Review Ch16

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KINENAMITCS OF RIGID

BODY

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GENARAL

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PLANAR KINEMATICS OF A RIGID BODY

We will now start to study rigid body motion. The analysis

will be limited to planar motion.

For example, in the design of gears, cams, and links in

machinery or mechanisms, rotation of the body is an

important aspect in the analysis of motion.

There are cases where an object cannot be treated as a

particle. In these cases the size or shape of the body must be

considered. Also, rotation of the body about its center of

mass requires a different approach.

A body is said to undergo planar motion when all parts of

the body move along paths equidistant from a fixed plane.

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PLANAR RIGID BODY MOTION

Translation: Translation occurs if every line segment on the

body remains parallel to its original direction during the

motion. When all points move along straight lines, the motion

is called rectilinear translation. When the paths of motion are

curved lines, the motion is called curvilinear translation.

There are three types of planar rigid body motion.

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General plane motion. In this case, the

body undergoes both translation and

rotation. Translation occurs within a plane and rotation occurs about an axis

perpendicular to this plane.

Rotation about a fixed axis. In this case, all

the particles of the body, except those on

the axis of rotation, move along circular paths in planes perpendicular to the axis of

rotation.

PLANAR RIGID BODY MOTION (continued)

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PLANAR RIGID BODY MOTION (continued)

The piston (C) undergoes rectilinear translation since it is

constrained to slide in a straight line. The connecting rod (D)

undergoes curvilinear translation, since it will remain horizontal asit moves along a circular path.

The wheel and crank (A and B)

undergo rotation about a fixed axis. In this case, both axes of

rotation are at the location of the pins and perpendicular to the plane

of the figure.

An example of bodies undergoing

the three types of motion is shown

in this mechanism.

The connecting rod (E) undergoes general plane motion, as it will

both translate and rotate.

C

E

A

D

B

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TRANSLATION AND

ROTATION ABOUT A

FIXED AXIS

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RIGID-BODY MOTION: TRANSLATION

The positions of two points A and B

on a translating body can be related by

r B = r A + r B/Awhere r A & r B are the absolute position

vectors defined from the fixed x-y

coordinate system, and r B/A is the

relative-position vector between B and A.

Note, all points in a rigid body subjected to translation

move with the same velocity and acceleration.

The velocity at B is vB = vA+ dr B/A/dt .

Now dr B/A/dt = 0 since r B/A is constant. So, vB = vA, and by

following similar logic, aB = aA.

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RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS

The change in angular position, d θ , is called the

angular displacement, with units of either

radians or revolutions. They are related by

1 revolution = 2π radians

When a body rotates about a fixed axis, any

point P in the body travels along a circular

path. The angular position of P is defined by θ .

Angular velocity, ω, is obtained by taking thetime derivative of angular displacement:

ω = d θ /dt (rad/s) +Similarly, angular acceleration is

α = d2θ/dt2 = dω/dt or α = ω(dω/dθ) + rad/s2

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If the angular acceleration of the body is

constant, α αα α = α αα α C, the equations for angular

velocity and acceleration can be integratedto yield the set of algebraic equations

below.

θO and ωO are the initial values of the body’sangular position and angular velocity. Note

these equations are very similar to the

constant acceleration relations developed for

the rectilinear motion of a particle.

RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS

(continued)

ω = ωO + αCt

θ = θO + ω Ot + 0.5 αCt2

ω2 = (ωO)2 + 2 αC (θ – θO)

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In the vector formulation, the magnitudeand direction of v can be determined

from the cross product of ω ωω ω and r p .

Here r p

is a vector from any point on the

axis of rotation to P.

RIGID-BODY ROTATION: VELOCITY OF POINT P

˙

˙̇̇̇

Using polar coordinates, radial and

transverse components of velocity of P

can be expresses as

vr = r = 0 and vθ = r θ ˙̇̇̇˙̇̇̇

As θ = ω ; therefore,˙̇̇̇

v = ωr

v = ω ωω ω x r p = ω ωω ω x r

The direction of v is determined by the

right-hand rule.

