Dynamics for Mechatronics Engineers, Concepts and Examples DR. OSAMA - PowerPoint PPT Presentation

Dynamics for Mechatronics Engineers, Concepts and Examples DR. OSAMA M. AL-HABAHBEH MECHATRONICS ENGINEERING DEPARTMENT THE UNIVERSITY OF JORDAN 2017

Chapter 1: introduction Dynamic is concerned with bodies having acceleration motion. It is divided into two parts: KINEMATICS, which deals only with the geometry of motion, and KINETICS, which deals with the forces that cause the motion.

Chapter 2: Kinematics of Particles A particle has a mass but negligible size and shape this type of approach is used when the dimensions of the object are negligible. Examples of particles are rockets and chicles provided that only motion of mass center is considered and any rotation is neglected.

Rectilinear Motion particle moves along a straight-line path. The kinematics of the particle involves the position, velocity, and the acceleration.

P . s o s Position A single coordinate axis s describes the straight line path of the particle. The origin o is a fixed point. The position is a vector. The magnitude of the vector is the distance between zero and P. the direction of the vector is determined using the sign of s, (+ or -)

Dispala spalacement cement The displace,ent is the change of the particle's position ,when the particle moves from " p " to " p' ". The displacememnt ∆𝒕 = 𝒕 ′ − 𝒕 When moving to the right . ∆𝒕 𝒊𝒃𝒕 𝒃 𝒒𝒑𝒕𝒋𝒖𝒋𝒘𝒇 𝒕𝒋𝒉𝒐 But while moving to the left . ∆𝒕 𝒊𝒃𝒕 𝒃 𝒐𝒇𝒉𝒃𝒖𝒋𝒘𝒇 𝒕𝒋𝒉𝒐 ∆𝒕 𝒋𝒕 𝒃𝒎𝒕𝒑 𝒃 𝒘𝒇𝒅𝒖𝒑𝒔 . ∆𝒕 𝒅𝒑𝒗𝒎𝒆 𝒄𝒇 𝒆𝒋𝒈𝒈𝒇𝒔𝒇𝒐𝒖 𝒈𝒔𝒑𝒏 𝒖𝒊𝒇 𝒆𝒋𝒕𝒖𝒃𝒐𝒅𝒇 𝒅𝒑𝒘𝒇𝒔𝒇𝒆 , 𝒙𝒊𝒋𝒅𝒊 𝒋𝒕 𝒃𝒎𝒙𝒃𝒛𝒕 Positive.

Velocity Vavg= ∆𝑡\∆𝑢 ∆𝑡: 𝑒𝑗𝑡𝑞𝑚𝑏𝑑𝑓𝑛𝑓𝑜𝑢 ∆𝑢: 𝑢𝑗𝑛𝑓 𝑗𝑜𝑢𝑓𝑠𝑤𝑏𝑚 Vins= lim ∆𝑢→0 ∆𝑡\∆𝑢 =ds\dt ∆𝑢: 𝑏𝑚𝑥𝑏𝑧𝑡 𝑞𝑝𝑡𝑗𝑢𝑗𝑤𝑓 The sign of (V)carresponds to the sign of ds(to the right is always +ve,to the left always -ve) V:Is the vector with it's magnitude called speed.

