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.

• s

.

P

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: lim =

∆𝑤 ∆𝑢 ∆𝑢→∞

=

𝑒𝑤 𝑒𝑢 = 𝑏𝑑𝑑𝑓𝑚𝑓𝑠𝑏𝑢𝑗𝑝𝑜 𝑏𝑢 𝑢𝑗𝑛𝑓 𝑢.

 𝑤 =

𝑒𝑡 𝑒𝑢 → 𝑏 = 𝑒 𝑒𝑢 𝑒𝑡 𝑒𝑢

=

𝑒2𝑡 𝑒2𝑢

 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: 𝑏 =

𝑒𝑤 𝑒𝑢 → 𝑏𝑑 = 𝑒𝑤 𝑒𝑢 , 𝑏𝑡𝑡𝑣𝑛𝑓 𝑤 = 𝑤𝑝𝑏𝑢 𝑢 = 0

→ 𝑒𝑤 = 𝑏𝑑𝑒𝑢 → 𝑒𝑤 = 𝑏𝑑𝑒𝑢 → 𝑤]𝑤𝑝

𝑤 = 𝑏𝑑𝑢 ]0 𝑢 𝑢 𝑤 𝑤𝑝

→ 𝑤 − 𝑤𝑝 = 𝑏𝑑 𝑢 − 0 → 𝑤 − 𝑤𝑝 = 𝑏𝑑𝑢 → 𝒘 = 𝒘𝒑 + 𝒃𝒅𝒖 Position as function of time, 𝑡 𝑢 can be obtained by integration: 𝑤 =

𝑒𝑡 𝑒𝑢 = 𝑤𝑝 + 𝑏𝑑𝑢, 𝑏𝑡𝑡𝑣𝑛𝑓 𝑡 = 𝑡𝑝𝑏𝑢 𝑢 = 0

→ 𝑡 = 𝑤𝑝 + 𝑏𝑑𝑢 𝑒𝑢 → 𝑡 − 𝑡𝑝 = 𝑤𝑝 𝑢 − 0 + 𝑏𝑑(

𝑢2 2 − 0) 𝑢

→ 𝑡 = 𝑡𝑝 + 𝑤𝑝𝑢 +

1 2 𝑏𝑑𝑢2

Constant Acceleration

 Constant Acceleration:  Velocity as function of position, 𝑤 𝑡 : 𝑏𝑒𝑡 = 𝑤𝑒𝑤 , 𝑏𝑡𝑡𝑣𝑛𝑓 𝑤 𝑢 = 0 = 𝑤𝑝 𝑡 𝑢 = 𝑝 = 𝑡_𝑝 → 𝑤𝑒𝑤 = 𝑏𝑑𝑒𝑡 →

1 2 𝑤2 − 𝑤𝑝 2 = 𝑏𝑑 𝑡 − 𝑡𝑝 𝑡 𝑡𝑝 𝑤 𝑤𝑝

→ 𝑤2 − 𝑤𝑝

2 = 2𝑏𝑑 𝑡 − 𝑡𝑝 → 𝑤2 = 𝑤𝑝𝑏𝑑𝑢 → 𝒘 = 𝒘𝒑 𝟑 + 𝟑𝒃𝒅(𝒕 − 𝒕𝒑)

 Acceleration as function of time, 𝑏 = 𝑔 𝑢 can be obtained by: 𝑏 =

𝑒𝑤 𝑒𝑢 = 𝑤 = 𝑔 𝑢 → 𝑒𝑤 = 𝑔 𝑢 𝑒𝑢 → 𝑤 = 𝑤𝑝 + 𝑔 𝑢 𝑒𝑢 𝑢 𝑢 𝑤 𝑤𝑝

To obtain Position: v =

𝑒𝑡 𝑒𝑢 = 𝑡 → 𝑒𝑡 = 𝑤 𝑒𝑢 → 𝑡 = 𝑡𝑝 + 𝑤𝑒𝑢 𝑢 𝑢 𝑡 𝑡𝑝

Acceleration

 Acceleration as function of velocity, a = 𝑔 𝑤 : 𝑏 =

𝑒𝑤 𝑒𝑢 = 𝑤 = 𝑔 𝑤 → 𝑒𝑤 𝑔 𝑤 = 𝑒𝑢 → 𝑢 = 𝑒𝑤 𝑔 𝑤 → 𝑤 𝑢 → 𝑡(𝑢) 𝑤 𝑤𝑝 𝑢 𝑤 𝑤𝑝

 Acceleration as function of displacement, a = 𝑔 𝑡 : 𝑤𝑒𝑤 = 𝑏𝑒𝑡 = 𝑔 𝑡 𝑒𝑡 → 𝑤𝑒𝑤 = 𝑔 𝑡 𝑒𝑡 → 𝑤2 = 𝑤𝑝