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The acceleration of P is expressed in terms of

its normal (an) and tangential (at) components

The tangential component, at, represents thetime rate of change in the velocity's

magnitude. It is directed tangent to the path of

motion.

The normal component, an, represents the time

rate of change in the velocity’s direction. It is

directed toward the center of the circular path.

RIGID-BODY ROTATION: ACCELERATION OF POINT P

at = dv/dt and an =v2 / ρ

but ρ = r ; v = ω r and α = d ω /dt

Therefore, at = α r and an= ω 2r

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Using the vector formulation, the acceleration

of P can also be defined by differentiating the

velocity.

a = dv/dt = dω ωω ω /dt x r P + ω ωω ω x dr P/dt

= α αα α x r P + ω ωω ω x (ω ωω ω x r P)

It can be shown that this equation reduces to

RIGID-BODY ROTATION: ACCELERATION OF POINT P

(continued)

The magnitude of the acceleration vector is a = (at)2 + (an)

2

a = α αα α x r – ω 2r

= at + an

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ROTATION ABOUT A FIXED AXIS: PROCEDURE

• Establish a sign convention along the axis of rotation.

• Alternatively, the vector form of the equations can be used(with i , j , k components).

v = ω ωω ω x r P = ωωωω x r a = a

t

+ an

= α αα α x r P

+ ω ωω ω x (ω ωω ω x r P

) = α αα α x r – ω2r

• If α is constant, use the equations for constant angularacceleration.

• If a relationship is known between any two of the variables (α,ω, θ, or t), the other variables can be determined from the

equations: ω = dθ/dt α = dω/dt α dθ = ω dω

• To determine the motion of a point, the scalar equations v = ω r,

at = α r, an = ω2r , and a = (at)

2 + (an)2 can be used.

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EXAMPLE 1

Given: Starting from rest when s = 0, pulley

A (r A = 50 mm) is given a constant

angular acceleration, αA = 6 rad/s2.

Pulley C (r C = 150 mm) has an innerhub D (r D = 75 mm) which is fixed

to C and turns with it.

Find:The speed of block B when it has risen s = 6 m.

Plan: 1) The angular acceleration of pulley C (and hub D) can be

related to αA if it is assumed the belt is inextensible anddoes not slip.

2) The acceleration of block B can be determined by using

the equations for motion of a point on a rotating body.

3) The velocity of B can be found by using the constant

acceleration equations.

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EXAMPLE 1 (continued)

Solution:

Since C and D turn together, αD = αC = 2 rad/s2

1) Assuming the belt is inextensible and does not slip, it will have

the same speed and tangential component of acceleration as it

passes over the two pulleys (A and C). Thus,

at = αAr A = αCr C => (6)(50) = αC(150) => αC = 2 rad/s2

2) Assuming the cord attached to block B is inextensible and

does not slip, the speed and acceleration of B will be the same

as the speed and tangential component of acceleration along

the outer rim of hub D:

aB = (at)D = αDr D = (2)(0.075) = 0.15 m/s2

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3) Since αA is constant, αD and aB will be constant. The constantacceleration equation for rectilinear motion can be used to

determine the speed of block B when s = 6 m (so = vo = 0):

(vB)2 = (vo)

2 + 2aB(s – so) +

(vB)2 = 0 + 2(0.15)(6 – 0)

vB = 1.34 m/s

EXAMPLE 1 (continued)

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GENERAL PLANE MOTION

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Absolute Motion Analysis

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PROCEDURE FOR ANALYSIS

The absolute motion analysis method (also called the parametric method) relates the position of a point, P, on a

rigid body undergoing rectilinear motion to the angular

position, θ (parameter), of a line contained in the body.(Often this line is a link in a machine.) Once a relationship

in the form of sP = f(θ) is established, the velocity andacceleration of point P are obtained in terms of the angular

velocity, ω, and angular acceleration, α, of the rigid body bytaking the first and second time derivatives of the positionfunction. Usually the chain rule must be used when taking

the derivatives of the position coordinate equation.

Absolute Motion Analysis

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Given:Two slider blocks are connected

by a rod of length 2 m. Also,

vA = 8 m/s and aA = 0.