Acceleration Acc(avg)= ∆𝑤\∆𝑢 ∆𝑤: 𝑤𝑓𝑚𝑝𝑑𝑗𝑢𝑧 𝑒𝑗𝑔𝑔𝑓𝑠𝑓𝑜𝑑𝑓 𝑤 ′ − 𝑤 ∆𝑢: 𝑢𝑗𝑛𝑓 𝑗𝑜𝑢𝑓𝑠𝑤𝑏

Acceleration  Instantaneous Acceleration: ∆𝑤 𝑒𝑤 lim = = 𝑒𝑢 = 𝑏𝑑𝑑𝑓𝑚𝑓𝑠𝑏𝑢𝑗𝑝𝑜 𝑏𝑢 𝑢𝑗𝑛𝑓 𝑢. ∆𝑢 ∆𝑢→∞ 𝑒 2 𝑡 𝑒𝑡 𝑒 𝑒𝑡  𝑤 = 𝑒𝑢 → 𝑏 = 𝑒𝑢 = 𝑒 2 𝑢 𝑒𝑢  If: 𝑏 = 0 → 𝐷𝑝𝑜𝑡𝑢𝑏𝑜𝑢 𝑤𝑓𝑚𝑝𝑑𝑗𝑢𝑧   𝑏 > 0 → 𝐵𝑑𝑑𝑓𝑚𝑓𝑠𝑏𝑢𝑗𝑝𝑜  𝑏 < 0 → 𝐸𝑓𝑑𝑓𝑚𝑓𝑠𝑏𝑢𝑗𝑝𝑜

Constant Acceleration  Constant Acceleration: Velocity as function of time, 𝑤 𝑢 can be obtained by integration: 𝑒𝑤 𝑒𝑤 𝑏 = 𝑒𝑢 → 𝑏 𝑑 = 𝑒𝑢 , 𝑏𝑡𝑡𝑣𝑛𝑓 𝑤 = 𝑤 𝑝 𝑏𝑢 𝑢 = 0 𝑤 𝑢 𝑤 = 𝑏 𝑑 𝑢 ] 0 𝑢 → 𝑒𝑤 = 𝑏 𝑑 𝑒𝑢 → 𝑒𝑤 = 𝑏 𝑑 𝑒𝑢 → 𝑤] 𝑤 𝑝 𝑤 𝑝 0 → 𝑤 − 𝑤 𝑝 = 𝑏 𝑑 𝑢 − 0 → 𝑤 − 𝑤 𝑝 = 𝑏 𝑑 𝑢 → 𝒘 = 𝒘 𝒑 + 𝒃 𝒅 𝒖 Position as function of time, 𝑡 𝑢 can be obtained by integration: 𝑒𝑡 𝑤 = 𝑒𝑢 = 𝑤 𝑝 + 𝑏 𝑑 𝑢, 𝑏𝑡𝑡𝑣𝑛𝑓 𝑡 = 𝑡 𝑝 𝑏𝑢 𝑢 = 0 𝑢 2 𝑢 → 𝑡 = 𝑤 𝑝 + 𝑏 𝑑 𝑢 𝑒𝑢 → 𝑡 − 𝑡 𝑝 = 𝑤 𝑝 𝑢 − 0 + 𝑏 𝑑 ( 2 − 0) 0 1 2 𝑏 𝑑 𝑢 2 → 𝑡 = 𝑡 𝑝 + 𝑤 𝑝 𝑢 +

Constant Acceleration  Constant Acceleration:  Velocity as function of position, 𝑤 𝑡 : 𝑏𝑒𝑡 = 𝑤𝑒𝑤 , 𝑏𝑡𝑡𝑣𝑛𝑓 𝑤 𝑢 = 0 = 𝑤 𝑝 𝑡 𝑢 = 𝑝 = 𝑡_𝑝 𝑤 𝑡 2 𝑤 2 − 𝑤 𝑝 2 = 𝑏 𝑑 𝑡 − 𝑡 𝑝 1 → 𝑤𝑒𝑤 = 𝑏 𝑑 𝑒𝑡 → 𝑤 𝑝 𝑡 𝑝 2 = 2𝑏 𝑑 𝑡 − 𝑡 𝑝 → 𝑤 2 = 𝑤 𝑝 𝑏 𝑑 𝑢 → 𝒘 = 𝒘 𝒑 𝟑 + 𝟑𝒃 𝒅 (𝒕 − 𝒕 𝒑 ) → 𝑤 2 − 𝑤 𝑝  Acceleration as function of time, 𝑏 = 𝑔 𝑢 can be obtained by: 𝑒𝑤 𝑤 𝑢 𝑢 𝑏 = 𝑒𝑢 = 𝑤 = 𝑔 𝑢 → 𝑒𝑤 = 𝑔 𝑢 𝑒𝑢 → 𝑤 = 𝑤 𝑝 + 𝑔 𝑢 𝑒𝑢 𝑤 𝑝 0 0 To obtain Position: 𝑡 𝑢 𝑢 𝑒𝑡 v = 𝑒𝑢 = 𝑡 → 𝑒𝑡 = 𝑤 𝑒𝑢 → 𝑡 = 𝑡 𝑝 + 𝑤𝑒𝑢 𝑡 𝑝 0 0

