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ME 352 - M

Summer Sem

EXAM 2. O

Use the blanfigures provierror. Staple

Problem 1 (order kinema

3 9.369′′θ = −mechanism arolling witho(i) The first a(ii) The radiu(iii) The Carthe given XY

Machine Desi

mester 2013

OPEN BOOK

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25 Points). Fatic coeffici

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the paper one followed wnd of Problem

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Determine:

CX ,′ CY ,′ X

point C. Ex

chanism.

__________

___________

day, July 17

nly. You canwill assumedm 1.

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ME 352 - M

Summer Sem

Problem 2 (reciprocatingfollower sysdiameter of t

Cam

Tab

At the ca(i) The first-o(ii) The cooanswers in th(iii) The radi(iv) The pres

Machine Desi

mester 2013

(25 Points).g roller follostem is showthe roller fol

m Angle (deg

0° - 15° 15° - 105°

105° - 135135° - 195195° - 210210° - 300300° - 315315° - 360

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Figure 2

o,150= detercond-order kthe point o

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se and fall, ale 1. A sketciameter of te eccentricity

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ME 352 - M

Summer Sem

Problem 3 velocity of

3 110.8α = −

The radius oon link 2 is moments of

2GI 0.105=

link 2 and th(i) Draw thediagram and (ii) Determin(iii) Determi

Machine Desi

mester 2013

(25 Points)the input

25 k rad/sec

of the wheel 75 NCF i=

f inertia of l2N ms , and

e wheel is roe free body d

list all the une the torquene the coeffi

ign I

3

. For the mlink 2 is

, and the acc

is 3 5 mmρ =

N and there link 2 and

d 3GI 0.22=

olling withoudiagrams forunknown vare acting on lificient of fric

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celeration of

m, 2 3O G =

is an unknothe wheel a

225 N m s . Tut slipping or links 2 andriables. ink 2. Give tction between

Figure 3. A3

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Lab Sec

n the positiorad / s, the

f the center o

20 mm, andown torque about their m

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the magnitudn the wheel

A planar mech

of Student__

ction Numb

on shown ine angular

of mass of w

d 2O C 35 m=

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e center of td link. he three sca

de and the diand the grou

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n Figure 3, acceleration

wheel 3 is A

mm. The foon link 2. Trs are 2m =

the wheel is

alar equation

irection of thund.

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the constantn of the w

3GA 0.55= +

orce acting aThe masses a

15 kg,= 3m

sliding in th

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his vector.

_______

________

t angular wheel is

2i m / s .

at point C and mass

25 kg,=

he slot in

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ME 352 - M

Summer Sem

Problem 4 (s acting on

equilibrium, nto the pape

Performany force po(i) Clearly id(ii) Draw the(iii) Determi(vi) Determin

Figure

Machine Desi

mester 2013

(25 Points). n link 4 abo

a horizontaler and the eff

m a static forclygons that y

dentify the me free body dne magnitudne the magn

4. The mech

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3

For the meout the shaftl external fofects of frictce analysis oyou need to

members in thdiagrams for des, directionnitude and di

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echanism in ft 4O , that orce CF is action in the mof the mechadraw is 1 mmhe mechanislinks 2, 3, a

ns, and locatrection of th

atic equilibri4

Name o

Lab Sec

the position is, 14T = −cting at poin

mechanism caanism using m representssm (i.e., two-and 4. tions of all inhe external fo

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of Student__

ction Numb

shown in F5 k Nm. To

nt C on link an be neglecthe graphica

s 20 N. -force memb

nternal reactorce acting a

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ber________

Figure 4, the o hold the 3. Assume ted. al method. A

bers, three-fo

tion forces. at point C.

n full scale (

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1 mm = 1 m

_______

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nal torque in static is acting

scale for

rs, etc.).

mm).

S(

T

w

a

T

Solution to P(i) A set of v

The vector lo

where the veThe X an

and

The angles fo

Problem 1. vectors for a

oop equation

ector 7R is tnd Y compon

2R

2R

for the given

2 120θ = °

velocity and

Figure 1

n (VLE) can

the arm and tnents of the

2 2 3cos Rθ +

2 2 3sin Rθ +

position of t

,° 3θ

d acceleratio

1(a). Suitable

be written a

√ I √ ? √

2 3R R R+ +

the vector RVLE, see Eq

3 3 7cos Rθ +

3 7sin Rθ +

the mechani

210 ,= °

5

n analysis o

e vectors for

as

√ √ ? √ √

7 9R R R+ +

9R follows thquation (1),

7 7 9cos Rθ +

7 9sin R sθ +

ism are

9 270θ =

f the mechan

r the mechan

√

1R 0=

he path of piare

9 9 1cos Rθ +

9 1sin R siθ +

0 ,° and

nism are sho

nism.

in B.

1cos 0θ =

1in 0θ =

d 7θ =

own in Figur

180= °

re 1(a).

