Angular Momentum Linear Momentum. Always work from Principle! Ex: Work of gravity Principle: dW = F * ds Here: dW = - mg * dy mg y.

vmrHo

Angular Momentum

Linear Momentum

vmG

Always work from Principle!Ex: Work of gravity

Principle: dW = F * ds

Here: dW = - mg * dy

mg y

A

O

C Di

J

Always work from Principle!Ex: Slider arm Kinematics

Principle: x-Position of A:xA = b*cos (q t)

The velocity x-dot is the derivative: x-dot =

- b*w *sinq

aAvA

Accel x-ddot is the 2nd derivative: x-ddot = -

b*w2 *cosq

Chapter 16

Rigid Body Kinematics

16.1

16.3 Rot. about Fixed Axis Memorize!

Vector Product is NOT commutative!

Cross Product

xyyxzxxzyzzy

zyx

zyx

babababababa

bbb

aaa

ba

kji

ba

Derivative of a Rotating Vector

• vector r is rotating around the origin, maintaining a fixed distance

• At any instant, it has an angular velocity of ω

rωr

dt

d

rω

rω

rωr

dt

dv

Page 317:

at = a x r

an =w x ( w x r)

Rotation Kinematics

Similar to translation:

dt*

dt*

and dd **

General Motion = Translation + RotationVector sum vA = vB + vA/B

fig_05_006Important!

Memorize and Practice!

fig_05_007Any rigid body motion can be

viewed as a pure rotation about an “Instantaneous Center”

(Chapter 16.6)

fig_05_011

fig_05_011

fig_05_012

fig_05_012

fig_05_013

fig_05_013

16.4 Motion Analysis

16.4 Motion Analysis

http://www.mekanizmalar.com/fourbar01.html

http://iel.ucdavis.edu/chhtml/toolkit/mechanism/

http://courses.engr.illinois.edu/tam212/aml.xhtml

1. A body subjected to general plane motion undergoes a/an

A) translation.

B) rotation.

C) simultaneous translation and rotation.

D) out-of-plane movement.

2. In general plane motion, if the rigid body is represented by a slab, the slab rotates

A) about an axis perpendicular to the plane.

B) about an axis parallel to the plane.

C) about an axis lying in the plane.

D) None of the above.

1. The position, s, is given as a function of angular position, q, as s = 10 sin 2 q . The velocity, v, is

A) 20 cos 2q B) 20 sin 2q

C) 20 w cos 2q D) 20 w sin 2q

2. If s = 10 sin 2q, the acceleration, a, is

A) 20 a sin 2q B) 20 a cos 2q − 40 w2 sin 2q

C) 20 a cos 2qD) -40 a sin2 q

Approach1. Geometry: Definitions

Constants

Variables

Make a sketch

2a. Analysis (16.4) Derivatives (velocity and acceleration)

3. Equations of Motion

4. Solve the Set of Equations. Use Computer Tools.

2b. Rel. Motion (16.5)

Example

Bar BC rotates at constant wBC. Find the angular Veloc. of arm OA.

Step 1: Define the Geometry

Example

Step 1: Define the Geometry

A

i

JB

C

(t) (t)

vA(t)

O

Bar BC rotates at constant wBC. Find the angular Veloc. of arm OA.

16.5 Relative Motion Analysis

General Motion = Translation + RotationVector sum vB = vA + vB/A

Geometry: Compute all lengths and angles as f(q(t))

All angles and distance AC(t) are time-variant

A

i

JB

C

(t) (t)

vA(t)

O

Velocities: w = q-dot is given.

Vector Analysis: OA rA vCOLL BC rAC Mathcad does not evaluate cross products symbolically, so the LEFT andRIGHT sides of the above equation are listed below. Equaling the i- and j-terms yields two equations for the unknowns OA and vCOLL

Analysis: Solve the rel. Veloc. Vector equation

A

i

JB

C

(t)

vA(t)

O (t)


Seen from O: vA = wOA x OA

Analysis: Solve the rel. Veloc. Vector equation

Seen from C: vCollar + wBC x AC(t) A

i

JB

C

(t)

O (t)


wBC x AC(t)

vA,rel = vColl

Analysis: Solve the rel. Veloc. Vector equation numerically

A

i

JB

C

(t)

vA(t)

O (t)


Enter vectors:

OA

0

0

wOA

rA

l cos ( )

l sin ( )

0

BC

0

0

wBC

rAC

l cos ( )

l sin ( )

0

Analysis: Solve the rel. Veloc. Vector equation numerically

A

i

JB

C

(t)

vA(t)

O (t)

Vector Analysis: OA rA vCOLL BC rAC

LEFT_i l wOA sin ( ) RIGHT_i l wBC sin ( ) vcoll cos ( )

LEFT_j l wOA cos ( ) RIGHT_j l wBC sin ( ) vcoll sin ( )

Here: wBC is given as -2 rad/s (clockwise). Find wOA

Analysis: Solve the rel. Veloc. Vector equation numericallyA

i

JB

C

(t)

vA(t)

O (t)

Solve the two vector (i and j) equations :

Given

l wOA sin ( ) l wBC sin ( ) vcoll cos ( )

l wOA cos ( ) l wBC sin ( ) vcoll sin ( )

vec Find wOA vcoll( ) vec4.732

0.568

A

i

JB

C

(t)

vA(t)

O (t)

Recap: The analysis is becoming more complex.