Acceleration Acceleration as function of velocity, a = 𝑔 𝑤 :  𝑒𝑤 𝑤 𝑒𝑤 𝑢 𝑤 𝑒𝑤 𝑏 = 𝑒𝑢 = 𝑤 = 𝑔 𝑤 → 𝑔 𝑤 = 𝑒𝑢 → 𝑢 = 𝑔 𝑤 → 𝑤 𝑢 → 𝑡(𝑢) 𝑤 𝑝 0 𝑤 𝑝 Acceleration as function of displacement, a = 𝑔 𝑡 :  𝑤𝑒𝑤 = 𝑏𝑒𝑡 = 𝑔 𝑡 𝑒𝑡 → 𝑤𝑒𝑤 = 𝑔 𝑡 𝑒𝑡 → 𝑤 2 = 𝑤 𝑝 𝑤 𝑡 2 + 2 𝑔 𝑡 𝑒𝑡 𝑡 𝑤 𝑝 𝑡 𝑝 𝑡 𝑝 𝑡 𝑢 𝑡 𝒆𝒕 𝑒𝑡 𝑒𝑡  𝒘 = 𝒆𝒖 = 𝑕(𝑡) → 𝑕 𝑡 = 𝑒𝑢 → 𝑢 = 𝑔(𝑡) → 𝑢 𝑡 → 𝑡(𝑢) 𝑡 𝑝 0 𝑡 𝑝

Plane Curvilinear Motion  Describes the motion of a particle along a curved path that lies in a single plane.  Where: ∆𝑠 = 𝑊𝑓𝑑𝑢𝑝𝑠 𝐸𝑗𝑡𝑞𝑚𝑏𝑑𝑓𝑛𝑓𝑜𝑢 ∆𝑡 = 𝑇𝑑𝑏𝑚𝑏𝑠 𝐸𝑗𝑡𝑞𝑚𝑏𝑑𝑓𝑛𝑓𝑜𝑢 ∆𝑠 𝑤 𝑏𝑤𝑕 = ∆𝑢 ∆𝑠 𝑒𝑠 𝑤 𝑗𝑜𝑡 = lim ∆𝑢 = 𝑒𝑢 = 𝑠 ∆𝑢 →0 𝑒𝑡 Speed = 𝑤 = 𝑒𝑢 = 𝑡 ∆𝑤 𝑏 𝑏𝑤𝑕 = ∆𝑢 ∆𝑤 𝑒𝑤 𝑏 𝑗𝑜𝑡 = lim ∆𝑢 = 𝑒𝑢 = 𝑤 ∆𝑢 →0

Rectangular coordination 𝒔 = 𝒚𝒋 + 𝒛𝒌 𝒘 = 𝒔 = 𝒚𝒋 + 𝒛𝒌 𝒃 = 𝒘 = 𝒔 = 𝒚𝒋 + 𝒛𝒌 From the figure:- 𝒘 𝟑 = 𝒘𝒚 𝟑 + 𝒘𝒛 𝟑 𝒘𝒚 𝟑 + 𝒘𝒛 𝟑 𝒘 = 𝒘𝒛 tan 𝜾 = 𝒘𝒚 𝒃 𝟑 = 𝒃𝒚 𝟑 + 𝒃𝒛 𝟑 𝒃𝒚 𝟑 + 𝒃𝒛 𝟑 𝒃 = If x= 𝒈 𝟐 𝒖 & 𝒛 = 𝒈 𝟑 𝒖

Projectile motion Rectangular coordination are used for the trajectory analysis of a projectile motion.