(1)

(2a)

(2b)

(3)

6

Differentiating Equations (2) with respect to the input position 2θ gives

2 2 3 3 3 9 9R sin R sin R cos 0′ ′− θ − θ θ + θ = (4a) and

2 2 3 3 3 9 9R cos R cos R sin 0′ ′+ θ + θ θ + θ = (4b) Writing Equations (4) in matrix form gives

3 3 9 3 2 2

3 3 9 9 2 2

R sin cos R sinR cos sin R R cos

′− θ + θ θ + θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′+ θ + θ − θ⎣ ⎦⎣ ⎦ ⎣ ⎦

(5)

Substituting the known data into Equation (5) gives

3

9

400sin(210 ) mm cos(270 ) rad 500sin(120 ) mmR400cos(210 ) mm sin(270 ) rad 500cos(120 ) mm′θ− ° ° + °⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′+ ° ° − °⎣ ⎦ ⎣ ⎦⎣ ⎦ (6a)

which simplifies to

3

9

200 mm 0 rad 433.01 mmR346.41 mm 1 rad 250 mm′θ+ +⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′− − +⎣ ⎦ ⎣ ⎦⎣ ⎦ (6b)

The determinant of the coefficient matrix in Equation (6b) is

DET ( 200 mm)( 1 rad) 200 mm= + − = − (7a) Using Cramer’s rule, the first-order kinematic coefficient for the angle of link 3 is

3433.01 mm 2.165 rad / rad

200 mm−′θ = = +−

(8)

The positive sign indicates that link 3 is rotating counterclockwise for a counterclockwise rotation of the input link 2.

From Cramer’s rule, the first-order kinematic coefficient for point B is

29

200 250 433.01 346.41 mmR 999.995 mm / rad200 mm

+ × + ×′ = = −−

(9a)

Therefore, the first-order kinematic coefficient for point B is

9R 1.0 m / rad′ = − (9b)

The negative sign indicates that the length of the vector 9R is becoming shorter for a counterclockwise rotation of the input link 2, that is point B is moving downward.

The rolling contact constraint between link 4 and the ground link 1 can be written as

9 4 4 7( )R ρ θ θ± Δ = Δ −Δ (10)

7

The correct sign is positive because as the wheel 4 rotates clockwise the magnitude of the vector 9R becomes shorter. Also, the change in the angular position of the vector 7R is zero, that is, 7 0,θΔ = therefore, Equation (10) can be written as

9 4 4R ρ θ+ Δ = Δ (11)

Differentiating Equation (11) with respect to the input position 2θ gives

9 4 4R ρ θ′ ′+ = (12a) Then rearranging this equation, the first-order kinematic coefficient for link 4 can be written as

94

4

Rθρ′

′ = + (12b)

Substituting Equation (9b) and the radius of link 4 into Equation (12b), the first-order kinematic coefficient for link 4 is

4( 999.995 mm) 7.143 rad/rad

140 mmθ −′ = + = − (13)

The negative sign indicates that link 4 is rotating clockwise as the input link 2 rotates counterclockwise.

Differentiating Equations (4) with respect to the input position θ2 gives

22 2 3 3 3 3 3 3 9 9cos cos sin cos 0R R R Rθ θ θ θ θ θ′ ′′ ′′− − − + = (14a)

and 2

2 2 3 3 3 3 3 3 9 9sin sin cos sin 0R R R Rθ θ θ θ θ θ′ ′′ ′′− − + + = (14b) Then writing Equations (14) in matrix form gives

23 3 9 3 2 2 3 3 3

23 3 9 9 2 2 3 3 3

sin cos cos coscos sin sin sin

R R RR R R R

θ θ θ θ θ θθ θ θ θ θ

′′− ⎡ ′ ⎤+⎡ ⎤ ⎡ ⎤= ⎢ ⎥⎢ ⎥ ⎢ ⎥′′ ′+⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(15a)

Substituting the known values into Equation (15a) gives

( )( )

23

29

500cos(120 ) 400cos(210 ) 2.165 mm400sin(210 ) mm cos(270 ) radR400cos(210 ) mm sin(270 ) rad 500sin(120 ) 400sin(210 ) 2.165 mm

⎡ ⎤′′ ° + ° +θ− ° ° ⎡ ⎤⎡ ⎤ ⎢ ⎥=⎢ ⎥⎢ ⎥ ′′+ ° ° ⎢ ⎥⎣ ⎦ ⎣ ⎦ ° + ° +⎣ ⎦ (15b)

which simplifies to

3

9

200 mm 0 rad 1873.70 mmR346.41 mm 1 rad 504.40 mm′′θ+ −⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′′− − −⎣ ⎦ ⎣ ⎦⎣ ⎦ (15c)

The determinant of the coefficient matrix in Equation (15c) can be written as

200 0200 mm

346.41 1DET

+= = −− −

(16)

Note that the determinant in Equation (16) is the same as the determinant in Equation (7a).

8

Using Cramer’s rule, the second-order kinematic coefficient of link 3 can be written as

2

1873.70 0504.40 1 1873.7 9.3685 rad/rad3 200DET

θ

−− − +′′ = = = −

− (17a)

Similarly, the second-order kinematic coefficient between links 3 and 4 can be written as

29

200 1873.7346.41 504.4 749954.0 3749.77 mm/rad

200R

DET

+ −− − −′′ = = = +

− (17b)

From Equation (12b), the first-order kinematic coefficient for link 4 can be written as

94

4

Rθρ′′

′′ = + (18)

Substituting Equation (17b) and the radius of link 4 into Equation (18), the second-order kinematic coefficient for link 4 is

4( 3749.77 mm) 26.784 rad/rad

140 mmθ +′′ = + = + (19)

(i) 15 Points. The vector equation for point C, see Figure 1(b), can be written as

??2 33I

R R RC =√ √√

+ (1a) or as

??71 9 333R R R R RC

√√ √√ √√ √√= − − + (1b)

From the right-angle triangle ABC, the distance AC is

2 233AC R AB BC= = − (2a)

Substituting the given data into Equation (2a), the distance is

2 233 400 200 200 3 mmAC R= = − = (2b)

Also, the angle

33 180oθ = (2c) The X and Y components of Equation (1a) are

2 2 33 33cos cos 596.41 mmCX R Rθ θ= + = − (3a) and

2 2 33 33sin sin 433.01 mmCY R Rθ θ= + = + (3b)

