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Principles Learn The Method Principles Basics should be automatic Memorize and Practice! The angular velocity of the disk (r = 1m) in rad/s is (A) 6 k J 5.129 i (B) (C) (D) (E) -6 k 6i - 3k 3k wDisk = vo / r = -3/1 = -3 rad/s The angular accel. of the disk (r = 2 m) in rad/s2 is (A) 5 k J i (B) - 5 k (C) - 10 k (D) 10k (E) 2.5 k aDisk = ao / r = 5/2 = 2.5 rad/s2 5.129 Arm AB with Length L moves the slotted rod BD in x-direction only. B J L AB i A (t) D vD(t) X() =L*cos How do we determine the velocity vD of point D as a function of angle (t)? w angular _ vel Arm AB with Length L moves the slotted rod BD in x-direction only. B J L AB i A (t) D vD(t) X() =L*cos How do we determine the velocity vD of point D as a function of angle (t)? w angular _ vel X-dot() =-L*sin*_dot Chapter 14 Energy Methods dW F dr Scalar _ Pr oduct Only Force components in direction of motion do WORK The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law. 2. If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is _____ A) 3.8 m/s. B) 6.9 m/s. C) 14.7 m/s. D) 21 m/s. The work done is mgh ½*m*v2 = mgh or V = sqrt(2gh) = sqrt(14.7) Power E dE Units of power: P t dt J/sec = N- m/sec = Watts 1 hp = 746 W The potential energy V is defined as: V - W - F * dr Conservative Forces T1 + V1 = T2 + V2 Potential Energy Potential energy is energy which results from position or configuration. An object may have the capacity for doing work as a result of its position in a gravitational field. It may have elastic potential energy as a result of a stretched spring or other elastic deformation. Potential Energy elastic potential energy as a result of a stretched spring or other elastic deformation. Potential Energy Potential Energy y Ex. Problem 14.26 Procedure 1. Frame, start and end points 2. Constraint equations? 3. Apply the Energy Principle (only ext. Forces!) 2 𝑥𝑐 + 𝑥𝐴 = 𝐿 T1 = 0, start from rest 0 + F*xc = 0.5*m*(vA)2 Chapter 16 Rigid Body Kinematics 16.1 16.3 Rot. about Fixed Axis Memorize! Vector Product is NOT commutative! Cross Product i a b ax j ay k az bx by bz a b a y bz a z by a z bx a x bz a x by a y bx Derivative of a Rotating Vector • vector r is rotating around the origin, maintaining a fixed distance • At any instant, it has an angular velocity of ω dr ωr dt r ω r ω Page 317: dr v ωr dt an = w x ( w x r) at = a x r General Motion = Translation + Rotation Vector sum vA = vB + vA/B 16.4 Motion Analysis Approach 1. Geometry: Definitions Constants Variables Make a sketch 2a. Analysis (16.4) Derivatives (velocity 2b. Rel. Motion (16.5) and acceleration) 3. Equations of Motion 4. Solve the Set of Equations. Use Computer Tools. Example Bar BC rotates at constant wBC. Find the angular Veloc. of arm BC. Step 1: Define the Geometry Example B J vA(t) A AC O A Bar BC rotates at constant wBC. Find the ang. Veloc. of arm BC. Step 1: Define the Geometry (t) O i (t) (t) C Geometry: Compute all lengths and angles as f((t)) All angles and distance AC(t) are time-variant B J vA(t) A Velocities: w = -dot is given. O A AC (t) O Vector Analysis: wOA rA i (t) (t) C vCOLL wBC rAC Rigid Body Acceleration Chapter 16.7 Stresses and Flow Patterns in a Steam Turbine FEA Visualization (U of Stuttgart) General Procedure A a (t) B J AB BD i (t) = 45 deg D vD(t)= const 1. Compute all velocities and angular velocities. 2. Start with centripetal acceleration: It is ALWAYS oriented inward towards the center. General procedure A a (t) B J AB BD i (t) = 45 deg D vD(t)= const 1. Compute all velocities and angular velocities. 2. Start with centripetal acceleration: It is ALWAYS oriented inward towards the center. 3. The angular accel is NORMAL to the Centripetal acceleration. General Procedure A a (t) B J AB BD i (t) = 45 deg D vD(t)= const 1. Compute all velocities and angular velocities. 2. Start with centripetal acceleration: It is ALWAYS oriented inward towards the center. 3. The angular accel is NORMAL to the Centripetal acceleration. The direction of the angular acceleration is found from the mathematical analysis. wExample 1. Find all v and w (Ch. 16.5) AB = -11.55k i i HIBBELER 16-125 wBC = -5k 2. View from A: aB = aABXrB – wAB2*rB 3. View from D: aB = aC + aBCXrB/C – wBC2*rB/C Centripetal Terms: We know magnitudes and HIB 16-125 directions We now can solve two simultaneous vector equations for wAB and wBC – wBC2*rB/C – wAB2*rB aD aABXrB – wAB2*rB = aC + aBCXrB/C – wBC2*rB/C HIB 16-125 aABXrB – wAB2*rB = aC + aBCXrB/C – wBC2*rB/C 16.8 Relative Motion aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE fig_05_11 16.8 Relative Motion aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE 16.8 Relative Motion From Ch. 12.8 radial aP = 𝒓*ur + ωx(ωxr) + (α x r) + 2 (ω x 𝒓) tangential End of Review Chapters 14 and 16