2 + 2 𝑔 𝑡 𝑒𝑡 𝑡 𝑡𝑝 𝑡 𝑡𝑝 𝑤 𝑤𝑝

 𝒘 =

𝒆𝒕 𝒆𝒖 = 𝑕(𝑡) → 𝑒𝑡 𝑕 𝑡 = 𝑒𝑢 → 𝑢 = 𝑒𝑡 𝑔(𝑡) → 𝑢 𝑡 → 𝑡(𝑢) 𝑡 𝑡𝑝 𝑢 𝑡 𝑡𝑝

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 𝑤𝑏𝑤𝑕 =

∆𝑠 ∆𝑢

𝑤𝑗𝑜𝑡 = lim

∆𝑢 →0 ∆𝑠 ∆𝑢 = 𝑒𝑠 𝑒𝑢 = 𝑠

Speed = 𝑤 =

𝑒𝑡 𝑒𝑢 = 𝑡

𝑏𝑏𝑤𝑕 =

∆𝑤 ∆𝑢

𝑏𝑗𝑜𝑡 = lim

∆𝑢 →0 ∆𝑤 ∆𝑢 = 𝑒𝑤 𝑒𝑢 = 𝑤

Rectangular coordination

𝒔 = 𝒚𝒋 + 𝒛𝒌 𝒘 = 𝒔 = 𝒚𝒋 + 𝒛𝒌 𝒃 = 𝒘 = 𝒔 = 𝒚𝒋 + 𝒛𝒌 From the figure:- 𝒘𝟑 = 𝒘𝒚𝟑 + 𝒘𝒛𝟑 𝒘 = 𝒘𝒚𝟑 + 𝒘𝒛𝟑 tan 𝜾=

𝒘𝒛 𝒘𝒚

𝒃𝟑 = 𝒃𝒚𝟑 + 𝒃𝒛𝟑 𝒃 = 𝒃𝒚𝟑 + 𝒃𝒛𝟑 If x= 𝒈𝟐 𝒖 & 𝒛 = 𝒈𝟑 𝒖

Projectile motion

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

𝒃𝒚 = 𝟏 𝒃𝒛 = −𝒉 Integration 𝒘𝒚 = (𝒘𝒚 )𝟏 𝒘𝒛 = (𝒘𝒛 )𝟏 – 𝒉𝒖 Integration X=𝒚𝟏 + (𝒘𝒚 )𝟏 t 𝒘𝒛𝟑 = (𝒘𝒛 )𝟏 −𝟑𝒉(𝒛 − 𝒛𝟏 ) Y=𝒛𝟏 + (𝒘𝒛 )𝟏 +

𝟐 𝟑 𝒉𝒖𝟑

t: tangent , n: normal

Normal and tangential coordinates ds = ρ df V=

𝑒𝑡 𝑒𝑢= ρ 𝑒𝑔 𝑒𝑢

𝑊=V 𝑓 t 𝑊= ρβ 𝑓 t 𝑏 =

𝑒𝑤 𝑒𝑢 = 𝑒(𝑤𝑓 t ) 𝑒𝑢

= v (et)`+(v)` et

• Direction of det is given by en :
• det=en dβ →

𝑒et 𝑒β =en

• 𝑒𝑓𝑢

𝑒𝑢 =( 𝑒β 𝑒𝑢)en → (𝑓

t)` =β en

• V=ρβ → 𝑏

=v (et)`+(v)` et

• v (et)`=v (β en)
• β =

𝑤 ρ

• V(

𝑤 ρ 𝑓n)= 𝑤2 ρ en

• 𝑏

=

𝑤2 ρ en +𝑤 𝑓

t

Page 9 :

2 2 2 2 2

at an a s v a v p p v p p v a

t n

                      

  



     r v v r r

a v a

t n

     . v = v

2 2

Circular motion

Ex : problem 2/3 page 27

Velocity of a particle is given by : Evaluate : position

2 3

5 4 2 t t v   

     ? ? ? ) ( a v s

m s s t when s t at 3 3

0 

   

m s m s c s c t t t t s c t t t dt t t vdt s 2 . 22 3 2 . 31 18 6 3 ) 3 ( 2 ) 3 ( 2 ) 3 ( 2 ) 3 ( 3 ) ( 2 2 2 ) ( 8 16 2 4 2 ) 5 4 2 ( 2

2 5 2 2 5 2 2 5 2 2 3

                         



s m a t t dt dv a s m v / 99 . 8 4 ) 3 ( 2 15 ) 3 ( 4 ) ( 2 15 ) 2 3 )( 5 ( 4 / 98 . 15 ) 3 ( 5 ) 3 ( 4 2 ) 3 (

2 1 2 1 2 1 2 3

             

Example 2-15. Page 29#

 A particle is fired vertically with 𝒘𝒑 = 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/𝒕𝟑. Sol: 𝒕 = 𝒕𝒑 + 𝒘𝒑𝒖 +

