Lecture 2 - 2020 Dynamics MECH ENG 1007 PDF

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Lecture 2 - 2020 Dynamics MECH ENG 1007 University of Adelaide Notes...

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Curvilinear Motion of Particles This course restricts itself to the description of motion in a plane, i.e. 2-D motion. Three co-ordinate systems will be investigated for the purpose of describing planar motion:  Cartesian (Or x-y)  Polar (Or Radial-Transverse, or r-)

Position:

r(t)

Velocity:

v(t) =

 Tangential-Normal (Or t-n, or n-t) The motion of many machine parts in mechanical devices is planar, although there are significant examples where 3-D motion occurs.

Acceleration:

dr dt dv a(t) = dt

(2.1)

The standard convention for vectors in Physics and Engineering texts is to embolden the symbol. In your own work, indicate vectors with either an overhead arrow, or squiggle underneath. For r instance, r or r . ~

2.1

2.2

*

v(t) is a vector tangent to the path s(t). Moreover, its magnitude v is: v = ds dt

v = lim t0

r ( t + t ) - r ( t ) t

a(t) has both normal and tangential components to the position path s(t).

2.3

2.4

Cartesian Coordinates

x  r=    y   vx v =   vy

. vx = x

. vy = y

(2.2)

. .. ax = vx = x

.. . ay = vy = y

(2.3)

r = xiˆ + y jˆ

  

v = vx iˆ + vy ˆj

 ax   a =   ay 

a = ax iˆ + ay jˆ

If ax is independent of y and vy, and ay is independent of x and vx , then the problem can simply be solved by its components of motion.

where iˆ and jˆ are unit vectors in the x and y directions respectively

2.5

2.6

x Component

Projectile Motion (No air resistance)

ax = 0

v x   ax dt   0 dt  C C is a constant of integration C = v x (0 ) = x& (0) = v0 cos 

 0: Launch angle

 

Differential Equations: .. .. ax = x = 0 ay = y = -g

v x  v0 cos θ0

x(t) =  v x d t =

v

0

cos θ0 dt = v0 cos  t + D

x(0) = D = 0 (for co-ordinate system shown)

Initial conditions: x(0) = 0

y(0) = 0

x&(0) = v0 cos 

y&(0) = v0 sin 



x  v0 cos θ0 t

Initial values for x and y depend critically on the placement of the origin of your Cartesian co-ordinate system, so you need to be certain of your co-ordinate system definition. Note:  in these expressions is the launch angle. 2.7

2.8

y Component

In summary

a y = -g

v y   a y dt    g dt   gt  C

x(t) = v0cost

C is a constant of integration

1 y(t) =  gt 2  v 0 sin  t 2

C = v y ( 0) = y& (0) = v0 sin 

(2.4)

(when x(0) = y(0) = 0)

 

v y   g t  v 0 sinθ0

In this example the launch point was

y(t)

=  v y dt =

 - g t  v

0

sin θ 0  dt

(x0, y0) = (0, 0) because our co-ordinate system was chosen so that it was so.

1 =  gt 2  v0 sin θ 0 t  D 2 y(0) = D = 0 (for co-ordinate system shown)



1 y   gt 2  v 0 sin θ 0 t 2

Also, the moment of launch was designated as t0 = 0. In general we need to define a a spatial origin for our coordinate system, and a temporal origin for time (t). The choices for space and time coordinates give us the simplest analysis, although is not always possible to do (e.g. A problem where 2 projectiles are launched from different places and/or at different times).

2.9

2.10

Rotating Unit Vector

e& =

θ& 2 sin 2 θ + θ& 2 cos2 θ

e& = θ&

(2.6)

Note also that e  e&  0 , which means that the unit vector and its time derivative are perpendicular. eˆ = 1

x = cos()

vx = - θ& sin()

y = sin()

& cos() vy = θ

& sinθ   -θ de = e& =   dt   θ& cosθ 

(2.5)

2.11

2.12

**

Tangential and Normal Coordinates

The time derivative e& of a unit vector e is also a vector d e = e& = θ& n dt

n is a unit vector that is normal to e and points in the sense of rotation of the e.