•To succeed: TryClear Organization from the start

•Mathcad

•Vector Equation = 2 simultaneous equations, solve simultaneously!

fig_05_011

fig_05_011Relative Velocity

vA = vB + vA/B

Relative Velocity

vA = vB + vA/B

= VB (transl)+ vRot

vRot = w x r

Seen from O:vB = wo x rSeen from A:

vB = vA + wAB x rB/A

Seen from O:vB = wo x rSeen from A:

vB = vA + wAB x rB/A

Visualization

http://www.mekanizmalar.com/fourbar01.html

http://iel.ucdavis.edu/chhtml/toolkit/mechanism/

http://courses.engr.illinois.edu/tam212/aml.xhtml

fig_05_012Mathcad Examples

Crank and Slider Pin part 1Geometry

Example

Given are: BC wOA 6 (counterclockwise), Geometry: l triangle with

OA 4 inches. OC 12 inches. Angle 30

180

Collar slides rel. to bar BC.

GuessValues:(outwardmotion ofcollar ispositive)

wBC 1

vcoll 1

gamma 1

AC 1

Vector Analysis: OA rA vCOLL BC rAC Mathcad does not evaluate cross products symbolically, so the LEFT andRIGHT sides of the above equation are listed below. Equaling the i- and j-terms yields two equations for the unknowns BC and vCOLL Step 1: Geometry: Find length AC and angle gamma at C.Law of cosines and law of sines:

Given

AC2

OA2

OC2 2 OA OC cos ( )

AC sin gamma( ) OA sin ( )

vec Find AC gamma( ) vec8.767

0.23

gamma_deg vec1

180

gamma_deg 13.187

AC vec0

AC 8.767 gamma vec1


Crank and Pinpart 2:Solving the vector equations

LEFT_i OA wOA sin ( ) RIGHT_i AC wBC sin gamma( ) vcoll cos gamma( )

LEFT_j OA wOA cos ( ) RIGHT_j AC wBC sin gamma( ) vcoll sin gamma( )

Solve the two vector (i and j) equations :

Given

OA wOA sin ( ) AC wBC sin gamma( ) vcoll cos gamma( )

OA wOA cos ( ) AC wBC cos gamma( ) vcoll sin gamma( )

vec Find wBC vcoll( ) vec1.996

16.425

wBC = 1.996 rad/s (cw).The pin moves radially outward at vcoll = 16.425 in/s

Vector Analysis Concepts: Always start from default assumptions, i.e. assume positiverotations and velocities. While magnitudes and signs are not initially known, all vector anglesare known from the given geometry.


Crank and slider Pinpart 3Graphical Solution

BC

rAC X BC

OA X rOA

RIGHT ARMBC: VA =BC X rAC

Left ARM OA:VA = OA X rOA

Pin slides rel. to Arm BC atvelocity vColl. The angle of

vector vColl is = 13o

Vector Analysis: OA rA vCOLL BC rAC Mathcad does not evaluate cross products symbolically, so the LEFT andRIGTHT sides of the above equation are listed below. Equaling the i- and j-terms yields two equations for the unknowns OA and vCOLL

OA

L

B

A

i

J

vA = const

Given Velocity V_A = const as shown at

leftThe velocity of Point B is

(A) constant, same as V_A(B) constant, but different

from V_A(C)VB(t) is variable (D) None of the above

Find: vB and AB

i

J

B

A

vA = const

Counterclockw.

vB

Given: Geometry andVA

B

A

vA = const

Find: vB and AB

In order to connect points Aand B, we define vector rB/A.

(we use the symbol 'r' as short notation)

r defines the position ofpoint B relative to point A.

i

J


vB = vA + vB/A

Find: vB and AB

Graphical Solution Veloc. of Bi

J

B

AvA = const

Counterclockw.

vB


vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

vB = ?

Find: vB and AB


J

B

AvA = const

Counterclockw.

vB


vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

vB = ?

AB rx denotes arotation about A

AB rx

Find: vB and AB


J

B

AvA = const

Counterclockw.

vB


vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

vB = ?

AB rx

AB rx

Find: vB and AB

i

J

B

AvA = const

Counterclockw.

vB


vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

Solution:vB = vA + AB X r

AB rx

AB rx

16.6 INSTANTANEOUS CENTER OF ZERO VELOCITY

Today’s Objectives:Students will be able to:1. Locate the instantaneous center of

zero velocity.2. Use the instantaneous center to

determine the velocity of any point on a rigid body in general plane motion.

Rigid Body AccelerationChapter 16.7

Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)

Rigid Body AccelerationConceptual Solution

Using Vector Graphics

Propulsion Mechanism of a Baldwin Steam LocomotiveBaldwin Locomotive Works, Philadelphia, 1926

Baldwin Locomotive 60,000

Q: Is this a Freight or Passenger Locomotive ?

A: We can tell from the wheel diameter.