𝒘 𝒚 = (𝒘 𝒚 ) 𝟏 𝒃 𝒚 = 𝟏 𝒃 𝒛 = −𝒉 Integration 𝒘 𝒛 = (𝒘 𝒛 ) 𝟏 – 𝒉𝒖 Integration X= 𝒚 𝟏 + (𝒘 𝒚 ) 𝟏 t 𝒘𝒛 𝟑 = (𝒘 𝒛 ) 𝟏 −𝟑𝒉(𝒛 − 𝒛 𝟏 ) 𝟐 𝟑 𝒉𝒖 𝟑 Y= 𝒛 𝟏 + (𝒘 𝒛 ) 𝟏 +

t: tangent , n: normal

Normal and tangential coordinates d s = ρ df 𝑒𝑡 𝑒𝑔 V= 𝑒𝑢 = ρ 𝑒𝑢 𝑊 =V 𝑓 t 𝑊 = ρ β 𝑓 t t ) 𝑒𝑤 𝑒(𝑤𝑓 𝑏 = 𝑒𝑢 = = v ( e t )`+(v)` e t 𝑒𝑢

• Direction of de t is given by e n : 𝑒 e t • de t =e n d β → 𝑒β =e n 𝑒𝑓𝑢 𝑒β 𝑒𝑢 )e n → (𝑓 t)` = β e n • 𝑒𝑢 =( • V= ρ β → 𝑏 =v ( e t )`+(v)` e t • v ( e t )`=v ( β e n) 𝑤 • β = ρ 𝑤 2 𝑤 ρ 𝑓 n )= • V( ρ e n 𝑤 2 • 𝑏 = ρ e n + 𝑤 𝑓 t

Page 9 : 2   2 v v           2 a p p v   n   p p      a v s t   2 2 a an at

Circular motion   v = v 2 2 v .      a r  v n r       a v r t

Ex : problem 2/3 page 27 3    2 Velocity of a particle is given by : v 2 4 t 5 t Evaluate : position  ( ) ? s   v ?   a ? at  t 3 s when    0  t 0 s s 3 m

5   2 2 4 t 16 t 3         2 s 2 vdt ( 2 4 t 5 t ) dt 2 t c 2 8 5      2 2 s ( t ) 2 t 2 t 2 t c    s ( 0 ) c s 3 m 0 5     2 2 s ( 3 ) 2 ( 3 ) 2 ( 3 ) 2 ( 3 ) 3      6 18 31 . 2 3 22 . 2 m

3     ( 3 ) 2 4 ( 3 ) 5 ( 3 ) 2 15 . 98 / v m s dv 3 15 1 1        2 2 a 4 ( 5 )( ) t ( t ) 4 2 2 dt 15 1    2 a ( 3 ) ( 3 ) 4 8 . 99 m / s 2

Example 2-15. Page 29# A particle is fired vertically with 𝒘 𝒑 = 𝟑𝟏 m/s. What is the max altitude h reached by the projectile and  the time after firing for it to return to ground. Neglect air drag and take g as constant at 9.81 m/ 𝒕 𝟑 . Sol: 𝟐 𝟑 𝒃𝒖 𝟑 𝒕 = 𝒕 𝒑 + 𝒘 𝒑 𝒖 + 𝟑 𝒉𝒖 𝟑 = 𝟏 + 𝟑𝟏𝟏 ∗ 𝒛 + 𝟐 𝟐 𝟑 𝟘. 𝟗𝟐 𝒖 𝟑 𝒊 = 𝒊 𝒑 + 𝒘 𝒑 𝒖 + 𝒃𝒖 𝒖𝒑𝒒, 𝒘 = 𝟏 , 𝒘 = 𝒘 𝒑 + 𝒃𝒖 = 𝟏 = 𝟑𝟏𝟏 + 𝟘. 𝟗𝟐 𝒖 𝒖 = 𝟑𝟏. 𝟒𝟗 𝒕 𝟑 𝟘. 𝟗𝟐 ∗ 𝟑𝟏. 𝟒𝟗 𝟑 = 𝟑𝟏𝟒𝟘 𝒏 𝟐 𝒊 = 𝟑𝟏𝟏 ∗ 𝟑𝟏. 𝟒𝟗 − 𝒖 𝒖𝒑𝒖𝒃𝒎 = 𝟑𝒖 = 𝟓𝟏. 𝟗 𝒕