Rigid Body Dynamics (MENG233) Instructor: Dr. Mostafa Ranjbar

Textbook Engineering Mechanics, Dynamics R. C. Hibbeler, 10 th Edition

Grading system 25% Attendance, Quizzes, Homework Assignments 30% Midterm Exam 45% Final Exam So, You Must be Active!

Homework Homework problems are pre-assigned in syllabus every week. All homework problems assigned during a given week are due in class on the following week unless stated otherwise. Late Homework will not be accepted Attendance will be checked during each lecture.

Course coverage Kinematics of a Particle. (Ch. 12) Kinetics of a Particle: Force and Acceleration. (Ch. 13) Kinetics of a Particle: Work and Energy. (Ch. 14) Kinetics of a Particle: Impulse and Momentum. (Ch. 15) Planar Kinematics of a Rigid Body. (Ch. 16) Planar Kinematics of a Rigid Body: Force and Acceleration. (Ch. 17) Planar Kinematics of a Rigid Body: Work and Energy. (Ch. 18) Planar Kinematics of a Rigid Body: Impulse and Momentum. (Ch. 19)

Areas of mechanics (section12.1) 1)Statics -Concerned with body at rest. 2)Dynamics -Concerned with body in motion 1.Kinematics: is a study the geometric aspect of the motion. 2.Kinetics: Analysis of forces that causing the motion

Review of Vectors and Scalars A Vector quantity has both magnitude and direction. A Scalar quantity has magnitude only.

Scalars (e.g)Scalars (e.g) –distance –speed –mass –temperature –pure numbers –time –pressure –area, volume –charge –energy Vectors (e.g.)Vectors (e.g.) –displacement –velocity –acceleration –force –weight (force) –momentum

Vectors Can be represented by an arrow (called the “vector”). Length of a vector represents its magnitude. Symbols for vectors: – (e.g. force) F, or F (bold type), or F2 F

Position Position : Location of a particle at any given instant with respect to the origin r : Displacement ( Vector ) s : Distance ( Scalar )

Distance & Displacement Displacement : defined as the change in position. r : Displacement ( 3 km ) s : Distance ( 8 km ) Total length For straight-line Distance = Displacement s = r  s  r Vector is direction oriented  r positive (left )  r negative (right)

DistanceDistance –Total length of path travelled –Must be greater than (or equal to) magnitude of displacement –Only equal if path is straight –Symbol d DisplacementDisplacement –Refers to the change in particle’s position vector –Direct distance –Shortest distance between two points –Distance between Start and End points –“as the crow flies” –Can be describe with only one direction –Symbol S –S = X final - X initial

Average Speed and Average Velocity

Velocity & Speed Velocity : Displacement per unit time Average velocity : V =  r  t Speed : Distance per unit time Average speed :   sp  s T  t ( Always positive scalar ) Speed refers to the magnitude of velocity Average velocity :  avg =  s /  t

Velocity (con.) Instantaneous velocity : For straight-line  r =  s

Average Speed and Average Velocity

# Problem A particle moves along a straight line such that its position is defined by s = (t 3 – 3 t ) m. Determine the velocity of the particle when t = 4 s. At t = 4 s, the velocity = 3 (4) 2 – 6(4) = 24 m/s

Acceleration Acceleration : The rate of change in velocity {(m/s)/s} Average acceleration : Instantaneous acceleration : If v ‘ > v “ Acceleration “ If v ‘ < v “ Deceleration”

Problem A particle moves along a straight line such that its position is defined by s = (t 3 – 3 t ) m. Determine the acceleration of the particle when t = 4 s. At t = 4 a(4) = 6(4) - 6 = 18 m/s 2

Problem A particle moves along a straight line such that its position is defined by s = (t 3 – 12 t t -20 ) cm. Describe the motion of P during the time interval [0,9] t02469 s v a Total time = 9 seconds Total distance = ( )= 145 meter Displacement = form -20 to 61 = 81 meter Average Velocity = 81/9= 9 m/s to the right Speed = 9 m/s Average speed = 145/9 = 16.1 m/s Average acceleration = 27/9= 3 m/s 2 to the right

Relation involving s, v, and a No time t Acceleration Velocity Position s

Problem A car starts from rest and moves along a straight line with an acceleration of a = ( 3 s -1/3 ) m/s 2. where s is in meters. Determine the car’s acceleration when t = 4 s. (Rest t = 0, v = 0)

For constant acceleration a = a c

Velocity as a Function of Time

Position as a Function of Time

Velocity as a Function of Position

Summary Time dependent accelerationConstant acceleration This applies to a freely falling object:

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