 62. Ch5: 14, 52, 54, 65, 67, 83, 87, 88, 91, 93,

More information

1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3

1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3 1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.

More information

Class XI Chapter 7- System of Particles and Rotational Motion Physics

Class XI Chapter 7- System of Particles and Rotational Motion Physics Page 178 Question 7.1: Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie

More information

where G is called the universal gravitational constant.

where G is called the universal gravitational constant. UNIT-I BASICS & STATICS OF PARTICLES 1. What are the different laws of mechanics? First law: A body does not change its state of motion unless acted upon by a force or Every object in a state of uniform

More information

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial

is acting on a body of mass m = 3.0 kg and changes its velocity from an initial PHYS 101 second major Exam Term 102 (Zero Version) Q1. A 15.0-kg block is pulled over a rough, horizontal surface by a constant force of 70.0 N acting at an angle of 20.0 above the horizontal. The block

More information

Chapter 8. Rotational Motion

Chapter 8. Rotational Motion Chapter 8 Rotational Motion The Action of Forces and Torques on Rigid Objects In pure translational motion, all points on an object travel on parallel paths. The most general motion is a combination of

More information

2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity

2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity 2007 Problem Topic Comment 1 Kinematics Position-time equation Kinematics 7 2 Kinematics Velocity-time graph Dynamics 6 3 Kinematics Average velocity Energy 7 4 Kinematics Free fall Collisions 3 5 Dynamics

More information

Slide 1 / 37. Rotational Motion

Slide 1 / 37. Rotational Motion Slide 1 / 37 Rotational Motion Slide 2 / 37 Angular Quantities An angle θ can be given by: where r is the radius and l is the arc length. This gives θ in radians. There are 360 in a circle or 2π radians.

More information

Mechatronics. MANE 4490 Fall 2002 Assignment # 1

Mechatronics. MANE 4490 Fall 2002 Assignment # 1 Mechatronics MANE 4490 Fall 2002 Assignment # 1 1. For each of the physical models shown in Figure 1, derive the mathematical model (equation of motion). All displacements are measured from the static

More information

Chapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity

Chapter 8. Rotational Equilibrium and Rotational Dynamics. 1. Torque. 2. Torque and Equilibrium. 3. Center of Mass and Center of Gravity Chapter 8 Rotational Equilibrium and Rotational Dynamics 1. Torque 2. Torque and Equilibrium 3. Center of Mass and Center of Gravity 4. Torque and angular acceleration 5. Rotational Kinetic energy 6. Angular

More information

Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Chapter 1: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics Translational vs Rotational / / 1/ m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv / / 1/ I

More information

DYNAMICS MOMENT OF INERTIA

DYNAMICS MOMENT OF INERTIA DYNAMICS MOMENT OF INERTIA S TO SELF ASSESSMENT EXERCISE No.1 1. A cylinder has a mass of 1 kg, outer radius of 0.05 m and radius of gyration 0.03 m. It is allowed to roll down an inclined plane until

More information

31 ROTATIONAL KINEMATICS

31 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 information

Two-Dimensional Rotational Kinematics

Two-Dimensional Rotational Kinematics Two-Dimensional Rotational Kinematics Rigid Bodies A rigid body is an extended object in which the distance between any two points in the object is constant in time. Springs or human bodies are non-rigid

More information

Chapters 10 & 11: Rotational Dynamics Thursday March 8 th

Chapters 10 & 11: Rotational Dynamics Thursday March 8 th Chapters 10 & 11: Rotational Dynamics Thursday March 8 th Review of rotational kinematics equations Review and more on rotational inertia Rolling motion as rotation and translation Rotational kinetic energy

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Common Quiz Mistakes / Practice for Final Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A ball is thrown directly upward and experiences

More information

CHAPTER 8: ROTATIONAL OF RIGID BODY PHYSICS. 1. Define Torque

CHAPTER 8: ROTATIONAL OF RIGID BODY PHYSICS. 1. Define Torque 7 1. Define Torque 2. State the conditions for equilibrium of rigid body (Hint: 2 conditions) 3. Define angular displacement 4. Define average angular velocity 5. Define instantaneous angular velocity

More information

if the initial displacement and velocities are zero each. [ ] PART-B

if the initial displacement and velocities are zero each. [ ] PART-B Set No - 1 I. Tech II Semester Regular Examinations ugust - 2014 ENGINEERING MECHNICS (Common to ECE, EEE, EIE, io-tech, E Com.E, gri. E) Time: 3 hours Max. Marks: 70 Question Paper Consists of Part- and

More information

1 2 Models, Theories, and Laws 1.5 Distinguish between models, theories, and laws 2.1 State the origin of significant figures in measurement

1 2 Models, Theories, and Laws 1.5 Distinguish between models, theories, and laws 2.1 State the origin of significant figures in measurement Textbook Correlation Textbook Correlation Physics 1115/2015 Chapter 1 Introduction, Measurement, Estimating 1.1 Describe thoughts of Aristotle vs. Galileo in describing motion 1 1 Nature of Science 1.2

More information

FINAL EXAM -- REVIEW PROBLEMS

FINAL EXAM -- REVIEW PROBLEMS Physics 10 Spring 009 George Williams FINAL EXAM -- REVIEW PROBLEMS A data sheet is provided. Table 10- from your text is assumed, and will be provided on the final exam. 1. A rock is thrown downward from

More information

DYNAMICS ME HOMEWORK PROBLEM SETS

DYNAMICS ME HOMEWORK PROBLEM SETS DYNAMICS ME 34010 HOMEWORK PROBLEM SETS Mahmoud M. Safadi 1, M.B. Rubin 2 1 safadi@technion.ac.il, 2 mbrubin@technion.ac.il Faculty of Mechanical Engineering Technion Israel Institute of Technology Spring

More information

NEWTON 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 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 information