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

Transcription

1 Kinematics, Dynamics, and Vibrations FE Review Session Dr. David Herrin March 7, 0

2 Example A 0 g ball is released vertically from a height of 0 m. The ball strikes a horizontal surface and bounces back. The coefficient of restitution between the surface and the ball is The height that the ball will reach after bouncing is most nearly KE PE vr e v 0 mv 0 mgh mgh 0 0 v mvr mgh h 5.6 m r 0 r

3 Example A 5 g mass is to be placed on a 50 cm diameter horizontal table that is rotating at 50 rpm. The mass must not slide away from its position. The coefficient of friction between the mass and the table is 0.. What is the maximum distance that the mass can be placed from the axis of rotation? R 5 g ω

4 R 5 g F ma µ mω R µ g R.07m ω r ω

5 Example 3 A truck of 4000 kg mass is traveling on a horizontal road at a speed of 95 km/h. At an instant of time its brakes are applied, locking the wheels. The dynamic coefficient of friction between the wheels and the road is 0.4. The stopping distance of the truck is most nearly (A) 55 m (B) 70 m (C) 85 m (D) 00 m KE mv Δx KE µ Δx mv k + W µ k 84.5 m

6 Example 4 A translating and rotating ring of mass kg, angular speed of 500 rpm, and translational speed of m/s is placed on a horizontal surface. The coefficient of friction between the ring and the surface is For an outside radius of 3 cm, the time at which skidding stops and rolling begins is most nearly ω v 0

7 µ mg a F x ma 0 F y µ ma mg M µ R v v o + Iα mr at ω ω αt 0 α Skidding Stops when v Rω

8 Example 5 A stationary uniform rod of length m is struck at its tip by a 3 kg rigid ball moving horizontally with velocity of 8 m/s as shown. The mass of the rod is 7 kg, and the coefficient of restitution between the rod and the ball is The velocity of the ball after impact is most nearly m v

9 ω r Before ω r After v v mvl mv' L e v' r v' v v r + 3 m L ω' r v '.88 m / s

10 Example 6 The D Alembert force is (A) Force of gravity in France. (B) Force due to inertia. (C) Resisting force due to static friction (D) Atomic force discovered by D Alembert.

11 Example 7 A 5 kg pendulum is swung on a 7 m long massless cord from rest at 5 from center. The time required for the pendulum to return to rest is most nearly ω n g L f π g L f Hz T f

12 Example 8 A kg mass is suspended by a linear undamped spring with a spring constant 3. k/m. The mass is given an initial velocity of 0 m/s from the equilibrium position. Example 8. Calculate the natural frequency of the mass. k ωn 40 rad / m s

13 Example 8. How long does it take for the mass to complete one complete cycle. f ω n π T f s

14 Example 8.3 What is the maximum deflection of the spring from the equilibrium position? x v ω 0 ( t) x cos( ω t) sin( ω t) 0 n + n n

15 Example 9 A 5 kg motor turns at 800 rpm, and it is mounted on a pad having a stiffness of 500 k/m. Due to an unbalanced condition, a periodic force of 85 is applied in a vertical direction, once per revolution. eglecting damping and horizontal movement, the amplitude of vibration is k ωn m ω 88.5 δ f pst F k o 66 rad / s rad / s.7e 4m β ω f ωn D βδ pst ω f + C ωn 0.04mm 0.4 m

16 Example 0 A uniform disk of 0 kg mass and 0.5 m diameter rolls without slipping on a flat horizontal surface, as shown. When its horizontal velocity is 50 km/h, the total kinetic energy of the disk is most nearly KE + mv o I o ω

17 Example A homogeneous disk of 5 cm radius and 0 kg mass rotates on an axle AB of length 0.5 m and rotates about a fixed point A. The disk is constrained to roll on a horizontal floor. Given an angular velocity of 30 rad/s in the x direction and -3 rad/s in the y direction, the kinetic energy of the disk is KE I ω x + I yωy + x I z ω z

18 Example The natural frequency of the system is designed to be ω n 0 rad/s. The spring constant k is half of k, and the mass is kg. The mass associated with the other components may be assumed to be negligible. For the given natural frequency, the spring constant k is most nearly m x + k x eq + k eq k 0 k ω n k eq m

19 Example 3 A two-bar linkage rotates about the pivot point O as shown. The length of members AB and OA are.0 m and.5 m, respectively. The angular velocity and acceleration of member OA is ω OA 0.8 rad/s CCW and α OA 0 rad/s. The angular velocity of member AB is ω AB. rad/s CW, and the acceleration of member AB is α AB 3 rad/s CCW. When the bars are in the position shown, the magnitude of the acceleration of point B is most nearly a a A B ωoa OA aa + ωab ω v AB ( ω r ) B / A A/ O ( ω r ) AB + a B / A + α B / A OA r + α AO AB r B / A +

20 Gears Quick Review Spur Gears" Helical Gears" Bevel Gears"

21 Types of Gears Worm Gear" Rack and Pinion"

22 Concepts Review

23 Terminology

24 Fundamental Law of Gearing Fixed Shafts θ Gear Ratio r r θ rθ r θ r r θ θ ω ω

25 Gear Compatibility USCS p r Units???" SI m r

26 Idlers 3 θ θ ω ω ω ω ω ω Gear Ratio Fixed Shafts ω ω θ θ r r θ

27 Double Reductions Gear Ratio Fixed Shafts 3 n n n n n n n

28 Double Reductions Fixed Shafts Gear Ratio n n output input product of product of driving teeth driven teeth n 4 n 3 4

29 Example - Speed Reducer n 50 n3 40 rev min

30 Example The compound gear train shown is attached to a motor that drives gear A at ω in clockwise as viewed from below. What is the expression for the angular velocity of gear H in terms of the number of teeth on the gears? What is the direction of rotation of gear H as viewed from below.

31 Gear Forces tanφ W W r t W t W r T T W t d m H H W t dω mω SI Units:, mm, kw

32 Classification of 4-Bar Linkages Grashof Mechanism One link can perform a full rotation relative to another link Grashof Criterion L + L < L + L max min a b

33 Crank-Rocker Mechanism L L min

34 Drag Link Mechanism L L 0 min

35 Double-Rocker Mechanism L L min

36 Change-Point Mechanism Also called Crossover-Position Mechanism L + L L + L max min a b

37 on-grashof Mechanism o link can rotate through 360 Double Rocker Mechanism of the nd Kind Triple-Rocker Mechanism L + L > L + L max min a b

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39 Crank-Slider Mechanism

40 Quick Return Mechanism

41 Geneva Mechanism