Mechanics of Machines Dr. Mohammad Kilani

1. Mechanics of MachinesDr. Mohammad Kilani Class 4 Velocity Analysis

2. DERIVATIVE OF A ROTATING VECTOR

3. Derivative of a Rotating Unit Vector • A unit vector in the θdirection, uθ, is a vector of unity magnitude and an angle θwith the x- axis. It is written as: • If uθ rotates, it angle θ changes with time, the time derivative of uθ is found by applying the standard differentiation rules on the expression of uθ above uθ sinθj θ cosθi

4. Derivative of a Rotating Unit Vector • Noting that • the time derivative of the vector uθ is uθ sinθj θ cosθi

5. Derivative of a Rotating Unit Vector • The derivative of a unit vector whose angle with the x-axis is θ, (θ changes in time), is a vector whose angle with the x-axis is θ+π/2 and whose magnitude is dθ/dt. • If we define a vector ωas a vector in the k direction of magnitude ω = dθ/dt, then duθ/dt uθ θ + π/2 θ

6. Derivative of a Rotating Vector • A vector rθ = r uθ, is a general vector of magnitude r pointing in the θ direction. The time derivative of rθ is found by applying the normal differentiation rules on the expression for rθ duθ/dt uθ θ + π/2 θ

7. Derivative of a Rotating Vector • Given a vector rθ = ruθwhich rotates relative to the reference coordinates, the time derivative of rθ has two components; a component in the direction of uθ and a component normal to uθ in the direction of uθ+π/2 • The magnitude of the component of drθ /dt in the direction of uθ is equal to dr/dt; that is the time derivative of the length of rθ. • The magnitude of the component of drθ /dt in the direction of direction of uθ+π/2is equal to ωr (dr/dt) uθ ω x rθ θ + π/2 rθ θ

8. VELOCITY ANALYSIS OF FOUR BAR MECHANISMS

9. Derivative of the Loop Closure Equation for a Four Bar Kinematic Chain • The loop closure equation of a 4-bar kinematic chain is written as • When all the links in the chain are of constant lengths, the equation above reduces to

10. Derivative of the Loop Closure Equation for a Four Bar Mechanism • For a four bar mechanism with link 1 fixed we have dθ1/dt =ω1 = 0. • The vector equation above contains two scalar equations and can be solved for two unknowns. Knowledge of dθ2/dtallows the calculation of dθ3/dtand dθ4/dt .

11. Derivative of the Loop Closure Equation for a Four Bar Mechanism • Eliminate dθ3/dtby carrying out a dot product with uθ3 on both sides of the equation • Alternatively, eliminate dθ4/dtby carrying out a dot product with uθ4 on both sides of the equation

12. Angular Velocity Ratio and Mechanical Advantage • The angular velocity ratio mV is defined as the output angular velocity divided by the input angular velocity. For a four bar mechanism with link 2 as the input and link 4 as the output this is expressed as • The efficiency of a four bar linkage is defined as the output power over the input power, • Assuming 100% efficiency, which is normally approached by four bar mechanisms, we have

13. Angular Velocity Ratio and Mechanical Advantage • The mechanical advantage is defined as the ratio between the output force to the input force

14. VELOCITY ANALYSIS OF SLIDER-CRANK MECHANISMS

15. Velocity Analysis of a Slider-Crank Mechanism • Given r2, r3, r4, θ1, θ2 , ω2 • Position analysis: • Find r3, θ3 • Velocity analysis: • Find dr1/dt, dθ3/dt • To eliminate ω3 dot product both sides by uθ3 r3 • To eliminate dr1/dtdot product both sides by u(θ1+π/2) r4 rp r2 r1

16. Velocity Analysis of an Inverted Slider-Crank Mechanism • Given r1 , r2, θ1, θ2 , ω2 • Position analysis: Find r3, θ3 • Velocity analysis: Find dr3/dt, dθ3/dt • To eliminate ω3 and find dr1/dtdot product both sides by uθ3 • To eliminate dr1/dt and find ω3 dot product both sides by u(θ3+π/2) θ3 r2 r3 r1 θ2 θ1

17. Velocity Analysis of an Inverted Slider-Crank Mechanism • Given r1 , r2, r4 , θ1, θ2 , ω2 • Position analysis: Find r3, θ3 , θ4 • Velocity analysis: Find dr3/dt, dθ3/dt , dθ4/dt r3 r4 r2 r1

18. Velocity Analysis of an Inverted Slider-Crank Mechanism r3 r4 r2 r1

19. Example r5 s b a r4 r2 r1 r5 b a s r4 r2 r1

20. Example r5 s b a r4 r2 r1 r5 b a s r4 r2 r1

21. HW#36-17, 6-18 (b+c), 6-21 (b), 6-44.