𝟐 𝟑 𝒃𝒖𝟑

𝒊 = 𝒊𝒑 + 𝒘𝒑𝒖 +

𝟐 𝟑 𝒉𝒖𝟑 = 𝟏 + 𝟑𝟏𝟏 ∗ 𝒛 + 𝟐 𝟑 𝟘. 𝟗𝟐 𝒖𝟑

𝒃𝒖 𝒖𝒑𝒒, 𝒘 = 𝟏 , 𝒘 = 𝒘𝒑 + 𝒃𝒖 = 𝟏 = 𝟑𝟏𝟏 + 𝟘. 𝟗𝟐 𝒖 𝒖 = 𝟑𝟏. 𝟒𝟗 𝒕 𝒊 = 𝟑𝟏𝟏 ∗ 𝟑𝟏. 𝟒𝟗 −

𝟐 𝟑 𝟘. 𝟗𝟐 ∗ 𝟑𝟏. 𝟒𝟗 𝟑 = 𝟑𝟏𝟒𝟘 𝒏

𝒖𝒖𝒑𝒖𝒃𝒎 = 𝟑𝒖 = 𝟓𝟏. 𝟗 𝒕

Example 2-25. Page 30#

 Distance s, when he overtake the blue car? 𝒘𝒏𝒑 = 𝟏 𝒘𝒅 = 𝟐𝟑𝟏 𝒍𝒏/𝒊 𝒃𝒏 = 𝟕𝒏/𝒕𝟑 𝒕𝒅 = 𝟕 𝒕 𝒘𝒏𝟐 = 𝟐𝟔𝟏 𝒍𝒏/𝒊 Sol: 𝒕𝒅 = 𝒕𝒅𝒑 + 𝒘𝒅𝒑𝒖 +

𝟐 𝟑 𝒃𝒅𝒖𝟑 = 𝟕𝟕. 𝟖 + 𝟐𝟑𝟏 ∗ 𝒖 + 𝟏

M C

𝑊𝑛2 = 𝑊

𝑛0 2+2am( Sm1 – Sm1 )

(150∗1000m

60∗60𝑡 )=0+2(6 m/𝑡2)(Sm1-0)=(12 m/𝑡2) Sm1

Sm1=(1736/12)=144.7m Sm2=Sm1+Vm1 t2=144.7+41.7(t-6.9 s) V=V˳+at → t=𝑊−𝑊

𝑏 tm1=𝑊 −𝑊 𝑏

tm1= 41.7−0

6

= 6.94 s Sm1=41.7t-144.7m=Sc=(33.3)t+66.7m t(8.4)-211.4=0 → t=22.5 sec Sc=(33.3)(25.2)+66.7=905.9m Sm2=(41.7)(25.2)-144.7=906.1m

• Relative motion (translating axes)
• Fixed reference axes → absolute coordinates
• Morning reference axes → relative motion
• Translation of axes will be discussed for plane motion:

RELATIVE-MOTION OF TWO PARTICLES (CONTINUED)

 

CONSTRAINED MOTION OF CONNECTED PARTICLES

 The motion of particles are interrelated by the constraints of

interconnecting members

 Example:  x and y are position coordinates

CONSTRAINED MOTION OF CONNECTED PARTICLES (CONTINUED)

     It is a one degree of freedom system because one variable (x or y) is

enough to specify the positions of all parts of the system

1- polar coordinates (r-θ) 2-cylindrical coordinates (r- θ- z) 3-spherical coordinates ( r - θ - φ ) Example problem ( plane curvilinear motion ) What is min horizontal velocity ( u ) of a rock to clear obstruction B?

14

• Solution :

Ax=0 Ay=-g Vx=(vx)˳ u=constant Vy=(vy)˳ -gt = 0 –gt = -gt X= X˳ + (vx) ˳ t Y= y˳ + (vy) ˳ t- ⅟2 gt² Vy²=(vy)²˳-2g(y-y˳) = 0 +2(9.81m \ s²)+(10m + 0 ) = 196.2m²\s² vy =14 m\s T=

−𝑤𝑧 −𝑕 = 14𝑛\s 9.81𝑛\s² = 1.43sec

X = 0 + u (1.43sec) = 40 m u=27.97 m\s U≈28 m\s 14

Example: A race car C travels around the horizontal circular track that has a radius

• f 300 ft , if the car increases its speed at a constant rate of 7 ft/s²,

starting from rest , Determine : *the time needed for it to reach an acceleration of 8 ft/s² . *what is its speed at this instant .

Solution : a = ( at² + an² )½ at = 7 ft/s² an = v²/ƿ ** hint :- v = v˳+(at)c .t ** an = ( 7t )²/ 300 = 0.163t² ft/s² Time needed to reach 8 ft/s² :- a = ( at² + an² )½ 8 = ( 7² +(0.163t²) ² ) ½ 0.163t² = (8²+ 7²) ½ The speed at this instant :- V= 7t = 7*( 4.87 ) = 34.1 ft/s ²

t= 4.87 sec

EX:

• The motion of a particle is defined by the position vector r=A(cost+tsint)i+A(sint+tcost)j

t in sec Determine the value of t for which 𝑠 and 𝑏 are : a)Perpendicular b)parallel