Dc

a

Dc

a

(

R

Differentiatincoefficients o

and

Differentiatincoefficients o

and

(iii) 5 points

Recall the de

ng Equationof point C ar

ng Equationof point C ar

CX ′′ =

CY ′′ =

s. The radius

efinition

Fig

ns (3a) and (re

CX ′ =

CY ′ = +

ns (4a) and (4re

2 2cosR θ− −

2 2sinR θ= −

s of curvature

gure 1(b). Th

(3b) with re

2 2sinR θ= − −

2 2cosR θ+ +

4b) with resp

33 33cosR θ θ−

33 33sinR θ θ−

e of the path

Cρ =

CR′ =

9

he vectors fo

espect to the

33 33sinR θ θ−

33 33cosR θ θ+

pect to the i

23 33 sinRθ θ′ −

23 33 cosRθ ′ +

h of point C c

3C

C C C

RX Y Y X

′=

′ ′′ ′−

2C CX Y′± +

or point C.

e input posit

3 433.01θ ′ = −

3 999.98θ ′ = −

input positio

33 3 187θ θ ′′ = +

33 3s 28θ θ ′′ = +

can be writte

CX ′′

2CY ′

tion θ2, the

1 mm/rad

8 mm/rad

on θ2, the sec

73.7 mm/rad

812.5 mm/ra

en as

first-order k

cond-order k

2d

2d

kinematic

(4a)

(4b)

kinematic

(5a)

(5b)

(6)

(7a)

10

where the positive sign must be used here because the input angular velocity is shown as positive in the given figure (that is, counterclockwise). Substituting Equations (4a) and (4b) into Equation (7a) gives

2 2( 433.01) ( 999.98) 1089.7 mmCR′ = + − + − = + (7b)

Substituting Equations (4), (5), and (7b) into Equation (6), the radius of curvature of the path of point C can be written as

3( 1089.7 mm)( 433.01 mm)( 2812.5 mm) ( 999.98 mm)( 1873.7 mm)Cρ

+=

− + − − + (8a)

Therefore, the radius of curvature of the path of point C is

1973.1 mm 1.973 mCρ = + = + (8b) The positive sign indicates that the unit normal vector to the path of point C is pointing towards the center of curvature of the path of point C, see Figure 1(c).

The coordinates of the center of curvature of the path of point C can be written as

CC C

C

YXcc XR

ρ⎡ ⎤′

= − ⎢ ⎥′⎣ ⎦ (9a)

and

CC C

C

XYcc YR

ρ⎡ ⎤′

= + ⎢ ⎥′⎣ ⎦ (9b)

Substituting Equations (2), (4), (5), and (7b) into Equations (9), the coordinates of the center of curvature of the path of point C can be written as

999.98 mm596.41 mm ( 1973.1 mm)1089.7 mm

Xcc⎡ ⎤−

= − − + ⎢ ⎥+⎣ ⎦ (10a)

and 433.01 mm433.01 mm ( 1973.1 mm)1089.7 mm

Ycc⎡ ⎤−

= + + + ⎢ ⎥+⎣ ⎦ (10b)

which can be written as 596.41 mm 1810.65 mmXcc = − + (11a)

and 433.01 mm 784.04 mmYcc = + − (11b)

Therefore, the coordinates of the center of curvature of the path of point C are

1214.2 mm 1.214 mXcc = + = + (12a) and

351.0 mm 0.351 mYcc = − = − (12b) The Cartesian coordinates of the center of curvature of the path of point C are as shown in Figure 1(c).

Figure 11(c). The rad

dius and cent

11

ter of curvatture of the paath of point C.

12

The velocity of point C can be written as

2ˆ ˆV ( )C C CX i Y j ω′ ′= + (5a)

Substituting Equations (3) and the given angular velocity of the input link 2 into Equation (5a), the velocity of point C is

ˆ ˆ ˆ ˆV ( 433.01 mm 999.98 mm )( 50 rad / sec) 21651 49999 mm / secC i j i j= − − + = − − (5b) The magnitude and direction of the velocity of point C are

V 54485 mm/ secC = and V 246.6C∠ = ° (6)

The acceleration of point C can be written as

22 2

ˆ ˆ ˆ ˆA ( ) ( )C C C C CX i Y j X i Y jω α′′ ′′ ′ ′= + + + (7a) Substituting Equations (4) and the given angular velocity and acceleration of the input link 2 into Equation (7a), the acceleration of point C is

2ˆ ˆ ˆ ˆA ( 1873.7 mm 433.01 mm ) ( 50) ( 3678.9 mm 999.98 mm ) (0)C i j i j= + − + + − − (7b) or

2ˆ ˆA 4684300 9197200 mm / secC i j= + − (7c) The magnitude and direction of the acceleration of point C are

2A 10321000 mm/ secB = and A 63.01B∠ = − ° (8)

The unit tangent vector (which points in the direction of the velocity of point C) can be written as ˆ ˆ

ˆ C Ct

C

X i Y juR

′ ′+=

′ (9a)

Recall the definition

2 2C C CR X Y′ ′ ′= ± + (9b)

The positive sign must be used here since the input is positive. Substituting Equations (3a) and (3b) into Equation (9b) gives

2 2( 433.01) ( 999.98) 1089.7 mmCR′ = + − + − = + (9c) The positive sign must be used here because the input angular velocity is counterclockwise.

