Introduction to Dynamics

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Introduction to Dynamics

(formerly) Spinning & Shaking

Prof. Nicholas Zabaras

Centre for Predictive Modelling University of Warwick

Coventry CV4 7AL United Kingdom

Email: [email protected] URL: http://www.zabaras.com/

February 4, 2016

http://www.zabaras.com/Courses/Dynamics/Dynamics.html

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Introduction to Dynamics (N. Zabaras)

Kinematics of a Particle

2

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Rectilinear Motion: Position, Velocity & Acceleration

• Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed.

• Consider particle which occupies position P at time t and P’ at t+Dt,

t v x

t x

t D

 D

 D 0

lim Average velocity

Instantaneous velocity

• From the definition of a derivative, dt

dx t

v x

t

D 

 D

 D 0

lim e.g.

, 2

3 2

3 12 6

t dt t

v dx

t t

x









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Rectilinear Motion: Position, Velocity & Acceleration

• Consider particle with velocity v at time t and v’ at t+Dt,

Instantaneous acceleration

t a v

t D

 D

 D 0

lim

dt t a dv

t t v

dt x d dt

dv t

a v

t

6 12

3 12 e.g.

lim

2

2 2 0









 D 

 D

 D

• From the definition of a derivative,

• Instantaneous acceleration may be:

- positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity

or increasing negative velocity.

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1

2

3

4

5

6

7 Velocity Acceleration Velocity Acceleration

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Example Problem

Determine:

• velocity and elevation above ground at time t,

• highest elevation reached by ball and corresponding time, and

• time when ball will hit the ground and corresponding velocity.

Ball tossed with 10 m/s vertical velocity from window 20 m above ground.

SOLUTION:

• Integrate twice to find v(t) and y(t).

• Solve for t at which velocity equals

zero (time for maximum elevation) and evaluate corresponding altitude.

• Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity.

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Example Problem

 ^dv ^dt ^v

 

^t ^v ^t

dt a dv

t t v

v

81 . 9 81

. 9

s m 81 . 9

0 0

2

0















 

t t

v 



 





 2

s 81m . s 9

10m

 

   

₀ ₂¹ ²

0

81 . 9 10

0

t t

y t

y dt

t dy

t dt v

dy

t t y

y















 

²

2

905 m . m 4

10 m

20 t t

t

y 



 







 





 SOLUTION:

• Integrate twice to find v(t) and y(t).

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Example Problem

• Solve for t at which velocity equals zero and evaluate corresponding altitude.

 

0

s 81m . s 9

10m

2  



 





 t

t v

s 019 .

1 t

 

   

²

2 2 2

s 019 . 1 s

905 m . 4 s

019 . s 1 10 m m

20

s 905 m . s 4

10 m m

20



 







 



 





 







 



 



y

t t

t y

m 1 .

 25 y

8

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Example Problem

• Solve for t at which altitude equals zero and evaluate corresponding velocity.

 

⁰

s 905 m . s 4

10m m

20 ²

2  



 







 





 t t

t y

 

s 28 . 3

s meaningles

s 243 . 1





 t t

 

  

³^.²⁸^s



s 81m . s 9

10m s

28 . 3

s 81 m . s 9

10m

2 2



 









 







v

t t

v

s 2m .

22

 v

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Velocity as a Function of Time

c

a dv

 dt

dv  a dt

c

o 0

v t

c v

dv  a dt

 

0 c

v   v a t

For constant acceleration

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Position as a Function of Time

0 c

v dx v a t

 dt  

0 0

( )

o

x t

c s

dx  v  a t dt

 

2 0 0

1 2 ^c x  x  v t  a t

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Relation involving x, v, & a (no time t)

a dv

 dt

dt dx

 v

v dx

 dt

dt dv

 a

dx dv v  a

adx  vdv

Acceleration Velocity Position x

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Velocity as a Function of Position

0 0

v x

c

v s

v dv  a dx

 

v dv  a dx

c

2 2

0 2 ( _c 0 )

v  v  a x  x

2 2

0 0

1 1

( )

2 v  2 v  a x

^c

 x

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Example of Constant Acceleration: Free Fall

 We throw two balls at the top of a cliff of height H. Ball A is thrown with an initial speed v₀ downwards and ball B with the same speed upwards (as shown).