Soln: 𝑠=A(cost+tsint)i+A(sint-tcost)j

𝑤 =

𝑒𝑠 𝑒𝑢=A(-sint+sint+tcost)i+A(cost-cost+tsint) =A(tcost)i+A(tsint)j

𝑏 =

𝑒𝑠 𝑒𝑢=A(cost-tsint)i+A(sint+tcost)j

(a) 𝑠& 𝑏 are perpendicular ⇒ 𝑠 . 𝑏 =0 A[(cost+tsint)i+(sint-tcost)j] . A[(cost-tsint)i+(sint+tcos)]=0 =𝐵2 [(cost+tsint)(cost-tsint)+(sint-tcost)(sint+tcost)]=0 (𝑑𝑝𝑡2t-𝑢2si𝑜2t)+(si𝑜2t- 𝑢2 𝑑𝑝𝑡2t)=0 1-𝑢2=0 ⇒t=1 s (b) 𝑠 & 𝑏 are parallel ⇒ 𝑠 X 𝑏 =0 A[(cost+tsint)i+(sint-tcost)j] X A[(cost-tsint)i+(sint-tcost)j]=0 𝐵2[(cost+tsint)(sint+tcost)-(sint-tcost)(cost-tsint)] k=0

Polar coordinates (r – θ)/Radial and transverse coordinates.

r = r er = (r er)

Radial & Transverse (r - θ) coordinates cont. velocity should be expressed in terms of er and eθ.

𝑏 =

𝑒𝑤 𝑒𝑢 = 𝑒 𝑒𝑢 (𝑠°𝑓

r+r𝜄°𝑓 𝜄)=𝑤 =𝑠 𝑓 𝜄+𝑠

𝑒𝑓𝑠 𝑒𝑢 +𝑠 𝜄 𝑓

𝜄+ 𝜄 𝑓 𝜄 + 𝜄

𝑒𝑓 𝑒𝑢

=𝑠 er+𝑠 𝜄 e+2𝑠 𝜄e+r𝜄 e+r𝜄 𝜖𝑓

𝜖𝑢

=𝑠 e+r𝜄 𝑒𝑓

𝑒𝑢+[2𝑠 𝜄 +r𝜄 ]𝑓

Example:

The park ride shown consists of a chair that is rotating in a horizontal circular path of radius r such that the arm OB has an angular velocity Ӫ and angular acceleration Ӫ ,determine the radial and transverse components of velocity and acceleration of the passenger .

Solution : r = r ṙ = 0 = 0

Space curvilinear motion , z) Ө Cylindrical coordinates (r ,

Spherical coordinates(R ,Ө,Ф):

• R=R*eR.
• V=R'* eR+R*Ө'*cos(Ф)* eӨ + R*Ф'* eФ .
• a=aR* eR +aӨ* eӨ +aФ* eӨ
• aR=R''-R*Ф'²-R*Ө'²*cos²(Ф).
• aӨ=cos(Ф)/R *d/dt(R^2*Ө'(-2*R* Ө'*Ф'*sin(Ф).
• aФ= 1/R * d/dt(R^2*Ф'(+R* Ө'²*sin(Ф)*cos(Ф).

Kinetic of particles:

• -Kinetic is the study of the relations between

unbalanced forces and the changes in motion they produce.

• -Kinetic problem can be solved using three

approaches:

• Newton's 2nd low (F=m*a).
• Work & Energy.
• Impulse & Momentum.

Newton’s 2nd law:

• F=m*a
• F :is the magnitude of the resultant force acting on

the particle.

• m: is a quantitative measure of inertia (mass) of the

particle.

• a:is the magnitude of the resulting acceleration of

the particle.

• There are two types of motion; constrained such as

atrain or unconstrained such as airplane.

• Motion in space has a three degrees of

freedom(DOF).

• Motion in a plane has a 2 DOF.

• Motion in a linear path has 1 DOF.
• DOF : is the #coordinates needed to describe

the position in vector newton F=m*a.

• Construction of a free body diagram (FBD) is

an important step in the solution of mechanics

• problems. The +ve sign directions should be

indicated cleary.

• For rectilinear motion in the x-direction:
• Σ Fx = m*a.
• ΣFy=0.
• ΣFz=0.

The three components of equation motion are

• :



when

EXAMPLE

• 75 KG man stands in an elevator.during the 1st 3 sec of motion from rest,tention (T)in the cable is 8300

N and the upward velocity V of the elevator at the end of the 3 sec total mass of elevator man and scale is 750k

Solution

 Since F=ma m is the coust during the 3 sec F=T = 8300 N =constant a is also

constant ….. For elevator scale and man the FBD gives :

  8300-7360=750 ay ……. ay=1257 m/ 𝑡2  Equ of motion for man  R-763=75*1257 …… R=830 N  V(3)=3*a ……….=3*1257 …………=377 m/s

Curvlinear motion (kinetics)

 In rectangular coordinates :

For normal and rectangular coordinates: For polar coordinates :

Example:



A rod OA is rotating in the horizontal plane such that ϴ= 𝑢3 rad.