Substituting Equations (3) and (9b) into Equation (9a), the unit tangent vector is

ˆ ˆ( 433.01) ( 999.98) ˆ ˆˆ 0.397 0.9181089.7ti ju i j− + −

= = − −+

(10)

Check: The unit tangent vector can also be written as

13

VˆV

Ct

C

u = (11a)

Substituting Equations (5b) and (6) into Equation (11a) gives

ˆ ˆ21651 49999 ˆ ˆˆ 0.397 0.91854485t

i ju i j− −= = − − (11b)

The unit normal vector (which is 90° counterclockwise from the unit tangent vector) can be written as

ˆ ˆˆ C C

nC

Y i X juR

′ ′− +=

′ (12a)

Substituting Equations (3) and (9b) into Equation (12a), the unit normal vector is

ˆ ˆˆ 0.918 0.397nu i j= + − (12b) The unit tangent and unit normal vectors to the path of point C are shown in Figure 1(c). The velocity and acceleration vectors of point C are also shown on the figure. Check: Recall that the unit normal vector to the path of point C must pass through the absolute instant center of the coupler link 3 (that is, 13I ), see Figure 1(c).

Figure 11(c). The vellocity and ac

14

cceleration vvectors of pooint C.

15

16

Solution to Problem 2. (i) 10 Points. The coordinates of the follower center when the cam is resting on the prime circle are

0 20 mmX ε= = (1) and

2 22 2

0150 30 20 87.75 mm

2 2D dY ε+ +⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (2)

The lift equation for full-return cycloidal motion, see Equation (6.13a), page 292, in the text book, is

* *1 2sin 2basey y L θ π θ

β π β−

⎛ ⎞= + ⎜ ⎟⎜ ⎟

⎝ ⎠ (3a)

where the starting height is

20 mmbasey = (3b) The first-order kinematic coefficient for full-rise cycloidal motion, see Equation (6.13b), is

*2 ' 1 cosLy π θβ β⎛ ⎞

= −⎜ ⎟⎜ ⎟⎝ ⎠

(3c)

Therefore, the second-order kinematic coefficient for full-rise cycloidal motion, see Equation (6.13c), is

2

*2 2 '' sinLy π π θβ β

= (3d)

Note that the third-order kinematic coefficient for full-rise cycloidal motion is

2

3

*4 2 cosLy π π θβ β

′′′ = (3e)

For the cam angle ,150oθ = the angle

* 15150 135 15 rads rads180 12

o o o π πθ = − = = = (4a)

and the angle

360 radso πβ = = (4b)

Therefore, the ratio

* 3 0.2512

θ ππβ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= = (4c)

Substituting the given data into Equation (3a), the lift for full-rise cycloidal motion is

* *1 2 120 35 sin 20 35 0.25 sin2 2 2

y θ π θ πβ π β π

⎛ ⎞ ⎛ ⎞= + − = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ (5a)

17

Therefore, the lift for full-rise cycloidal motion is

23.18 mmy = + (5b) Substituting the given data into Equation (3b), the first-order kinematic coefficient of the lift is

35 105 ' 1 cos 1 cos 33.42 mm/rad/ 3 2 2

y π ππ π

⎛ ⎞ ⎛ ⎞= − = − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6)

Substituting the given data into Equation (3c), the second-order kinematic coefficient of the lift is

22

2 (35) 210 '' sin sin 200.54 mm/rad( / 3) 2 2

y π π ππ π

= = = + (7)

(ii) 4 Points. The coordinates of the follower center (or the pitch curve) in the fixed reference frame are

20 mmPC oX X= = (8a) and

87.75 23.18 110.93 mmPC oY Y y= + = + = (8b) The coordinates of the follower center (or the pitch curve) in the moving reference frame are

5 5cos sin 20 cos 110.93 sin 38.14 cm6 6PC PC PCx X Y π πθ θ= + + = + = + (9a)

and 5 5sin cos 20sin 110.93 cos 106.07 mm6 6PC PC PCy X Y π πθ θ= − + = − + = − (9b)

The first-order kinematic coefficients of the follower center (or the pitch curve) can be written as

sin cos sinPC PC PCx X Y yθ θ θ′ ′= − + + (10a) and

cos sin cosPC PC PCy X Y yθ θ θ′ ′= − − + (10b) Substituting the given data into Equations (10a) and (10b), the first-order kinematic coefficients of the follower center (or the pitch curve) are

5 5 520sin 110.93cos ( 33.42)sin 89.36 mm/rad6 6 6PCx π π π′ = − + + + = − (11a)

and 5 5 520cos 110.93sin ( 33.42)cos 67.09 mm/rad6 6 6PCy π π π′ = − − + + = − (11b)

Recall the definition

2 2 2 2( 89.36) ( 67.09) 111.74 mmPC PC PCR x y′ ′ ′= + + = + − + − = + (11c) The second-order kinematic coefficients of the follower center (or the pitch curve) are

cos sin 2 'cos "sin′′ = − − + +PC PC PCx X Y y yθ θ θ θ (12a)

18

and sin cos 2 'sin "cosPC PC PCy X Y y yθ θ θ θ′′ = + − − + (12b)

Substituting the given data into Equations (12a) and (12b), the second-order kinematic coefficients of the follower center (or the pitch curve) are

25 5 5 520cos 110.93sin 2( 33.42)cos 200.54sin 4.23 mm/rad6 6 6 6PCx π π π π′′ = − − + + + = + (13a)

and 25 5 5 520sin 110.93cos 2( 33.42)sin 200.54cos 101.02 mm/rad

6 6 6 6PCy π π π π′′ = + − − + + = − (13b)

The coordinates of the point of contact between the follower and the cam, in the moving reference frame, can be written as

