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DEC 1013 ENGINEERING SCIENCEs

DEC 1013 ENGINEERING SCIENCEs. 2. KINEMATICS AND KINETICS. NAZARIN B. NORDIN nazarin@icam.edu.my. What you will learn:. Distance and displacement Speed and velocity Acceleration Distance vs. time graph: interpret slope as speed

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DEC 1013 ENGINEERING SCIENCEs

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  1. DEC 1013 ENGINEERING SCIENCEs 2. KINEMATICS AND KINETICS NAZARIN B. NORDIN nazarin@icam.edu.my

  2. What you will learn: • Distance and displacement • Speed and velocity • Acceleration • Distance vs. time graph: interpret slope as speed • Velocity vs. time graph: interpret slope as acceleration; area under graph as distance moved • Rotational kinematics • Angular velocity and angular acceleration

  3. Distance & Displacement • Distance is the actual length measured of a particular path taken. • Displacement is the length and direction of a straight line drawn from the start to finish.

  4. Cont.. • Distance and displacement are two quantities which may seem to mean the same thing, yet they have distinctly different meanings and definitions. • Distance (d) is a scalar quantity which refers to "how far an object has moved" during its motion. • Displacement(d) is a vectorquantity which refers to "how far an object is from its original position"; it is the object's change in position.

  5. Displacement

  6. Speed & Velocity

  7. Illustration of distinction between velocity and speed

  8. Acceleration

  9. Acceleration is the rate of change of velocity. Example 2-1: Average acceleration. A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration?

  10. Motion along x-axis

  11. Interpreting Graph • Distance vs time graph

  12. Interpreting graph • Velocity vs time graph

  13. Unit Conversion

  14. Rotational kinematics

  15. Rotational kinematics

  16. Angular motion • Angular velocity

  17. Angular motion • Angular acceleration

  18. Angular motion • ROTATIONAL MOTION UNDER CONSTANT ANGULAR ACCELERATION

  19. Problem solving

  20. Problem solving

  21. Summary of Chapter 2 (p.1) • Kinematics is the description of how objects move with respect to a defined reference frame. • Displacement is the change in position of an object. • Average speed is the distance traveled divided by the time it took; average velocity is the displacement divided by the time. • Instantaneous velocity is the average velocity in the limit as the time becomes infinitesimally short.

  22. Summary of Chapter 2 (p.2) • Average acceleration is the change in velocity divided by the time. • Instantaneous acceleration is the average acceleration in the limit as the time interval becomes infinitesimally small. • The equations of motion for constant acceleration are given in the text; there are four, each one of which requires a different set of quantities.

  23. Topic 2.6/2.7 Uniform Circular Motion

  24. Ө = angular position (angle) ω = angular velocity (rate at which an angle changes) α = angular acceleration Easy ways of relating angular kinematics to rotational kinematics: s = r Ө v = r ω a = r α Also, v=

  25. Linear Kinematics: x = ½at2 + v0t + x0 v = at + v0 v2 = 2a∆x + v02 Look similar? They are. Rotational Kinematics: θ = ½αt2 + ω0t + θ 0 ω = αt + ω0 ω2 = 2α∆ θ + ω02

  26. From the results of the lab, we can find that Fc=

  27. Whenever an object is moving in a circle, the forces acting on it cause the object to move in a circle. As such, the net force is the centripetal force. ΣF = Fc

  28. Example A 1500kg car drives over the top of a hill at a constant 20m/s. If the radius of curvature for the hill is 60m, • Sketch a free-body diagram of this situation. • Solve for the centripetal force on the car. • What is the magnitude of the normal force on the car?

  29. 2) Fc= = 3) ΣF = Fc = Fg– FN FN= mg – FC FN = 14,700N - 10,000N FN= 4,700N We know the centripetal force is downward because the center of the circle is below the car. The only forces acting on the car are gravity and the normal force. The mg is greater than the normal force, and the arrows show it.

  30. Torque τ = r x F Or for those who may not know cross-products, τ = rF sin(Ө) τ (tau) stands for torque. It is equal to the radius from the fulcrum times the force times the sine of the angle between the radius and the torque. Units for torque are Nm Clockwise is negative and counterclockwise is positive (just like angles)

  31. Torque Example You apply a force of 20N to a 10cm long wrench at a30 degree angle. Solve for the torque. τ = rF sin (Ө) τ = (0.1m)(20N) sin (60) τ = 1.73 N*m

  32. Torque and Force ∑F = Fnet For a non-accelerating system, ∑F = Fnet= 0 At the same time, ∑τ= τ net For a non-rotating system, ∑τ = τ net = 0

  33. Example A meter stick has a fulcrum placed at the 50cm mark. An object that weighs 20N is hung from it 20cm from the center. What force would you have to apply on the other side of the meter stick 30cm from the fulcrum to keep the meter stick from rotating? ∑τ = τ person - τweight= 0 F(0.3m)-(0.2m)(20N)=0 F=13.3N

  34. Rotational Dynamics The equation describing most situations in this section is: ∑ τ = I α The sum of torques is equal to the moment of inertia times angular acceleration.

  35. Moment of Inertia Moment of inertia of an object tells you how difficult it is to make the object rotate. In general, I= mr2 That is true for each individual mass, but an entire system of masses can be more complicated.

  36. Some commonly used moments of inertia are: I = mr2 for a point mass I = ½ mr2 for a disc I = 1/3 mr2 for a rod rotating around its end

  37. You exert a 60N force at the end of a 3m long rod. The rod has a mass of 20kg. Assuming the rod is rotating about its end (the circle in the picture), • Calculate the moment of inertia of the rod. • Calculate the net torque on the rod. • What is the angular acceleration of the rod?

  38. a) This is a rod rotating about its end, so I= 1/3 mr2 . = 1/3(20kg)(3m)2 = 60 kg*m2 b) τ = rF sin (Ө) = (3m)(60N)sin(90) =180Nm c) ∑ τ = I α 180Nm = 60(kgm2)α α = 30 rad/s2

  39. Angular Momentum Angular momentum is another quantity that is conserved in physics. Always: L = Iω For each individual point, L = mrv. Depending on the shape of the entire rotating object, there may be a coefficient.

  40. Example A ball is tied to a pole with a string that is 2m long and is orbiting the pole. The initial speed of the ball is 6m/s. What is the distance of the ball from the pole when its speed is 20m/s? L1 = L2 I1 ω1 = I2ω2 or mr1v1 = mr2v2 mr12(v1/r1) = mr22(v2/r2) r1v1 = r2v2 (2m)(6m/s) = r2 (20m/s) r2 = 0.6m/s

  41. Linear Kinematics: x = ½at2 + v0t + x0 v = at + v0 v2 = 2a∆x + v02 ∑ F = ma p = mv Ek = ½ mv2 Rotational Kinematics: θ = ½αt2 + ω0t + θ 0 ω = αt + ω0 ω2 = 2α∆ θ + ω02 ∑ τ = I α L = I ω Ek = ½ I ω 2

  42. THANK YOU

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