The internal forces (accelerations) in the piston mechanism limit the maximum speed (10 m/s max. Piston velocity).

Find: aB and AB

B

A

vA = const

Counterclockw.

vB

Given: Geometry andVA,aA, vB, AB

i

J

First: Find all velocities.

Rigid Body Acceleration

G iven: G eom etry andV A ,aA , vB , A B

Law : a B = a A + a B/ATransl + C entripeta l +

angular accelFind: a B and AB

L

i

J

B

A

vA = const

C o u n te rc lo c k w .

vB

The re la tive m otion equation a B = a A + a B/Aconnects the unknown acce lera tion a t B to the known(given) acce lera tion a t A .

B

A

vA = const

Find: a B and AB

C entrip . r* AB 2

In order to connect points Aand B , w e define vector rB/A.

(w e use the symbol 'r' as short notation)

r defines the position ofpoint B re lative to point A .r

i

J


aB = aA + aB/A ,centr+ aB/A ,angular

r* AB2 r* +

Find: a B and AB

Look at the Accel. o f B re la tive to A :i

J

B

AvA = const

C o u n te rc lo ck w

.

vB


r


r* AB2 r* +

r

Find: a B and AB

Look at the Accel. o f B re la tive to A :

W e know:

1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)

i

J

B

AvA = const

C entrip .r* AB

2



r* AB2 r* +

Find: aB and AB

Look at the Accel. of B relative to A:

We know:

1. Centripetal: magnitude r2 anddirection (inward). If in doubt, computethe vector product x(*r)

2. The DIRECTION of the angular accel(normal to bar AB)

i

J

B

AvA = const

Centrip. r* AB 2

r*

Given: Geometry andVA,aA, vB, AB

aB = aA + aB/A,centr+ aB/A,angular

r* AB2 r* +

Find: a B and AB

Look at the Accel. o f B re la tive to A :

W e know:

1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)

2. The DIRECTION o f the angular accel(norm al to bar AB)

3. The DIRECTION o f the accel o f po int B(horizonta l a long the constra int)

i

J

B

AvA = const

Centrip. r* AB 2

Angular r*



r* AB2 r* +

aB

r

B

A

vA = const


Find: a B and AB

C entrip . r* AB 2

aB

Angular r*

r is the vector from reference

point A to point B

r

i

J

W e can add graphically:S tart w ith C entipeta l


W e know : a A =0

B

A

vA = const

C entrip . r* AB 2

aB

Angular r*


point A to point B

r

i

J




Find: a B and AB

B

A

vA = const

C entrip . r* AB 2

aB

Angular r*


point A to point B

r

i

J




Find: a B and AB

B

A

vA = const

C entrip . r* AB 2

aB

Angular r*


point A to point B

r

i

J




Find: a B and AB

B

A

vA = const

C entrip . r* AB 2

aB

Angular r*


point A to point B

r

i

J



The vectors form a trianglew ith aB as the hypotenuse.

W e can therefore determ inethe m agnitudes and d irections

of both a B and r*


Find: a B and AB



aB

r* r* AB

2

Result: is < 0 (c lockwise)

aB is negative (to theleft)

B

AvA = const

C entrip . r* A B2

r is the vector from

reference point A to point B

r

i

J

N owC om plete the

Triangle:


Find: a B and AB

A

i

JAB

B

D

(t) = 45deg

(t)

vD(t)= const

General Procedure

1.Compute all velocities and angular velocities.

2.Start with centripetal acceleration: It is ALWAYS oriented inward towards the center.

A

i

JAB

B

D

(t) = 45deg

(t)

vD(t)= const

General procedure



3. The angular accel is NORMAL to theCentripetal acceleration.

A

i

JAB

B

D

(t) = 45deg

(t)

vD(t)= const

General Procedure



3. The angular accel is NORMAL to theCentripetal acceleration. The direction of the angular acceleration is found from the mathematical analysis.

Example HIBBELER 16-1251. Find all vi and wi (Ch. 16.5)

2. aB = aABXrB – wAB2*rB

3. aB = aC + aBCXrB/C – wBC2*rB/C

wAB = -11.55k

wBC = -5k

HIB 16-125Centripetal Terms: We know magnitudes and directions

aABXrB – wAB2*rB = aC + aBCXrB/C – wBC

2*rB/C

– wBC2*rB/C

– wAB2*rB

aC

We now can solve two simultaneous vector equations for aAB and aBC

HIB 16-125

aABXrB – wAB2*rB = aC + aBCXrB/C – wBC

2*rB/C

fig_05_11

fig_05_01116.8 Relative Motion

aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE

fig_05_01116.8 Relative Motion

aA = aB + aA/B,centr+ aA/B,angular + aA,RELATIVE

Midterm #2 Preparation

Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)

Posted:• Collection of Problems• Practice exam #2 • Powerpoint Slides• Four questions will be on

Chapter 16, 2Q. on Ch. 14

http://session.masteringengineering.com/myct/assignment?assignmentID=1030137

Angular Momentum Linear Momentum. Always work from Principle! Ex: Work of gravity Principle: dW = F * ds Here: dW = - mg * dy mg y.

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