We know (*) that

v = vet = ds e t dt

2.13

(2.7)

2.14

Definition

By definition (2.1)

 is the instantaneous radius of curvature. d (ve t ) a = dv = dt dt so that

a=

dv de t (by the product rule) (e t ) + v dt dt

ds = d

However, by (**)

Hence ds = v =  d θ dt dt

d et dθ = (e n ) dt dt

and using (2.8) we have which gives

a=

dv dθ (e ) (e t ) + v dt dt n

v2 dv (en ) a= (et ) +  dt

(2.8) 2.15

(2.9)

2.16

Circular Motion

Remarks

- In t-n coordinates If a point P moves in a circular path of radius R then s = R (circular path)

(2.11)

The unit vector en points toward the concave side of the path. The radius of curvature is given by 3

  d y 2 2 1 +     dx   =  d2y dx2

Here  = R = constant

(2.10)

2.17

2.18

The angular velocity  is defined by

It follows from (*) that dθ v = ds = R dt dt

 θ&

(2.12)

(circular path)

The angular acceleration  is defined by

By (2.7) and (2.8) we have

v=R

α  ω&  &θ&

dθ e dt t

d2θ a = R 2 et + R dt

With these definitions (2.13) gives s = R

2

 dθ    en  dt 

(2.13) v = Ret

(circular path)

(2.14) a = Ret + R2 e n (Circular path) 2.19

2.20

Example 2.1 A satellite is in a circular orbit around the earth, a distance R from the earth centre. What is its velocity?

From example 1.2  = 0.  +

   

Note that here  =  =  Generally, by (2.9) d   +  = d   2.21

2.22

Equating the tangential component gives d d

Example 2.2 A particle P starts to move from rest at t = 0. Its Cartesian components of acceleration are

=0

Hence

a x = 0.6t  = 

a y = 1.80 - 0.36t

  

m/s2

At t = 4s

The normal component gives

=

m/s2

(i) What are the normal and tangential components of its acceleration?

ANS

(ii) What is its instantaneous radius of curvature?

2.23

2.24

Solution v x =  a x dt 

vx = 0.3t2 + A vx (0) = 0 = A



(“Rest” condition)

v=

vx = 0.3t2

2

2

vx + vy

v(4) = 6.46 m/s

Similarly

v   = arctan  y   vx 

vy = 1.8t + 0.18t2 At t = 4s

(4) = 42.0o

vx (4) = 4.80 m/s vy (4) = 4.32 m/s a x(4) = 2.4 m/s2 a y(4) = 0.36 m/s2 2.25

2.26

3) Associate the co-ordinates correctly between the two co-ordinate sets i.e. which co-ordinate maps onto which by a pure rotation? In this case: x  n;

In order to find the acceleration in t-n coordinates, a co-ordinate transformation must be carried out: 1) Identify the right-handed co-ordinate system we are transforming FROM – in this case Cartesian (x, y). 2) Identify the right-handed system we are transforming TO – in this case the tangential-normal (t, n) VERY IMPORTANT: In performing coordinate transformation by this method ensure that BOTH co-ordinate systems are right-handed. 2.27

y t

4) Deduce the rotation angle, going from the FROM co-ordinates to the TO coordinates, using the convention that anti-clockwise is positive. In this case: R  90     48.0 o 5) Apply the co-ordinate transformation matrix (setting up for conventional righthanded co-ordinates):  an      at 

 cos R    sin  R

sin R   a x   cos  R   a y

  

      sin 48   2.40  cos  48   0.36

an  cos  48 o  2.40  sin  48 o  0.36 at

o

o

2.28

Radial and Transverse Coordinates a t(4) = 2.02

m/s2

a n(4) = 1.34

m/s2

The radial and transverse coordinates,

ANS

sometimes called polar coordinates, are often used to solve central force

2 an = v 



v2 = an = 31.2

problems. In central force systems the acceleration of a particle is directed

2 = (6.46) 1.34

m

towards a given point.

ANS

2.29

2.30

Let r(t) be the magnitude of the position vector r(t). Then

r =rer

Position

(2.15)

By definition (2.1) The unit vector er points in the direction of the radial line from the origin to P. The unit vector e is perpendicular to er and points in the direction of increasing .

v = dr dt

der = dθ e  dt dt

and from (**)

In a conventional polar co-ordinate system is measured in a counter-clockwise sense from the x axis, although for certain dynamical problems a non-conventional system is more advantageous.

= d r er + r der dt dt

dθ v = dr e r + r e dt dt



= r& er + re   

2.31

 2.32

The acceleration is obtained by differentiating v

and de = -e r dt

a = dv dt dr & e  = d r er + r& de r + dt dt dt + r d  e  + r de  dt dt

Using (**) again

we have

d e r = e dt

a = [ r&& - r2 ] er + [r + 2 r& ] e

2.33





2.34

Circular Motion In Polar (or r-  coordinates)

a = ( &r& - r2)e r + (r + 2 r& )e  v=

dr e r + re dt

Here r = R and r& = r&& = 0

Here r = R = constant 

dr =0 dt





(circular motion)

 

a = -R2e r + Re 

v = Re 









(Circular Motion) 2.35

2.36

Example 2.3

Solution

Disk D rotates with constant angular velocity 0 . Particle P moves with constant speed v0 along a straight line on the disk. What are the velocity and acceleration of the particle when it is at a distance R from the centre of the disk?

