48640 lecture note week 2(2)

8/11/2019 48640 Lecture Note Week 2(2)

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APPLICATIONS

As the slider block A moves horizontally to the left with

vA, it causes the link CB to rotate counterclockwise. Thus

vBis directed tangent to its circular path.

Which link is undergoing general plane motion? Link AB

or link BC?

How can the angular velocity, , of link AB be found?

Planetary gear systems are used in

many automobile automatic

transmissions. By locking or releasing

different gears, this system can

operate the car at different speeds.

How can we relate the angular velocities of the

various gears in the system?


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Point A is called the base pointin this analysis. It generally

has a known motion. The x'- y frame translates with the

body, but does not rotate. The displacement of point B

ABAB ddd

/rrr

due to

rotation

aboutA

due to

translation

aboutA

due to translation

and rotation

x-y frame provides

significant convenience

to study the rotation of

point B about A.

Velocity Relationship between the velocities of pointsAand B

dt

d

dt

d

dt

dABAB /rrr

ABAB /vvv

Since the body AB is taken as rotating about A,

Here only has a kcomponent since the axis of rotation isperpendicularto the plane of translation.

or

ABABAB rdtrd

/// /

v

Sometimes it

may be

impossible to

adopt this

equationdirectly and

we have to go

back to


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When using the relative velocity equation, points A and B

should generally be points on the body with a known

motion. Often these points are pin connections in

linkages.

Point A on link AB must move along

a horizontal path, whereas point B

moves on a circular path.The directions of and areknown since they are always tangent

to their paths of motion.

Case A

vB= vA + rB/A

Furthermore, point B at the center of the wheel moves

along a horizontal path. Thus, has a known direction,e.g., parallel to the surface.

When a wheel rolls without slipping, point A is often

selected to be at the point of contact with the ground.

Since there is no slipping, point A has zero velocity.

ABAB /rvv

Case B


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3. If the solution yields a negative answer, the sense of

direction of the vector is oppositeto that assumed.

2. Express the vectors in Cartesian vector form(CVN) and

substitute them into vB = vA + rB/A. Evaluate thecross product and equate respective i and j

components to obtain twoscalar equations.

1. Establish the fixedx - ycoordinate directions and drawthe kinematic diagram of the body, showingthe vectors vA, vB, rB/A and . If the magnitudes areunknown, the sense of direction may be assumed.

Procedure for Analysis

VECTOR ANALYSIS

It is possible to determine the directions of some

parameters by observation even their magnitudes are

unknown.

It is OK to show the observed correct directions in the

kinematic diagram.

However, when we do vector analysis, we can always

assume the direction of these parameters that have

unknown parameters (no matter how we express them

in the kinematic diagram). The final answer from the

vector analysis can show the correct sense

automatically.

VECTOR ANALYSIS


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3. Write the scalar equations from the x and y

components of these graphical representations of the

vectors. Solve for the unknowns.

1. Establish the fixedx-ycoordinate directions and draw akinematic diagramfor the body. Then establishthe magnitude and direction of the relative velocity

vector /.

2. Write the equation vB = vA + vB/A. In the kinematic

diagram, represent the vectors graphically by showing

their magnitudes and directions underneath each

term.

Procedure for Analysis

SCALAR ANALYSIS

Kinematic Diagram(s)

Must Be Presented

in Solution


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READING QUIZ

1. When a relative-motion analysis involving two sets of coordinate

axes is used, thex- ycoordinate system will

A) be attached to the selected point for analysis.

B) rotate with the body.

C) not be allowed to translate with respect to the fixed frame.

D) None of the above.

2. In the relative velocity equation, vB/A is

A) the relative velocity of B with respect to A.

B) due to the rotational motion.

C) rB/A.

D) All of the above.

The link is guided by two blockAand B, which move in the

fixed slots. If the velocity of A is 2 m/s downward,

determine the velocity of B at the instant= 45.

Example I (Example 16.6)


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Solution

Vector Analysis

Kinematic DiagramSince points A and B are restricted to move and vA is

directed downward, vB must be directed horizontally to

the right.

This motion causes the link to

rotate CCW

Apply the velocity equation

to pointsAand Bin order to

solve for the two unknown

magnitudes vBand

ABAB/

vvv

SolutionVector Analysis

Velocity Equation

We have

Equating theiandjcomponents gives

ijji

jikji

rvv

45cos2.045sin2.02

)]45cos2.045sin2.0([2

/

B

B

ABAB

v

v

m/s2,rad/s1.1445sin2.020

45cos2.0

B

B

v

v


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The collar Cis moving downward with a velocity of 2 m/s.

Determine the angular velocity of CBat this instant.