At the same time , the rollar B is sliding out word along oA so that



𝑠 = 100𝑢2 mm , if in both cases t is in seconds , determine the velocity and acceleration of the roller when t= 1s



Solution :



𝑠 = 100𝑢2|@t = 1s = 100 mm .



ϴ = 𝑢3|@t= 1s = 1 rad = 57.3°.



𝑠 =200t |@t=1s =200 mm\s.



ϴ = 3𝑢2|@t= 1s = 3 rad\s.



r**= 200|@t=1s = 200mm\s .



ϴ = 6t |@t=1s = 6 rad\s .



𝑤 = 𝑠 𝑓 𝑠 + r ϴ 𝑓ϴ= 200𝑓 𝑠 + 100 𝑓ϴ (3) = 200𝑓 𝑠 + 300 𝑓ϴ mm\s



𝑤 = 2002 + 3002 = 361 mm\s

 Ș=tan−1 300

200 = 56.3°

 Ș + 56.3 = 114°  Acceleration:  𝑏

= 𝑠 − rϴ2 𝑓 𝑠 + rϴ + 2𝑠 ϴ 𝑓ϴ

 = (200 – 100 (32)) 𝑓

𝑠 + (100 (6) + 2 (200) 3) 𝑓ϴ.



𝑏 = 7002 + 18002 =1930 mm\𝑡2.

 Ø = tan−1 1800

700 = 68.7°

 (180° – Ø) + 56.3°= 169° 

Example:

Passengers of plane A observe plane B passing under plane A in a horizontal flight. Plane B apears to the passengers in plane A as moving away from A at 𝟕𝟏°. Find 𝐖𝐂 ( Velocity of plane B). Solution: Moving net axes attached to A 𝑾𝑪 = 𝑾𝑩 + 𝑾𝑪/𝑩 Direction √ Magnitude ? 1) Graphical method: 2) Trigonometric method: 𝑾𝑪 sin 60° = 𝑾𝑩 sin 75° 𝑾𝑪 = 𝟗𝟏𝟏 sin 𝟕𝟏° sin 𝟖𝟔° = 𝟖𝟐𝟖 𝒍𝒏/𝒊 3) Vector algebra method: 𝑾𝑩 = 𝟗𝟏𝟏𝒋 𝒍𝒏/𝒊 𝑾𝑪 = 𝑾𝑪 cos 𝟓𝟔 𝒋 + 𝑾𝑪 sin 𝟓𝟔 𝒌 𝑾𝑪/𝑩 = 𝑾𝑪

𝑩

cos 𝟕𝟏 −𝒋 + 𝑾𝑪

𝑩

sin 𝟕𝟏 𝒌 𝑾𝑪 = 𝑾𝑩 + 𝑾𝑪/𝑩 𝑾𝑪 cos 𝟓𝟔 𝒋 = 𝟗𝟏𝟏 − 𝑾𝑪

𝑩

𝒅𝒑𝒕 𝟕𝟏 𝒋 𝑾𝑪 sin 𝟓𝟔 𝒌 = 𝑾𝑪

𝑩

sin 𝟕𝟏 𝒌 𝑾𝑪/𝑩 = 𝟔𝟗𝟕 𝒍𝒏/𝒊 𝑾𝑪 = 𝟖𝟐𝟖 𝒍𝒏/𝒊

D.O.Maaitah

27

Example:

𝒃𝑪 = 𝒃𝑩 + 𝒃𝑪/𝑩 To be continued.. Plane A is flying straight, while plane B flies in circular path as shown in the figure. 𝑾𝑩 = 𝟖𝟏𝟏 𝒍𝒏/𝒊 𝑾𝑪 = 𝟕𝟏𝟏 𝒍𝒏/𝒊 Find 𝑾𝑪/𝑩 , 𝒃𝑪/𝑩. Solution: 1. Velocity (+↑) 𝑾𝑪 = 𝑾𝑪/𝑩 + 𝑾𝑩 𝟕𝟏𝟏 = 𝟖𝟏𝟏 + 𝑾𝑪/𝑩 𝑾𝑪/𝑩 = −𝟐𝟏𝟏 𝒍𝒏 𝒊 = 𝟐𝟏𝟏 𝐥𝐧 𝐢 ↓ 2. Acceleration (𝒃𝑪)𝒐= 𝑾𝑪

𝟑

𝝇 = (𝟕𝟏𝟏 𝒍𝒏/𝒊)𝟑 𝟓𝟏𝟏 𝒍𝒏 = 𝟘𝟏𝟏𝒍𝒏/𝒊𝟑

D.O.Maaitah

28

(continued)

900i – 100j =50j + aB/A aB/A = { 900i – 150j } 𝑙𝑛/𝑖2 |aB/A|= 9002 + 1502 = 912 𝑙𝑛/𝑖2 𝜄 = tan−1(

150 900) = 9.46°

(aB)n= 900 𝑙𝑛/𝑖2

𝜄

150𝑙𝑛/𝑖2

example

Q: Car A and car B travelling at VA=18 m/s VB=12 m/s Car A speed is decreasing by 2 𝑛/𝑡2 Car B speed is increasing by 3 𝑛/𝑡2 Find VB/A and aB/A ?