2PC

PCPC

camydx xR

⎛ ⎞′= + ⎜ ⎟′⎝ ⎠

(14a)

and

2PC

PCPC

camxdy yR

⎛ ⎞′= − ⎜ ⎟′⎝ ⎠

(14b)

Substituting the given data into Equations (14a) and (14b), the coordinates of the point of contact between the follower and the cam, in the moving body reference frame, are

30 67.0938.14 29.14 mm2 111.74camx⎛ ⎞−

= + = +⎜ ⎟+⎝ ⎠ (15a)

and 30 89.36106.07 94.07 mm2 111.74camy⎛ ⎞−

= − − = −⎜ ⎟+⎝ ⎠ (15b)

(iii) 5 Points. The radius of curvature of the pitch curve can be written as

3PC

PCPC PC PC PC

Rx y y x

ρ′

=′ ′′ ′ ′′−

(16a)

Substituting the given data into Equation (16a), the radius of curvature of the pitch curve can be written as

3 3

2

111.74 1395166 mm( 89.36)( 101.02) ( 67.09)( 4.23) 9311 mmPCρ + +

= =− − − − + +

(16b)

Therefore, the radius of curvature of the pitch curve is

149.83 mmPCρ = + (16c)

Note that the positive sign indicates that the cam profile is concave and this may cause design problems. A more detailed study of the cam design would be required.

19

The radius of curvature of the cam profile can be written as

2PCcamdρ ρ= + (17a)

Therefore, the radius of curvature of the cam profile is

30149.83 164.83 mm2camρ = + + = + (17b)

Note that the radius of curvature of the cam profile is a smaller negative value than the radius of curvature of the of the pitch curve. (iv) 5 Points. The pressure angle of the cam can be written as

cos cos sinPC PC

PC PC

x yR R

φ θ θ′ ′

= −′ ′

(18a)

Substituting the given data into Equation (18a), the pressure angle of the cam can be written as

89.36 5 67.09 5cos cos sin111.74 6 111.74 6

π πφ⎛ ⎞ ⎛ ⎞− −

= −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ (18b)

or as cos 0.9928φ = + (18c)

Therefore, the pressure angle of the cam for the given input position is

6.90φ = ° (19) The pressure angle is inside the acceptable range, that is, the pressure angle is less than 30°.

S(

T

w

a

T

D

a

Solution to P(i) A set of v

The vector lo

where the ve

and

The angles fo

Differentiatin

and

Problem 3. vectors for a

Figure 3

oop equation

ector 7R is t

for the given

ng Equation

velocity and

3(a). Suitable

n (VLE) can

the arm. The

23R co

23R sin

position of t

23θ = θ

s (2) with re

2R′+

2R′+

d acceleratio

e vectors for

be written a

? I √ C 23 7R R+ −

e X and Y co

23 7s R coθ +

23 7n R sinθ +

the mechani

2 30θ = − °

espect to the

23 23cos Rθ −

23 23sin Rθ +

20

n analysis o

r a kinematic

as

? √ √ √ 9 1R R 0− − =

omponents o

7 9os R coθ −

7 9n R sinθ −

ism are

and

input positio

23 23R sinθ −

23 23R cosθ −

f the mechan

c analysis of

0

f Equation (

9 1os R coθ −

9 1n R sinθ −

θ

on 2θ gives

9 9R cos′− θ =

9 9R sin′− θ =

nism are sho

f the mechan

(1) are

1os 0θ =

1n 0θ =

9 0θ = °

0=

0=

own in Figur

nism.

re 3(a).

(1)

(2a)

(2b)

(3)

(4a)

(4b)

21

Writing Equations (4) in matrix form gives

23 9 23 23 23

23 9 9 23 23

cos cos R R sinsin sin R R cos

′+ θ − θ + θ⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′+ θ − θ − θ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(5)

Substituting the known data into Equation (5) gives

23

9

Rcos( 30 ) cos(0 ) 20sin( 30 ) mmRsin( 30 ) sin(0 ) 20cos( 30 ) mm′+ − ° − ° + − °⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′+ − ° − ° − − °⎣ ⎦ ⎣ ⎦⎣ ⎦ (6a)

which simplifies to

23

9

R0.866 1 10 mmR0.5 0 17.32 mm′+ − −⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′− −⎣ ⎦ ⎣ ⎦⎣ ⎦ (6b)

The determinant of the coefficient matrix in Equation (6b) is

DET ( 0.5 )( 1) 0.5= − − − = − (7a) Using Cramer’s rule, the first-order kinematic coefficient for the angle of link 3 is

2317.32 mmR 34.64 mm / rad

0.5−′ = = +

− (8)

The positive sign indicates that the length of the vector 23R is getting longer for a counterclockwise rotation of the input link 2.

From Cramer’s rule, the first-order kinematic coefficient for point B is

920 mmR 40.0 mm / rad

0.5−′ = = +−

(9)

The positive sign indicates that the length of the vector 9R is becoming longer for a counterclockwise rotation of the input link 2.