Find: Angular velocity, ω, andangular acceleration, α, of therod when θ = 60°.

Plan: Choose a fixed reference point and define the position of

the slider A in terms of the parameter θ. Notice from the position vector of A, positive angular position θ ismeasured clockwise.

EXAMPLE 2

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Relative motion analysis

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RELATIVE MOTION ANALYSIS: DISPLACEMENT

When a body is subjected to general plane motion, it undergoes acombination of translation and rotation.

=

dr B = dr A + dr B/A

Disp. due to translation and rotation

Disp. due to translation

Disp. due to rotation

Point A is called the base point in 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 can be written:

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The velocity at B is given as : (dr B/dt) = (dr A/dt) + (dr B/A/dt) or

vB = vA + vB/A

RELATIVE MOTION ANALYSIS: VELOCITY

= +

Since the body is taken as rotating about A,vB/A = dr B/A/dt = ω ωω ω x r B/A

Here ω ωω ω will only have a k component since the axis of rotation

is perpendicular to the plane of translation.

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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.Here both points A and B have

circular motion since the disk and

link BC move in circular paths.

The directions of vA and vB are

known since they are always

tangent to the circular path of

motion.

vB = vA + ω ωω ω x r B/A

RELATIVE MOTION ANALYSIS: VELOCITY (continued)

(16-16)

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Furthermore, point B at the center of the wheel moves along a

horizontal path. Thus, vB has a known direction, e.g., parallel

to the surface.

vB = vA + ω ωω ω x r B/A

RELATIVE MOTION ANALYSIS: VELOCITY (continued)

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.

ROLLING WITHOUT SLIPPING

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PROCEDURE FOR ANALYSIS

The relative velocity equation can be applied using either aCartesian vector analysis or by writing scalar x and y component

equations directly.

Vector Analysis:

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

direction of the vector is opposite to that assumed.

2. Express the vectors in Cartesian vector form and substitute

into vB = vA + ω ωω ω x r B/A. Evaluate the cross product and

equate respective i and j components to obtain two scalar

equations.

1. Establish the fixed x-y coordinate directions and draw the

kinematic diagram of the body, showing the vectors vA, vB,

r B/A and ω ωω ω . If the magnitudes are unknown, the sense of

direction may be assumed.

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PROCEDURE FOR ANALYSIS (continued)

3. Write the scalar equations from the x and y components of

these graphical representations of the vectors. Solve forthe unknowns.

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

kinematic diagram for the body. Then establish the

magnitude and direction of the relative velocity vector vB/A.

Scalar Analysis:

2. Write the equation vB = vA + vB/A and by using the kinematic

diagram, underneath each term represent the vectorsgraphically by showing their magnitudes and directions.

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Example 3: Determining Velocities

and Angular Velocities

• Question

Bar AB in the figure below rotates with aclockwise angular velocity of 10 rad/s.

Determine the angular velocity of bar BC

and the velocity of point C.

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• Strategy

Bar AB rotates about the fixed point A witha known angular velocity, so we can

determine the velocity of B. Then, by

expressing the horizontal velocity of C in

terms of the velocity of B and the angular

velocity of bar BC, we can obtain twoequations in terms of the velocity of C and

the angular velocity of bar BC.

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and Angular Velocities• Solution

Because the angular velocity of bar AB isgiven and point A is stationary, we canuse Eq. 16-16 to determine the velocity ofpoint B. From Fig a, the position vector ofB relative to A is r B/A = 0.4i + 0.4j (m).

Fig a

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The angular velocity vector of bar AB is ω AB

= -10k (rad/s), so the velocity of B is

m/s j4i4

04.04.0

10000

r vv /

−=

−+=

×+=

k ji

A B A B ω

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and Angular VelocitiesWe can use Eq. 16-16 to express the velocity of

point C in terms of the velocity of point B. Let ωBCbe the counterclockwise angular velocity of bar

BC (Fig b), so that the angular velocity vector of

bar BC is ωBC = ωBC k. Because point C is

moving horizontally, we can write its velocity asvC = vC i.