60

y x

+

Example solution :

VB = VA + VB/A

• 12 j =(-18cos 60𝑗 − 18 sin 60𝑘) + VB/A

VB/A ={9i + 3.588j} m/s |VB/A|= 92 + 3.5882 = 9.69 m/s tan 𝜄 =

𝑊𝐶/𝐵 𝑧 𝑊𝐶/𝐵 𝑦 = 3.588 9

So 𝜄 = 21.7°

𝑧

aB=3 𝒏/𝒕𝟑 aA=2 𝒏/𝒕𝟑

VB VA

𝑦 60°

• Acceleration :

𝑏

= 𝑤 𝑓𝑢 +

𝑊2 𝑞

𝑓𝑜 𝑏𝐵 = 𝑤 𝑓𝑢 +

𝑊2 𝑞

= 2𝑓𝑢 +

6𝑛 𝑡 2

28.28𝑛 𝑓𝑜

= 2𝑓𝑢 + 1.273𝑓𝑜 m/𝑡2 𝑏 = 22 + 1.2732 = 2.37

𝑛 𝑡2

𝛸 = tan−1

2 1.273 = 57.5°

57.5° − 45 = 12.5° 𝑏 = 2.37 𝑛/𝑡2

𝑏𝐵 12.5°

A disk rotates about z with w= 𝛴 =

𝜌 3 rad/sec, Arm OB is elevated

at 𝛸

=

2𝜌 3

at t=0 : ϴ=0 and Φ=0 P slides along the rod according to R(in mm)= 50 + 200𝑢2 Find: 𝑊 , 𝑏 , 𝑏𝑞 at t=

1 2 𝑡 ?

Solution: 𝑊

𝑆 = 𝑆 = 400𝑢|𝑢=1

2 = 200 mm/s = 0.2 m/s

𝑊

𝛴= 𝑆𝛴 𝑑𝑝𝑡Φ

𝑆 𝑢 = 0.5 = 50 + 200 0.5 2 = 100 𝑛𝑛 = 0.1𝑛 Φ(t=0.5)= (0.5)(

2π 3 ) = π 3

V𝚺 = 0.1

π 3 cos 𝜌 3 = 0.052 m/s

V𝛸 = 𝑆𝛸 = 0.1

π 3

= 0.209 m/s V = 𝑊

𝑆 2 + 𝑊 𝜮 2 + 𝑊 𝛸 2

= 0.2 2 + 0.052 2 + 0.209 2 = 0.294 m/s aR = 𝑆 − 𝑆𝚾 𝟑 − Rϴ cos 𝚾 𝑆 = 400 mm = 0.4m

𝑏𝑆 = 0.4 − 0.1 2𝜌 3

2

− 0.1 𝜌 3

2

cos (𝜌 3)2 = 0.4 − 0.44 − 0.0274 = −0.0674 𝑛\𝑡2 𝑏∅ = cos ∅ 𝑆 𝑆2𝜄 + 2𝑆𝑆 ∅ − 2𝑆𝜄 ∅ sin ∅ = cos 𝜌 3 0.1 0.1 2 0 + 2 0.1 0.2 𝜌 3 − 2 0.1 𝜌 3 2𝜌 3 sin (𝜌 3) = 8.66 0.042 − 0.38 = −0.0163 𝑛\𝑡2 𝑏∅ = 1 𝑆 𝑆2𝜄 + 2𝑆𝑆 ∅ + 𝑆𝜄 2 sin ∅ cos ∅ = 1 0.1 0.1 2 0 + 2 0.1 0.2 2𝜌 3 + 0.1 𝜌 3

2

sin 𝜌 3 cos 𝜌 3 = 0.84 + 0.0475 = 0.887𝑛\𝑡2 𝑏𝑞 = 𝑏𝑆2 + 𝑏𝜄2 + 𝑏∅2 = −0.0674 2 + −0.0163 2 + 0.887 2 = 0.9𝑛\𝑡2 Any questions on the exam material? 34

𝑥𝐵 = 10 𝐿𝑂 𝑥𝑕 = 20 𝐿𝑂 𝑤°𝐵 = 2 𝑛\𝑡 𝜈𝑙 = 0.2 Find 𝑤𝐵 when A has moved 4 meters. Take 𝑕 = 10 𝑛\𝑡2 Solution: FBD for A: 𝐺

𝑦 = 𝑛𝐵𝑏𝐵 = 𝑈 − 𝜈𝑙𝑂

𝐺

𝑧 = 0 = 𝑂 − 𝑛𝐵𝑕 → 𝑂 = 𝑛𝐵𝑕

𝑥𝐵 = 𝑛𝐵𝑕 = 𝑂 = 10𝐿𝑂 𝑏𝐵 = 𝑈 − 𝜈𝑙𝑂 𝑛𝐵 = 𝑈 − (0.2)(10𝐿𝑂) 𝑛𝐵 = 𝑈 − 2000 𝑛𝐵 = 𝑏𝐵 FBD for B: 𝐺