Example II (Example 16.8)

Only know

Two unknown

and

so two equations expected

SolutionVector Analysis

Kinematic Diagram

CBandABrotate counterclockwise.

Velocity Equation

Link CB(general plane motion):

m/s2andrad/s10

2.020

2.0

2.02.02

)2.02.0(2

/

BCB

CB

CBB

CBCBB

CBB

CBCBCB

v

v

v

v

ijji

jikji

rvv


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Solution

Velocity Equation

Link AB(rotation about a fixed axis)

rad/s10

2.02

)2.0(2

AB

AB

AB

BABB

jki

rv

1. Which equation could be used to findthe velocity of the center of the gear, C,if the velocity vAis known?

A) vB= vA+ gearrB/A B) vA= vC+ gearrA/C

C) vB= vC+ gearrC/B D) vA= vC+ gearrC/A

vA

2. If the bars velocity at A is 3 m/s, whatbase point (first term on the RHS of the

velocity equation) would be best used to

simplify finding the bars angular velocity

when = 60?

A) A B) B

C) C D) No difference.

A

4 m

B

C

CONCEPT QUIZ


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Given:

Roller A is moving to the right at 3m/s.

Find:

The velocity of B at the instant

= 30.

1. Establish the fixed x-y directions and draw a kinematic

diagram of the bar and rollers.

2. Express each of the velocity vectors for A and B in

terms of their i,j, kcomponents and solve

Example III

ABAB /rvv

Plan:

Equating the iandjcomponents gives:

0 = 30.75

-vB=1.299

Express the velocity vectors in CVN

vB= vA + rB/A

-vBj= 3 i + [ k

(-1.5cos30i+1.5sin 30j)]

-vBj= 3 i1.299 j0.75 i

Solution:

Solving: = 4 rad/s or = 4 rad/s k

vB= 5.2 m/s or vB= -5.2 m/sj

y

Kinematic diagram:

y


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Given:

Crank rotates OA with an

angular velocity of 12 rad/s.

Find:

The velocity of piston B and the

angular velocity of rod AB.

Notice that point A moves on a circular path. The

directions of vA

is tangent to its path of motion.

Draw a kinematic diagram of rod AB and use

vB= vA+AB rB/A

Example IV

Plan:

By comparing the i, jcomponents:

i: 0 = -3.6 + 0.3 AB AB = 12 rad/s

j: vB = 0.5196AB vB = 6.24 m/s

Rod AB. Write the relative-velocity equation:

vB= vA+ AB rB/A

Since crank OA rotates with an

angular velocity of 12 rad/s, the

velocity at Awill be:

vA= -0.3(12) i= -3.6 im/s

Kinematic diagram of AB

vBj = -3.6 i+ AB k (0.6cos30 i 0.6sin30j)

vBj = -3.6 i+ 0.5196 ABj+ 0.3 AB i


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GROUP PROBLEM SOLVING

Given:

The shaper mechanism is designed togive a slow cutting stroke and a quick

return to a blade attached to the

slider at C. The link AB is rotating at

AB = 4 rad/s.

Notice that link AB rotates about a fixed point A. Thedirections of vB is tangent to its path of motion. Draw a

kinematic diagram of rod BC. Then, apply the relative

velocity equations to the rod and solve for unknowns.

Find:

The velocity of slider block C when = 60.

Plan:

Solution:

Draw a kinematic diagram of rod BC

Since link AB is rotating at AB = 4 rad/s,

the velocity at point B will be:

vB= 4 (300) = 1200 mm/s

At= 60, vB= -1200 cos 30i+1200 sin 30j

= (-1039i+600j) mm/s

Notice that the slider

block C has a

horizontal motion.xy

vB

vC

rC/BBC

45

30

B

C


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Apply the relative velocity equation in order to find the

velocity at C.

vC= vB+

BC rC/B

BC =4.8 rad/s

vC= 1639 mm/s = 1639 mm/sx

y

vB

vC

rC/BwBC

45

30

B

C

vCi= (- 1039i+600j)

+BCk (-125 cos 45i+125 sin 45j)

vCi= (1039125 BC sin 45)i+(600 125 BC cos 45)j

Equating the iandjcomponents yields:

vC= 1039125 BC sin 45

0 = 600 125 BC cos 45

1. If the disk is moving with a velocity at

point O of 15 m/s and = 2 rad/s,

determine the velocity at A.

A) 0 m/s B) 4 m/s

C) 15 m/s D) 11 m/s

2. If the velocity at A is zero, then determine the

angular velocity, .

A) 30 rad/s B) 0 rad/s

C) 7.5 rad/s D) 15 rad/s

2 mV=15 m/s

A

O

ATTENTION QUIZ

48640 lecture note week 2(2)

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