Fig b

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( ) ( ) j8.04i4.04

04.08.0

00

k ji

j4i4iv

r vv /

BC C

BC C

BC BC BC

ω ω

ω

ω

−−+=

−+−=

×+=

Therefore, we have

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Equating the i and j components in this

equation, we obtain

BC

BC C v

ω

ω

8.040

,4.04

−=

+=

Solving it, we have vC = 6 m/s and ωBC = 5 rad/s.

the angular velocity of bar BC is ωBC

= 5k (rad/s),

and the velocity of point C is vC = 6i m/s.

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INSTANTANEOUS CENTER(IC) OF ZERO VELOCITY

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INSTANTANEOUS CENTER OF ZERO VELOCITY

For any body undergoing planar motion, there always exists a

point in the plane of motion at which the velocity is

instantaneously zero (if it were rigidly connected to the body).

This point is called the instantaneous center of zero velocity,

or IC. It may or may not lie on the body!

If the location of this point can be determined, the velocity

analysis can be simplified because the body appears to rotate

about this point at that instant.

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LOCATION OF THE INSTANTANEOUS CENTER

To locate the IC, we can use the fact that the velocity of a point

on a body is always perpendicular to the relative position vector

from the IC to the point. Several possibilities exist.

First, consider the case when velocity vA of a

point A on the body and the angular velocity ω ωω ω

of the body are known.

In this case, the IC is located along the line

drawn perpendicular to vA at A, a distance

r A/IC = vA/ω from A. Note that the IC lies upand to the right of A since vA must cause a

clockwise angular velocity ω ωω ω about the IC.

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A second case is when the

lines of action of two non-

parallel velocities, vA and

vB, are known.

First, construct line

segments from A and B perpendicular to vA and vB.

The point of intersection of

these two line segments

locates the IC of the body.


(continued)

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(continued)

A third case is when the magnitude and direction of two

parallel velocities at A and B are known.

Here the location of the IC is determined by proportionaltriangles. As a special case, note that if the body is

translating only (vA = vB), then the IC would be located at

infinity. Then ω equals zero, as expected.

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VELOCITY ANALYSIS

The velocity of any point on a body undergoing general plane

motion can be determined easily once the instantaneous center

of zero velocity of the body is located.

Since the body seems to rotate about the

IC at any instant, as shown in this

kinematic diagram, the magnitude of

velocity of any arbitrary point is v =ω

r ,where r is the radial distance from the IC

to the point. The velocity’s line of action

is perpendicular to its associated radial

line. Note the velocity has a sense ofdirection which tends to move the point in

a manner consistent with the angular

rotation direction.

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Example 4: Linkage Analysis by

Instantaneous Centers

• Question

Bar AB in the Figure rotates with acounterclockwise angular velocity of 10

rad/s. what are the angular velocities of

bars BC and CD?

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• Strategy

Because bars AB and CD rotate aboutfixed axes, we know the directions of

motion of points B & C so we can locate

the instantaneous center of bar BC.

Beginning with bar AB, we can use the

instantaneous centers of the bars todetermine both the velocities of the points

where they are connected and their

angular velocities.

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• Solution

The velocity of B due to the rotation of bar AB about A (Fig a) is

( )( ) m/s20rad/s10m2 == B

v

Fig a

E l 4 Li k A l i b

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Drawing lines perpendicular to the

directions of motion of B & C, we locate theinstantaneous center of bar BC (Fig b). The

velocity of B is equal to the product of its

distance from the instantaneous center ofbar BC and the angular velocity ωBC:

Hence, ωBC = 10 rad/s with bar BC rotating

in the clockwise direction.

( ) C Bv ω m2m/s20 ==

E l 4 Li k A l i b

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Fig b

E l 4 Li k A l i b

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Using the instantaneous center of bar BC

and its angular velocity ωBC, we candetermine the velocity of point C:

m/s810m8 == C C v ω

Lastly is to use the velocity of C to determine the

angular velocity of bar CD about point D (Fig c).