𝑧 = 𝑛𝐵𝑏𝐵 = −2𝑈 + 𝑛𝐶𝑕 = −2𝑈 + 𝑥𝐶 = −2𝑈

+ 20𝐿𝑂 = −2𝑈 + 20000 𝑏𝐶 = −2𝑈 + 20000 𝑛𝐶 𝑥𝐵 = 𝑛𝐵𝑕 → 𝑛𝐵 = 𝑥𝐵 𝑕 = 10000𝑂 10 𝑛\𝑡2 = 1000𝐿𝑕 𝑥𝐶 = 𝑛𝐶𝑕 → 𝑛𝐶 = 𝑥𝐶 𝑕 = 20000𝑂 10 𝑛\𝑡2 = 2000𝐿𝑕 33

𝑏𝐵 = 𝑈 − 2000 1000 = 𝑈 1000 − 2 𝑏𝐶 = −2𝑈 + 20,000 2000 = −2𝑈 2000 + 20,000 2000 = −𝑈 1000 + 10 Let L=

𝑈 1000 , 𝑏𝐵 = 𝑀 − 2 , 𝑏𝐶 = −𝑀 + 10

𝑇𝐵 = 2 ∗ 𝑇𝐶 = 𝐷𝑝𝑜𝑡𝑢 → 𝑏𝐵 + 2𝑏𝐶 = 0 → 𝑏𝐵 = −2𝑏𝐶 𝑀 − 2 = −2 −𝑀 + 10 = 2𝑀 + 20 = 𝑀 − 2 → 𝑀 2 − 1 = −2 + 20 = 𝑀 = 18 , T = 1000L = 1000 ∗ 18 = 18000 𝑂 𝑏𝐵 = 𝑀 − 2 = 18 − 2 = 16 𝑛 /𝑇2 𝑊

𝐵 2 = 𝑊 𝐵. 2 + 2 ∗ 𝑏𝐵 𝑌𝐵 − 𝑌𝐵. = (2 𝑛/𝑇)2+2(16𝑛/𝑇2) *(4m)

= 4 𝑛2/𝑡2 +128 𝑛2/𝑡2 = 132 𝑛2/𝑡2 → 𝑊

𝐵 = 11.5𝑛/𝑇

35

EX:

Find min stopping distance S from a speed of (70 Km/h) with constant deceleration if the box is not to slip ..

SOL: 𝜈𝑡 = 0.3

𝑊

. = 70 Km/h a= constant 𝐺 = 𝑛𝑏

𝐺 = 𝑛𝑏 = −𝐺𝐺𝑠 = −𝜈𝑡 𝑊

𝐺 2 − 𝑊 . 2 = 2𝑏𝑡 = 2 −𝜈𝑡𝑕 𝑇 → 𝑇 = −𝑊

,2

−2𝜈𝑡𝑕

S =

(70Km/h)2 2∗0.3∗9.81 𝑛/𝑡2 = 64.2𝑛

36

37

𝐹𝑌

𝑛𝐵 = 40𝐿𝑕 𝜈𝑒 = 0.4 𝑏𝐵= ? (Neglect masses of pulleys) Solution: 𝐺𝐶𝐸 𝑝𝑔 𝐵 𝐺

𝑧 = 0

𝑂 − 𝑛𝑕 + 2𝑈𝑡𝑗𝑜 30 = 0 𝑂 = 𝑛𝑕 − 2𝑈𝑡𝑗𝑜 30 = 40 Kg 9.81𝑛/𝑡2 − 2 100N 0.5 𝑂 = 292.4 𝑂 𝐺

𝑦 = 𝑛𝑏𝑦

2𝑈𝑑𝑝𝑡 30 − 𝑂𝜈𝑒 = 𝑛𝑏 → 𝑏𝐵 = 2 100 cos 30 − (292.4)(0.4) 40 = 1.406 𝑛/𝑡2

Work and energy Work: 38 dr is differential displacement ,Work done by force F during dr is 𝑒𝑉 = F. dr = 𝐺. 𝑒𝑡 cos (𝛽) 𝛽 ,is angle between F and dr. 𝐺𝑢 = 𝐺𝑑𝑝𝑡 𝛽 , is force component along the displacement. 𝑒𝑡 cos 𝛽 , is displacement component along the force. 𝐺

𝑜 = F𝑡𝑗𝑜 𝛽 does no work.