Differentiating Equations (4) with respect to the input position θ2 gives

23 23 23 23 23 23 23 23 9 9cos sin sin cos cos 0R R R R Rθ θ θ θ θ′′ ′ ′ ′′+ − − − − = (10a) and

23 23 23 23 23 23 23 23 9 9sin cos cos sin sin 0R R R R Rθ θ θ θ θ′′ ′ ′ ′′+ + + − − = (10b) Then writing Equations (10) in matrix form gives

23 9 23 23 23 23 23

23 9 9 23 23 23 23

cos cos 2 sin cossin sin 2 cos sin

R R RR R R

θ θ θ θθ θ θ θ

′′ ′− − + +⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′′ ′+ − − +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(11a)

Substituting the known values into Equation (11a) gives

23

9

Rcos( 30 ) cos(0 ) 2(34.64)sin( 30 ) 20cos( 30 ) mmRsin( 30 ) sin(0 ) 2(34.64) cos( 30 ) 20sin( 30 ) mm′′+ − ° − ° + − ° + − °⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′′+ − ° − ° − − ° + − °⎣ ⎦ ⎣ ⎦⎣ ⎦ (11b)

22

which simplifies to

3

34

0.866 1 17.32 mmR0.5 0 70.00 mm′θ+ − −⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′− −⎣ ⎦ ⎣ ⎦⎣ ⎦ (12c)

The determinant of the coefficient matrix in Equation (12c) can be written as

DET ( 0.5 )( 1) 0.5= − − − = − (13) Note that the determinant in Equation (13) is the same as the determinant in Equation (7a).

Using Cramer’s rule, the second-order kinematic coefficient of vector R23 can be written as

2

17.32 170 0 70 140 mm/rad23 0.5

RDET

− −− −′′ = = = +

− (14a)

Similarly, the second-order kinematic coefficient of the vector R9 can be written as

29

0.866 17.320.5 70

138.56 mm/radRDET

−− −

′′ = = + (14b) The vector equation for the mass center G3 can be written as

3

?? ?23G

IR R= (15)

The X and Y components of Equation (15) are

3 23 23cosGX R θ= (16a) and

3 23 23sinGY R θ= (16b) Differentiating Equations (16a) and (16b) with respect to the input position θ2, the first-order kinematic coefficients of the mass center G3 are

3 23 23 23 23cos sinGX R Rθ θ′ ′= + − (17a) and

3 23 23 23 23sin cosGY R Rθ θ′ ′= + + (17b) Differentiating Equations (17a) and (17b) with respect to the input position θ2, the second-order kinematic coefficients of the mass center G3 are

3 23 23 23 23 23 23cos 2 sin cosGX R R Rθ θ θ′′ ′′ ′= + − − (18a) and

3 23 23 23 23 23 23sin 2 cos sinGY R R Rθ θ θ′′ ′′ ′= + + − (18b) Substituting Equations (8) and (14a) and the given angular velocity and acceleration of the input link 2 into Equations (18), the second order kinematic coefficients of the mass center G3 are

23

2

3 138.56 mm/radGX ′′ = + and 23 0 mm/radGY ′′ = (19)

The acceleration of the mass center G3 can be written as

2

3 3 3 2 3 3 2ˆ ˆ ˆ ˆA ( ) ( )G G G G GX i Y j X i Y jω α′′ ′′ ′ ′= + + + (20a)

Substituting Equation (19) and the given angular velocity and acceleration of the input link 2 into Equation (20a), the acceleration of the mass center G3 is

23

ˆ ˆA ( 138.56 mm 0 mm ) ( 2) 0G i j= + − + + (20b) or

23

ˆA 554.3 mm / secG i= + (20c)

The rolling contact constraint between link 3 and the ground link 1 can be written as

9 3 3 7( )R ρ θ θ± Δ = Δ −Δ (21) The correct sign is negative because as the wheel 3 rotates counterclockwise the magnitude of the vector

9R becomes shorter. Also, the change in the angular position of the vector 7R is zero, that is, 7 0,θΔ = therefore, Equation (21) can be written as

9 3 3R ρ θ− Δ = Δ (22)

Differentiating Equation (22) with respect to the input position 2θ gives

9 3 3R ρ θ′ ′− = (23a) Then rearranging this equation, the first-order kinematic coefficient for link 3 can be written as

93

3

Rθρ′

′ = − (23b)

Substituting Equation (9) and the radius of link 3 into Equation (23b), the first-order kinematic coefficient for link 3 is

3( 40.0 mm) 8.0 rad/rad

5 mmθ +′ = − = − (24)

The negative sign indicates that link 3 is rotating clockwise as the input link 2 rotates counterclockwise.

From Equation (23b), the second-order kinematic coefficient for link 3 can be written as

93

3

Rθρ′′

′′ = − (25)

Substituting Equation (14b) and the radius of link 3 into Equation (25), the second-order kinematic coefficient for link 3 is

23

( 138.56 mm) 27.71 rad/rad5 mm

θ +′′ = − = − (26)

T

Sl

T2

(

o

S

o

The angular

Substituting link 3 is

The negative2 rotates cou

(i) 5 Points.

The sum

or as

Similarly, th

or as

acceleration

Equation (2

e sign indicaunterclockwi

The free bod

Fig

of the extern

e sum of the

n of link 3 ca

6) and the in

α

ates that the ise.

dy diagram f

gure 3(b). Th

nal forces in

e external for

an be written

nput angular

3 0 ( 2α = + −

angular acce

for link 2 is

he free body

n the X-direc

F∑

12 XF +

rces in the Y

∑

12YF +

24

n as

3 3 2α θ α θ′= +

r velocity int

27.71)( 2)+ =

eleration of

shown in Fi

diagram for

ction can be

2 2X G XF m A=

32 X CF F+ + =

Y-direction c

2 2Y G YF m A=

32 2YF W+ − =

23 2θ ω′′

to Equation

110.85 ra= −

link 3 is rot

igure 3(b).

r link 2.

written as

0=

can be writte

Y

0=

(27), the ang

2ad/sec

tating clockw

n as

gular acceler

wise as the i

(27)

ration for

(28)

input link

(1a)

(1b)

(2a)

(2b)

T

o

Ea

(

o

S

o

The sum of t

or as

Equations (1and Y compo

(ii) 5 Points.

