Principles

Learn The Method

Principles

Basics should be automaticMemorize and Practice!

The angular velocity of the disk (r = 1m) in rad/s is

(A) 6 k(B) -6 k(C) 6 i(D) - 3k(E) 3 k

5.129

i

J

wDisk = vo / r = -3/1 = -3 rad/s

The angular accel. of the disk (r = 2 m) in rad/s2 is

(A) 5 k(B) - 5 k(C) - 10 k(D) 10k(E) 2.5 k

5.129

i

J

aDisk = ao / r = 5/2 = 2.5 rad/s2

A

i

J

B

D (t) vD(t)

velangular_

Arm AB with Length L moves the slotted rod BD in x-direction only.

How do we determine the velocity vD of point D as a function of angle q(t)?

L

X(q) =L*cosq

A

i

J

B

D (t) vD(t)

velangular_

Arm AB with Length L moves the slotted rod BD in x-direction only.

How do we determine the velocity vD of point D as a function of angle q(t)?

L

X(q) =L*cosq

X-dot(q) =-L*sin * _q q dot

Chapter 14

Energy Methods

Only Force components in direction of motion do WORK

oductScalar

rdFdW

Pr_

The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.

2. If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is _____

A) 3.8 m/s. B) 6.9 m/s.

C) 14.7 m/s. D) 21 m/s.

The work done is mgh½*m*v2 = mgh orV = sqrt(2gh) = sqrt(14.7)

Power

P E

t

dE

dt

Units of power:

J/sec = N-m/sec = Watts

1 hp = 746 W

The potential energy V is defined as:

dr*F- W - V

Conservative Forces

T1 + V1 = T2 + V2

Potential Energy

Potential energy is energy which results from position or configuration. An object may have the capacity for doing work as a result of its position in a gravitational field. It may have elastic potential energy as a result of a stretched spring or other elastic deformation.

Potential Energy

elastic potential energy as a result of a stretched spring or other elastic deformation.

Potential Energy

Potential Energy

y

Procedure

1. Frame, start and end points

1.2.xc

3. Apply the Energy Principle (only ext. Forces!)

2. Constraint equations?

xA

T1 = 0, start from rest0 + F*xc = 0.5*m*(vA)2

Ex. Problem 14.26

2 𝑥𝑐+𝑥𝐴=𝐿

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)

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

fig_05_006

fig_05_007

16.4 Motion Analysis

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 BC.

Step 1: Define the Geometry

Example

Bar BC rotates at constant wBC. Find the ang. Veloc. of arm BC.Step 1: Define the Geometry

A

i

JB

C

(t) (t)

vA(t)

O

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 = g-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

Rigid Body AccelerationChapter 16.7

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

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

1.Compute all velocities and angular velocities.

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

3. The angular accel is NORMAL to theCentripetal acceleration.

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.

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. View from A: aB = aABXrB – wAB

2*rB

3. View from D: 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

aD

We now can solve two simultaneous vector equations for wAB and wBC

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

fig_05_01116.8 Relative Motion

aP = *ur + ωx(ωxr) + (α x r) + 2 (ω x )

radial

tangential

From Ch. 12.8

End of Review Chapters 14 and 16