→ 𝑒𝑤 = 𝐺𝑢𝑒𝑡 If 𝐺𝑢 is in the direction of the displacement → Work is +ve If 𝐺𝑢 is in the opposite direction of the displacement → Work is -ve Work is a scalar quantity 𝑉 = F. dr = 𝐺

𝑦𝑒𝑦 + 𝐺 𝑧𝑒𝑧 + 𝐺 𝑨𝑒𝑨 = 𝐺𝑢 𝑒𝑡

• Mass System :
• Spring

Tension Compression

𝑽 = − 𝑮. 𝒆𝒚

𝒚𝟑 𝒚𝟐

= work done on the body by the spring. because the direction of force is opposite to the displacement] negative Work done is [

− 𝑮. 𝒆𝒚

𝒚𝟑 𝒚𝟐

= − 𝒍𝒚. 𝒆𝒚

𝒚𝟑 𝒚𝟐

= −

𝟐 𝟑 k [ (𝒚𝟑)𝟑 −(𝒚𝟐)𝟑 ] = Area Under The Curve

if the directions of force and displacement are the positive Work done on the body by the spring would be same (e.g: relaxing tension or compression). F=k =k x is a linear static relationship which holds when the mass of the spring itself is relatively small and not accounted for.

Work done on a particle:

Work done by during a finite movement of the particle from 1 to 2 is :

F

U1-2 = =



2 1

.dr F



2 1 S S

Ftds

Final point Initial point

Substitute = m

F

a

U1-2

= = atds



2 1

m



2 1

.dr a m

Recall that and

v dt ds 

a dt dv 

dt ds a av dt dv v  

vdv = ads = atds

U1-2

= atds = = ½ m(v22-v12)



2 1 m



1 2 v v mvdv

Kinetic energy (T):

T= it’s the total work required to bring the particle from rest to velocity v . it’s a scalar and always +ve !(why)? V1-2=T2-T1= Work –Energy equ. For a particle total work done on the particle= corresponding change in kinetic energy of the particle. could be +ve , -ve , or zero. with enrgy approach , no need to calculate acceleration.

2 ^ 2 1 mv

T 

power (p):

• it’s the time rate of doing work . it’s a scalar quantity

p = →p= F.V Efficiency (em): Mechanical Efficiency (em) is the ratio of the work done by a machine to the work done on the machine em=

• because of friction loss

dt dr F dt dr F dt dv . .  

1   em Pinput Poutput

ee is electrical efficiency →electrical energy loss et is thermal efficiency→ thermal loss e is total efficiency e= em + ee + et

Example:-

• A Satellite of mass (M) is orbiting the earth at A where h1= 500

Km, v1=30000 Km/h ,and h2= 1200 Km , Find v2 .

Solution:-

• V1-2 = -∫F.dr = -m.g.R² ∫dr/r²

= -mgR²(-1/r)│ =m.g. R²(1/r2²-1/r1²) V1-2=ΔT = 2mg R²(1/r2²-1/r1²)= 0.5.m(V2²-V1²) V2=V1 + 2g R²(1/r2-1/r1)- (30,000/3.6)+2(9.81).[(6,371)*10 ] ² = 58.73*10⁶ m²/s² V2= 7663 m/s = 27,590 Km/h Sketch in next slide

r1 r2 r1 r2 3

Potential Energy :-

1. Gravitational Potential Energy :- it is the work against the gravitational field to elevate the particle a distance (h) above an arbitrary reference plane Vg=m.g.h Going from (h1) level to a higher level (h2) ΔVg=mg(h2-h1) = mg Δh The earth’s gravitational force is (mgR²/r²) , therefore the earth’s gravitational potential energy Vg = - (mgR²/r) (R:- earth’s radius , r:- distance from earth center) the change in Vg from r1 to r2 is equal to ΔVg = mgR²(1/r1-1/r2)

• 2. Elastic Potential Energy :- Work done to deform an elastic body such

as a spring is stored in the body and is called its elastic potential energy (Ve) Ve = ∫ kx.dx = 0.5 *k*x² , For a deformation of x x

Going from (h1) level to a higher level (h2) ΔVg = mg(h2 – h1) = mg Δh

The earth’s gravitational force is therefore, the Earth’s gravitational potential energy is: R: Earth’s Radius r: Distance from Earth’s center Vg: for large altitudes

The change in Vg from r1 to r2 is:

B) Elastic Potential Energy: Work done to deform an elastic body such as a spring is stored in the body and is called its elastic potential energy Ve . The change in Ve from x1 to x2 is:

C) Work Energy Equation: Recall that: U1-2 = is the work of all external forces.

For gravitational and elastic energies the path

is not important. Only the end points (start and end) are important.

In the mass-spring system it is hard to evaluate U1-2 Therefore, it is easier to use the previous equ. where Total mechanical Energy Of the particle and the spring

The net work done on the system by all forces except gravitational and elastic forces equals the change in the total mechanical energy of the system

• Example :

Slider :A: moves up the guide M= 10 Kg ; and neglect friction; and the spring stiffness = 60 N/m ******************* Spring is stretched 0.6 M in position A neglect pulley resistance at B find V of the slider at C ? ********************** Note : slider released from rest at A

Δve = ½ K(X2 – X1) = ½ (60 N/m)([1.2m+0.6m] – [0.6m] )= 86.4 J Alternative work-energy equ: 150J=5V+58.9J+86.4J V=0.974 m/s