 The speed of the two balls when they hit the ground are v_A and v_B respectively. Which of the following is true:

(a) v_A < v_B (b) v_A = v_B (c) v_A > v_B

v₀ v₀

B A

H v_A v_B

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Motion of Several Particles: Relative Motion

• For particles moving along the same line, time should be recorded from the same

starting instant and displacements should be measured from the same origin in the same direction.





 _B _A

A

B x x

x relative position of B

with respect to A

A B A

B x x

x  





 _B _A

A

B v v

v relative velocity of B

with respect to A

A B A

B v v

v  





 _B _A

A

B a a

a relative acceleration of B with respect to A

A B A

B a a

a  

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The passenger aircraft B is flying east with velocity v

_B

= 800 km/h. A jet is traveling south with velocity v

_A

= 1200 km/h. What velocity does A appear to a passenger in B ?

B A B

A V V

V  



Solution

_v ₈₀₀_iˆ

B 



jˆ 1200 v_A  

x y

B

vA



B

vA

iˆ 800 jˆ

1200   



j i

V_A _B  800ˆ 1200ˆ

Relative Velocities using Vectors

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Position

 How to establish the position coordinate

 Reference from a fixed point (O) or fixed datum.

 Measures along each inclined plane in the direction of motion

 Has positive sense from C to A and from D to B.

 Establish a total cord length equation

A CD B T

s  l  s  l

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Velocity

 Velocity is the time derivative of the position

 From the total cord length equation

 Velocity

A CD B T

s  l  s  l

A B

0

B A

ds ds

dt  dt  or    

A B T CD

s  s   l l  const

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Acceleration

 Acceleration is the time derivative of the velocity

 From the velocity equation

 Acceleration

B A

   

B A

a   a

B A

d d

dt dt

  



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Curvilinear Motion: Position, Velocity & Acceleration

• Particle moving along a curve other than a straight line is in curvilinear motion.

• Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle.

• Consider particle which occupies position P

defined by at time t and P’ defined by at t + Dt,

r

r



D 

 D



D 

 D

 D

dt ds t

v s

dt r d t

v r

t t

0 0

lim lim



 

instantaneous velocity (vector)

instantaneous speed (scalar)

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Curvilinear Motion: Position, Velocity & Acceleration



D 

 D



D dt

v d t

a v

t



 

0

lim

instantaneous acceleration (vector)

• Consider velocity of particle at time t and velocity at t + Dt,

v v

• In general, acceleration vector is not

tangent to particle path and velocity vector.

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Rectangular Components of Velocity & Acceleration

• When position vector of particle P is given by its rectangular components,

k z j y i x

r     

• Velocity vector,

k v j v i v

k z j y i x dtk

j dz dt i dy dt v dx

z y

x

 





 













• Acceleration vector,

k a j a i a

k z j y i x k dt

z j d

dt y i d

dt x a d

z y

x

 





 



 



 



 











 2

2 2

2

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Rectangular Components of Velocity & Acceleration

• Rectangular components particularly effective when component accelerations can be integrated independently, e.g., motion of a projectile,

0

0     



 x a y g a z

a_x  _y  _z 

with initial conditions,

 

^,

 

^,

^{ }

⁰

0 ₀

0 0 0

0

0  y  z  v_x v_y v_z 

x

Integrating twice yields

   

 

₀⁰ ^

 

₀⁰ ^ ₂¹ ² ^ ⁰⁰







z gt

y v

y t v

x

v gt

v v

y x

z y

y x

x

• Motion in horizontal direction is uniform.

• Motion in vertical direction is uniformly accelerated.

• Motion of projectile could be replaced by two

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Horizontal motion is uniform motion

Notice that the horizontal motion is in no way affected by the vertical motion.

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At the peak of its trajectory, v

_y

= 0.

From

Time t

₁

to reach the peak Substituting into:

0 1

v

y

t  g

2 0

max

2 v

y

h y

  g

0

y y oy

v  v  gt  v  gt 

2 0

1

y

2 y  v t  gt

Maximum Height

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10 degrees 30 degrees 40 degrees 45 degrees 60 degrees 75 degrees

Angle that maximizes range

 _optimal = 45 degrees

Projection angle = 75 degrees

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Tangential and Normal Components

• Velocity vector of particle is tangent to path of particle. In general, acceleration vector is not.

Wish to express acceleration vector in terms of tangential and normal components.