The sum

or as

Similarly, th

or as

the external m

) through (3onents of the

. The free bo

Fig

of the extern

e sum of the

moments ab

(R

3) contain 4 ue contact for

ody diagram

gure 3(c). Th

nal forces in

e external for

out the mass

M∑

23 32 2X YR F R−

unknown varce between

32YF

for the whe

he free body

n the X-direc

F∑

13XF +

rces in the Y

∑

13YF +

25

s center of li

2 2 2G GM I α=

)23 32 0Y XF + −

ariables, namlinks 2 and 3

32 tan 6XF=

el is shown i

diagram for

ction can be

3 3X G XF m A=

23 3X GF m A+ =

Y-direction c

3 3Y G YF m A=

23 3YF W+ − =

ink 2 can be

2 12Y CR F T− +

mely: F12X, F3 are constra

60o

in Figure 3(c

r the wheel.

written as

3G X

can be writte

Y

0=

written as

2 0=

F12Y, (F32X, Fained by the

c).

n as

F32Y ), and T1equation

(3a)

(3b)

12. The X

(3c)

(4a)

(4b)

(5a)

(5b)

26

The sum of the external moments about the center of mass of the wheel can be written as

3 3 3G GM I α=∑ (6a) or as

3 13 3 30 Y X GR F I α− = (6b) Equations (4) through (6) contain 2 new unknown variables, namely: the internal forces F13X and F13Y. Therefore, there are a total of 6 equations and 6 unknown variables.

When the two components of the force between the wheel and the ground are obtained then the coefficient of friction between the wheel and the ground can be obtained from the constraint equation

13 13X YF Fμ= (7) (iii) 5 Points. The unknown variables will be obtained from Equations (1) through (7) using the Method of Inspection. The procedure is: (a) Determine the internal force F13X from Equation (6b), that is

3 313

3

GX

Y

IFRα

= − (8a)

Substituting the given data into Equation (8a), the internal force is

2 2

13(0.225 N ms )( 110.85 rad/s ) 4988.25 N

(5 mm) sin( 90 )XF −= − = −

− ° (8b)

(b) Determine the internal force F23X from Equation (4b), that is

23 3 3 13X G X XF m A F= − (9a) Substituting the given data and Equation (8b) into Equation (9a), the internal force is

23 25( 0.55) ( 4988.25) 5002.11 NXF = + − − = + (9b) (c) Determine the Y component of the contact force between links 2 and 3 from Equation (3c), that is

32 32 tan 60oY XF F= (10a)

Substituting Equation (9b) into Equation (10a), the internal force is

32 5002.11 tan 60 8663.90 NoYF = − = − (10b)

(d) Determine the Y component of the contact force between links 1 and 2 from Equation (2b), that is

12 2 32Y YF W F= − (11a) Substituting Equation (10b) into Equation (11a), the internal force is

12 15(9.81) ( 8663.90) 8811.05 NYF = − − = + (11b) (e) Determine the Y component of the contact force between links 1 and 3 from Equation (5b), that is

27

13 3 23Y YF W F= − (12a) Substituting Equation (10b) into Equation (12a), the internal force is

13 25(9.81) ( 8663.90) 8418.65 NYF = − + = − (12b) (f) Determine the torque acting on the ground link from Equation (3c), that is

( )12 2 23 32 23 32Y C X Y Y XT R F R F R F= − − (13a) Substituting Equation (10b) into Equation (12a), the torque acting on the ground link is

( )12 0.035sin( 30 ) 0.02cos( 30 )( 8663.90) 0.02sin( 30 )( 5002.11) 186.96 NmT = − ° − − ° − − − ° − = + (13b) (iv) 5 Points. The coefficient of friction between the wheel and the ground from Equation (7), can be written as

13

13

X

Y

FF

μ = (14a)

Substituting Equation (8b) and (12a) into Equation (14a), the coefficient of friction between the wheel and the ground link is

4998.25 0.598418.65

μ −= =−

(14b)

S( S((( S

(oNo

oftcdt

Solution to P(i) 6 Points.

Step I. Identi(i) Link 2 is (iii) Link 3 is(iii) Link 4 h

Step II. Draw

(ii) 6 points.opposite, andNote that theor in compre

The FBDof the two fofigure. Thenthrough poincan be negledashed line therefore, lin

Problem 4. The general

ify each linka two-force s a three-forc

has two force

w the free bo

. The FBD od collinear the sense of thession).

D of link 3 isorces 23F an

from the dent E. This foected. Therein Figure 2

nk 3 is indeed

l procedure t

k from an inimember. Thce member. es and a torq

ody diagram

of link 2 is shen the line e two forces

Figure 1. T

s shown in Fnd CF must efinition of orce must alefore, the fo. Note that d sliding (an

to the graphi

itial inspectiohe forces are The forces a

que. The forc

for each link

shown in Figof action of s is not know

The free bod

Figure 2. Sincintersect at

static equilibso be perpenrce F43 can the line of

nd not tippin

28

ical method o

on of the me12F and 32F

are ,23F 43Fces are 14F a

k beginning

gure 1. Sincethe two forc

wn at this tim

dy diagram o

ce link 3 is aa unique po

brium of a tndicular to tbe drawn oaction of th

ng) inside lin

of static forc

echanism. Th.2

, and .CF and ,34F and

with the two

e the two forces is shownme (i.e., it is

of link 2. Ful

a three forceoint. This pothree force mthe line AC on the free bhe force 43Fnk 4.

ce analysis i

he initial ans

d the torque

o force mem

rces 12F andn by the dash not known

l scale.

e member thoint is denotmember, thesince frictio

body diagra3 is inside t

s:

swers are:

is 14T .