Here r = R, r& = v0 , r&& = 0,  = 0 and  = 0 

v = v0 e r + R0 e

ANS

a = ( &&r - r2 )er + (r + 2 r& )e



2.37

a = -R02 er + 2v00e 

ANS

2.38

Example 2.4

Solution In polar coordinates

v = r& e r + re r = 1 - 0.5cos(2t) r& = sin(2t)

m/s

= 0.5 - 0.2sin(2t) The robot arm is programmed so that the point P describes the path r = 1 - 0.5cos(2t) = 0.5 - 0.2sin(2t)

m

rad

 = -0.4cos(2t) rad/s

v = sin(2t) er + (1-0.5cos(2t) (-0.4cos(2t))e m/s

m rad

At t = 0.8s, determine the velocity of P in terms of polar and Cartesian co-ordinates. 2.39

v(0.8) = -2.99er – 0.328e m/s

ANS

2.40

(0.8) = 0.5 - 0.2sin(20.8)

Relative Motion (with respect to non-rotating observer)

= 0.690 rad = 39.5o

In the co-ordinate transformation o   θ 39 . 5 formulae, , since the Cartesian unit vectors are clockwise from the polar ones.

Position

rB = rA + rB/A

(2.20)

Velocity

vB = v A + vB/A

(2.21)

Acceleration

aB = a A + aB/A

(2.22)

v x = vrcos -39.5 o + v sin -39.5o v y = - vrsin -39.5 o + v cos -39.5 o

v = -2.09i - 2.16j

m/s

ANS 2.41

2.42

Example 2.5

Solution

A plane moving at 50m/s relative to the air is in a uniform wind blowing east at 20m/s. What should be the direction of the plane if it needs to fly north-west relative to the earth? What will be the resulting magnitude of the plane’s velocity relative to the earth?

2.43

vB = vA + vB/A

2.44

Sine Law

vB/A vB = o sin(135 ) sin( )

50sin(28.57o ) vB = = 33.8 m/s sin(135o )



Sine Law

v B/A vA o = sin(135 ) sin( )

 = arcsin



20sin(135 o ) 50

 = 180o – 135o - 16.43o = 28.57o

The direction of the plane should be =28.57o north of west to cause the plane to travel in the north-west direction relative to the earth. The speed of the plane relative to the earth is vB = 33.8 m/s.

ANS

ANS

 = 16.43o ANS

2.45

2.46

Example 2.6

Solution We use (2.21)

aP = aQ + a P/Q

The bars OQ and QP rotate with constant angular velocities as shown in the figure. In terms of the fixed coordinate system x-y, what is the acceleration of point P? r - r Q2)e r + (rQ + 2 r&Q )e  aQ = ( &&

= (0 - (2)(4)2 )er = -32i 2.47

m/s2 2.48

aP = aQ + aP/Q = -32i - 64i - 64j

ANS

= -96i -64j m/s2

a P/Q

= ( r&& - rP/Q2)er + (rP/Q + 2 r&  )e = (0 -

2 )(-8)2 )er + 0e

=-64( 2 )cos(45o)i - 64( = -64i - 64j

This example demonstrates an important use of the concept of relative motion. The motion of point P relative to point O is quite complicated. But because the motion of P relative to Q and the motion of Q relative to O are comparatively simple, we can take advantage of the equations of the relative motion and solve the problem.

2 )sin(45o)j

m/s

2.49

2.50

Example 2.7

Solution From geometry

The rigid rod of length L moves so that its ends, A and B remain in contact with the surfaces. Find the velocity of B if at the instant shown the velocity of A is 5m/s to the left.

2.51

Hint: To determine these angles, place a Cartesian coordinate system at the centre of the semi-circle and then find an expression for   (which is known to be 10 m).

2.52

We want to use

Let us now examine the velocity vB/A in detail. We attach polar coordinates to A as shown

vB = vA + vB/A The velocity vA is completely known.

vA = -5i m/s According to (*) the direction of vB is tangential to the path of motion, ie. vB acts in the direction.

vB/A = r& e r + re = 10e (since r& = 0 w.r.t the co-ordinate system attached to A)

The magnitude of vB is, however, unknown. 2.53

2.54

It is thus concluded that vB/A acts in the direction of e, ie.

vB = vA + v B/A

x direction 5= v B cos(50.13o ) + v B/A cos(79.66o) y direction

vB sin(50.13o ) = vB/A sin(79.66o)

2.55

2.56

v B = 6.40

m/s

ANS

Note also that vB/A = 5 m/s and, therefore, the angular velocity  of the rod at the position shown is  =

5 v B/A = 10 = 0.5 rad/s clockwise L

2.57...