We have

ckwisecounterclorad/s10

m8m/s810

=∴==

CD

CDC v

ω

ω

E l 4 Li k A l i b

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Fig c

ATTENTION QUIZ

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ATTENTION QUIZ

2. Point A on the rod has a velocity of 8 m/s to the right.

Where is the IC for the rod?

A) Point A.

B) Point B.

C) Point C.D) Point D.

• C

D •

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RELATIVE MOTION ANALYSIS:ACCELERATION

RELATIVE MOTION ANALYSIS ACCELERATION

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RELATIVE MOTION ANALYSIS: ACCELERATION

The equation relating the accelerations of two points on the

body is determined by differentiating the velocity equation

with respect to time.

The result is aB = aA + (aB/A)t + (aB/A)n

These are absolute accelerations

of points A and B. They are

measured from a set of fixed

x,y axes.

This term is the acceleration

of B with respect to A.

It will develop tangentialand normal components.

/+dt

dv AB

dt

dvA

dt

dvB =

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The relative normal acceleration component (aB/A)n is (-ω2 r B/A)

and the direction is always from B towards A.


= +

Graphically: aB = aA + (aB/A)t + (aB/A)n

The relative tangential acceleration component (aB/A

)t

is (α αα α x r B/A

)

and perpendicular to r B/A.


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Since the relative acceleration components can be expressed

as (aB/A)t =α ×

r B/A and (aB/A)n = - ω2

r B/A the relativeacceleration equation becomes

aB = aA + αααα × r B/A - ω2 r B/A

Note that the last term in the relative acceleration

equation is not a cross product. It is the product of a

scalar (square of the magnitude of angular velocity,ω2) and the relative position vector, r B/A.


(continued)

(16-18)

APPLICATION OF RELATIVE ACCELERATION

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APPLICATION OF RELATIVE ACCELERATION

EQUATIONIn applying the relative acceleration equation, the two points used in the

analysis (A and B) should generally be selected as points which have a

known motion, such as pin connections with other bodies.

Point C, connecting link BC and the piston, moves along a straight-line

path. Hence, aC is directed horizontally.

In this mechanism, point B is known to travel along a circular path, so

aB can be expressed in terms of its normal and tangential components.

Note that point B on link BC will have the same acceleration as point B

on link AB.

PROCEDURE FOR ANALYSIS: RELATIVE

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PROCEDURE FOR ANALYSIS: RELATIVE

ACCELERATION ANALYSIS1. Establish a fixed coordinate system.

2. Draw the kinematic diagram of the body.

3. Indicate on it aA, aB, ω ωω ω , α αα α , and r B/A. If the points A and Bmove along curved paths, then their accelerations should

be indicated in terms of their tangential and normal

components, i.e., aA = (aA)t + (aA)n and aB = (aB)t + (aB)n.4. Apply the relative acceleration equation:

aB = aA + α × r B/A - ω2 r B/A

5. If the solution yields a negative answer for an unknown

magnitude, it indicates the sense of direction of the vector

is opposite to that shown on the diagram.

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Example 5: Acceleration of a Point• Question

The rolling disk in the Figure below has

counterclockwise angular velocity ω and

counterclockwise angular acceleration α.

What is the acceleration of point A?

E l 5 A l ti f P i t

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Example 5: Acceleration of a Point

• Strategy

We know that the magnitude of the

acceleration of the center of the disk is theproduct of the radius and the angular

acceleration. Therefore, we can express

the acceleration of A as the sum of the

acceleration of the center of the disk and

the acceleration of A relative to the center.

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Example 5: Acceleration of a Point• Solution

First Method. In terms of the coordinate

system in Fig a, the acceleration of thecenter B is aB = -αRi. A’s motion in a

circular path of radius R relative to B

results in the tangential and normal

components of relative acceleration shown

in fig b:

ia 2/ R R B α ω +−=

E l 5 A l ti f P i t

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The acceleration of A is

ji jiiaaa

2

2

/

R R R R R R B A B A

α ω α α ω α

+−−=+−−=+=

Fig a Fig b

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Example 5: Acceleration of a PointSecond Method. The angular acceleration vector

of the disk is α = αk, and the position of A relative

to B is r A/B = Ri (Fig c).