mber.

d 32F must bhed line on thif link 2 is in

hen the lines ted as point e force F43 mon in the meam of link 3the length o

be equal, he figure. n tension

of action E on the

must pass echanism 3, see the of link 4,

fo

The FBDforces and a of the two fo

D of link 4 istorque, the t

orces are sho

Figure 2. T

s shown in Ftwo forces F

own by the d

Figure 3. T

The free bod

Figure 3. Fro14F and 34Fashed lines o

The free bod

29

dy diagram o

om the definmust be equon the figure

dy diagram o

of link 3. Ful

nition of statiual, oppositee.

of link 4. Ful

l scale.

ic equilibriu, and paralle

l scale.

um of a link el. The lines

with two of action

(

o

T

Si

T

fi

C

(iii) 9 points

or as

The distance

Substituting nternal react

The internal

force must bn Figure 4 a

Consider the

s. The sum o

e from the cr

the known tion force be

reaction forc

e as shown iand the magn

Figure

e free body d

f the momen

ankshaft O4

torque 14T =etween links

ce 34F must

in Figure 4. nitude of this

e 4. The inte

diagram of lin

nts about the

∑

d+

to the line o

d

5 Nm= and 4 and 3 is

1434

TFd

=

t oppose the

Then the ints force is

34F =

ernal reaction

nk 3, see Fig

30

e crankshaft

40OM∑ =

34 14 0d F T− =

of action of t

15.6 mmd =

d Equation (

5 Nm0.0156 m

=

action of the

ternal reactio

14 320.5F= =

n forces 34F

gure 2, repea

O4 can be w

0

the force 34F

(3) into Equ

320.5 Nm=

e external to

on force 14F

5 N

and 14F ac

ated here as

written as

4 , see Figure

uation (2b)

orque, that is

4 must act in

ting on link

Figure 5.

e 3, is measu

and rearran

s, the directio

n the directio

4.

(2a)

(2b)

ured as

(3)

ging, the

(4)

on of this

on shown

(5)

(

o

R

T

S

T

Figur

(iv) 4 Points

or as

Rearranging

The distance

Substituting

The positive The sum

re 5. The fre

s. The sum o

Equation (6

es m and n

Equations (4

sign indicatof the mom

e body diagr

f the momen

6b), the exter

are shown o

95.3m =

4), (5), and (

tes that the foments about p

ram of link 3

nts about pin

M∑

43m F−

rnal force ac

CmF =+

on the FBD o

mm

(8) into Equa

95.CF = +

force is actinpoint C can b

31

3 showing th

n A can be w

0AM =

0Cn F+ =

ting at point

43m Fn

of link 3, see

and

ation (7), the

3 (320.5)110

= +

g to the righbe written as

he distances

written as

t C on link 3

e Figure 5, an

110n =

e external fo

277.7 N+

ht, that is, in s

k, l, m, and

3 can be writ

nd are meas

0.0 mm

orce acting at

the positive

n. Full scale

tten as

ured as

t point C on

X-direction

e.

(6a)

(6b)

(7)

(8)

link 3 is

(9)

n.

o

R

T

S

T

t

or as

Rearranging

The distance

Substituting

The positive The mag

Fig

The FBDtwo forces ar

Equation (1

es k and l ar

Equations (5

sign indicatnitudes and

gure 6. The f

D of link 2 shre equal, opp

0b), the exte

re shown on

63.5 mk =

5), and (12)

tes that the rethe direction

free body dia

howing the iposite, and c

M∑

43l F+ −

ernal force a

23F =

n the FBD of

mm

into Equatio

2331F =+

eaction forcens of the two

agram of link

internal reacollinear then

32

0CM =

23 0k F− =

acting at poin

43l Fk

=+

f link 3, see F

and

on (11), the e

1.8 (320.5)63.5

=

e is acting upo forces CF

k 3 showing

ction forces n link 2 must

nt A on link

Figure 5, and

31.8l =

external forc

160.5 N= +

pward (that and 23F are

g the forces F

12F and 32Ft be in comp

3 can be wr

d are measur

8 mm

ce acting at p

is, in the poe shown on F

CF and 23F .

2 is shown inpression.

ritten as

red as

point A on li

sitive Y direFigure 6.

. Full scale.

n Figure 7. S

(10a)

(10b)

(11)

(12)

ink 3 is

(13)

ection).

Since the

C

f

Check: Using

Measurinforce at poin

g the force p

Fig

ng the lengthnt C can be w

Figure 7. T

polygon for l

gure 8. The f

hs of the vecwritten as

The free bod

link 3. The f

force polygo

ctors CF an

6980CF =

33

dy diagram o

force polygo

on for link 3.

nd 23F on th

439.3 69.30 80

F =

of link 2. Ful

on for link 3

(Scale: 1 m

he scaled dra

3320 277.2=

l scale.

is shown in

mm = 20 N).

awing, see F

2 N

Figure 8.

Figure 8, the

e external

(14)

34

Also, the force 23F can be written as

23 4340 40320 160.0 N80 80

F F= = = (15) and is acting vertically upward. Note that the answers given by Equations (14) and (15) are in good agreement with the answers given by Equations (9) and (13). The directions of the forces CF and 23F are shown in Figure 8, that is, the external force CF is acting to the right and the internal reaction force 23F is acting vertically upward. Therefore, the internal reaction force 32F is acting vertically downward.

Also, the internal reaction force between the ground link 1 and link 2 is equal and opposite to the internal reaction force between links 2 and 3, that is

12 32 23 160.0 NF F F= − = + = + (16) This force is acting vertically upward, that is, link 2 is in compression.