• are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, and

is the angle between them.

t

t e

e and 

t t

t e e

e    

 D D

 



d e e d

e e e

e

n t

n t n

t

 



 



D 

 D D D

D

 D

 D



D 2

2 lim sin

lim

2 sin

2

0 0

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Tangential and Normal Components

et

v v  

• With the velocity vector expressed as

the particle acceleration may be written as

t t

de de

dv dv dv d ds

a e v e v

dt dt dt dt d ds dt



     

but

dt v ds ds

d d e

e d

t  n    



 

After substituting,



2

2 v

dt a a dv

v e dt e

a  dv _t  _n _t  _n 

• Tangential component of acceleration reflects change of speed and normal component reflects change of direction.

• Tangential component may be positive or negative. Normal component always points toward center of path curvature.

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Radius of Curvature

 Find the radius of curvature of the parabolic path in the figure at x = 150 ft.

1 2

200 200

y  x 

/ 1

dy dx  100 x

2 2 1

/ 100

d y dx 

2 3/ 2

2 2

1 1

100 1 ( / )

585.9 ft / 1

100 dy dx x

d y dx



   

   

      

   

  

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Sample Problem

A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration rate.

Knowing that after 8 s the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes are applied.

SOLUTION:

• Calculate tangential and normal components of acceleration.

• Determine acceleration magnitude and direction with respect to tangent to curve.

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Sample Problem

ft/s 66 mph

45

ft/s 88 mph

60



SOLUTION:

• Calculate tangential and normal components of acceleration.

 

2 2 2

2

66 88 ft s ft

8 s 2.75s

88 ft s ft

2500 ft 3.10s

t

n

a v

t a v

 D 

   

D

  

• Determine acceleration magnitude and direction with respect to tangent to curve.

 

² ²

2

2   2.75 3.10

 a_t a_n

a ₂

s 14 ft .

 4 a

75 . 2

10 . tan 3

tan^¹  ^¹



t n

a

 a  _ ₄₈_.₄_

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Radial and Transverse Components

• When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP.

r er

d e e d

d e

d  

 





 

 ^ ^

dt e d

dt d d

e d dt

e

d _r _r  

 ^

 





dt e d dt

d d

e d dt

e d

r



^

  







 



 e

r e

r

dt e r d dt e

dr dt

e r d dt e

e dr dt r

v d

r

r r r

r

 



 















• The particle velocity vector is

• Similarly, the particle acceleration vector is

  ^ ^



 



e r r

e r

r

dt e d dt r d e

dt r d dt e

d dt dr dt

e d dt e dr

dt r d

dt e r d dt e

dr dt a d

r r r

r

 

 

 

 



 



2 2

2 2 2

2













 



 

 er

r r  

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Sample Problem

Rotation of the arm about O is defined by  = 0.15t²where  is in radians and t in seconds. Collar B slides along the arm such that r = 0.9 - 0.12t² where r is in meters.

After the arm has rotated through 30^o, determine (a) the total velocity of the collar, (b) the total

acceleration of the collar, and (c)

the relative acceleration of the collar

SOLUTION:

• Evaluate time t for  = 30^o.

• Evaluate radial and angular positions, and first and second derivatives at time t.

• Calculate velocity and acceleration in cylindrical coordinates.

• Evaluate acceleration with respect to arm.

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Sample Problem

SOLUTION:

• Evaluate time t for  = 30^o.

0.15 2

30 0.524 rad 1.869 s t

t

 

    

• Evaluate radial and angular positions, and first and second derivatives at time t.

2 2

s m 24 . 0

s m 449 . 0 24

. 0

m 481 . 0 12

. 0 9 . 0

















r

t r



2 2

s rad 30 . 0

s rad 561 . 0 30

. 0

rad 524 . 0 15

. 0







 t

t

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Sample Problem

• Calculate velocity and acceleration.

  

r r

r

v v v

v v

r v

s r

v

 







1 2

2 tan

s m 270 . 0 s rad 561 . 0 m 481 . 0

m 449 . 0

 















 0.524m s  31.0 v

  

    ^ ^ ^

r r

r

a a a

a a

r r

a

r r a

 







1 2

2

2 2

2 2 2

tan s

m 359 . 0

s rad 561 . 0 s m 449 . 0 2 s

rad 3 . 0 m 481 . 0

2

s m 391 . 0

s rad 561 . 0 m 481 . 0 s

m 240 . 0

 































 



 



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Sample Problem

• Evaluate acceleration with respect to arm.

Motion of collar with respect to arm is rectilinear and defined by coordinate r.

s2

m 240 .

0



 r a_B _OA 

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Introduction to Dynamics