Fig c

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From equation of acceleration, the

acceleration of A is

( ) ( ) ( )

jR i

iik i

r r aa

2

2

/2

/

α ω α

ω α α

ω α

+−−=

−×+−=

−×+=

R R

R R R

B A B A B A

Example 6: Angular Accelerations of

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p g

Members of a Linkage• Question

Bar AB in the figure below has a

counterclockwise angular velocity of 10 rad/sand a clockwise angular acceleration of 300rad/s2. What are the angular accelerations ofbars BC and CD?


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p g

Members of a Linkage

• Strategy

Since we know the angular velocity of bar AB, we can determine the velocity of point

B. We can apply Eq. 16-16 to points C & D

to obtain an equation for vC in terms of theangular velocity of bar CD. We can repeat

the procedure to points B & C to obtain an

equation for vC in terms of the angular

velocity of bar BC.


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p g


By equating the two expressions for vC, we

will obtain a vector equation in twounknowns: the angular velocities of bars

BC and CD. Then, by following the same

sequence of steps, but using formula for

acceleration (Eq. 16-18) we can obtain the

angular accelerations of bars BC and CD.


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p g

Members of a Linkage• Solution

The velocity of B is

( ) ( )

( )m/si20

j2k 100

r vv /

−=

×+=×+= A B AB A B ω

Fig a


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p g


Fig b


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g

Members of a LinkageLet ωCD be the unknown angular velocity of

bar CD (Fig b). The velocity of C in terms ofthe velocity of D is

j2i2

022

00

k ji

0

r vv /

CDCD

CD

DC CD DC

ω ω

ω

ω

−−=

−

+=

×+=

Example 17.6: Angular Accelerations

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of Members of a Linkage

Fig c


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Members of a LinkageDefining the angular velocity of bar BC by

ωBC

(Fig c), we obtain the velocity of C in

terms of the velocity of B:

( ) ( ) j2i20

i2k i20

r vv /

C

BC

BC BC BC

ω ω

ω

+−=

×+−=

×+=

Equating both eqns for vC yields

j2i20 j2i2 BC CDCD ω ω ω +−=−−


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Members of a LinkageEquating the i and j components, we obtain

ωCD = 10 rad/s and ωBC = -10 rad/s.Using the same sequence to determine the

angular accelerations, the acceleration of B is

(Fig a) is

( ) ( ) ( ) ( )(m/s) j200i600

j210 j2k 3000

r r aa

2

/2

/

−=−×−+=

−×+= A B AB A B AB A B ω α


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Members of a LinkageThe acceleration of C in terms of the

acceleration of D is (Fig b)

( ) ( )

( ) ( ) (m/s) j2200i2200

j2i210

022

00k ji

0

r r aa

2

/2

/

CDCD

CD

DC CD DC CD DC

α α

α

ω α

+−−=

+−−

−

+=

−×+=


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Members of a LinkageThe acceleration of C in terms of the

acceleration of B is (Fig c)

( ) ( ) ( ) ( )( ) (m/s) j2200i400

i210i2k j200i600

r r aa

2

/2

/

BC

BC

BC BC BC BC BC

α

α

ω α

−−=−−×+−=

−×+=


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Members of a LinkageEquating the expressions for aC, we obtain

( ) ( ) ( ) j2200i400 j2200i2200 C CDCD α α α −−=+−−

Equating i and j components, we obtain theangular accelerations αBC = 100 rad/s

2 and αCD =

-100 rad/s2 .

ATTENTION QUIZ

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1. Two bodies contact one another withoutslipping. If the tangential component of

the acceleration of point A on gear B is

100 cm/sec2, determine the tangential

component of the acceleration of point

A’ on gear C.

A) 50 cm/sec2 B) 100 cm/sec2

C) 200 cm/sec2 D) None of above.

2. If the tangential component of the acceleration of point A on

gear B is 100 cm/sec2, determine the angular acceleration of

gear B.

A) 50 rad/sec2 B) 100 rad/sec2

C) 200 rad/sec2 D) None of above.

2 m1 m

Review Ch16

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