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ME 352 - Machine Design I Name_______________________________ Fall Semester 2009 Lab. Div.___________________________ EXAM 1. OPEN BOOK AND CLOSED NOTES. Wednesday, September 30th, 2009

Please use the blank paper provided for your solutions. Write on one side of the paper only. Where necessary, you can use the figures that are provided on the exam to show vectors and instantaneous centers of velocity. Any work that cannot be followed will assume to be wrong.

Problem 1 (25 Points).

Part I. (13 Points). (i) Clearly number each link and label the lower pairs and the higher pairs on the mechanism shown in Figure 1(a). Then determine the mobility of this mechanism. (ii) Define vectors that are suitable for a complete kinematic analysis of the mechanism. Label and show the direction of each vector on Figure 1(a). (iii) Write the vector loop equation(s) for the mechanism and clearly identify: (a) suitable input(s) for the mechanism; and (b) the known variables, the unknown variables, and any constraints. (c) If you identified constraints in part (b) then write the constraint equation(s).

Figure 1(a). A Planar Mechanism.

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ME 352 - Machine Design I Name___________________________________ Fall Semester 2009 Lab. Div.________________________________ Problem 1 (continued).

Part II. (12 Points). Consider the four-bar linkage in the position shown in Figure 1(b). The angle of the input link 2 is o

2 60 ,θ = measured counterclockwise from the ground link which is coincident with the fixed X-axis. The lengths of the four links are 1 2 4R O O 7 cm,= = 2 2R O A 8 cm,= =

3R AB 5 cm,= = and 4 4R O B 10 cm.= = Use Freudenstein's equation to determine the angular position of the output link 4θ .

Figure 1(b). A Planar Four-Bar Linkage.

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ME 352 - Machine Design I Name___________________________________ Fall Semester 2009 Lab. Div.________________________________

Problem 2 (25 Points). For the mechanism in the position shown in Figure 2, the input link 2 is rotating counterclockwise with a constant angular velocity 2 30 rad / s.ω = Also for this position, the coupler link 3 is horizontal and is sliding along the vertical output link 4. The known link lengths are

2 4O O 30 cm,= 2O A 15 cm,= and AB 22.5 cm.=

(i) Write a vector loop equation that would be suitable for a complete kinematic analysis of this mechanism. Indicate the input, the known variables, the unknown variables, and any constraints. Draw your vectors clearly on Figure 2.

(ii) Determine the first-order kinematic coefficients for the mechanism from your vector loop equation.

(iii) Determine the angular velocities of links 3 and 4. Give the magnitudes and the directions.

(iv) Determine the velocity of point B fixed in link 3 relative to the velocity of the coincident point B fixed in link 4. Give the magnitude and the direction of this vector.

Figure 2. A Planar Mechanism.

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ME 352 - Machine Design I Name___________________________________

Fall Semester 2009 Lab. Div.________________________________

Problem 3 (25 points). For the mechanism in the position shown in Figure 3, the input link 2 is rotating clockwise with a constant angular velocity 2 15 rad / sω = . Link 2 is in contact with link 3 at point C and link 4 is in contact with the slot in link 3 at point E. The figure is drawn full scale, i.e., 1 in = 1 in. (i) List the primary instant centers and the secondary instant centers for the mechanism. (ii) Using the Kennedy circle, show the location of all the instant centers on Figure 3. Using the location of the instant centers, determine: (iii) The first-order kinematic coefficients of links 3 and 4. (iv) The magnitudes and directions of the angular velocities of links 3 and 4. (v) The magnitude and direction of the velocity of point B, and the magnitude and direction of the

slipping velocity at point E.

Kennedy Circle.

Figure 3. A Planar Mechanism. (Drawn Full Scale: 1 in = 1 in).

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ME 352 - Machine Design I Name___________________________________

Fall Semester 2009 Lab. Div.________________________________

Problem 4 (25 points). For the gear mechanism in the position shown in Figure 4, the input link 2 is rotating with an angular velocity 2 50 rad / sω = clockwise and an angular acceleration 2

2 15 rad / sα =

clockwise. Link 2 is pinned to the ground at 1O and is pinned to the center of gear 3 at point A. The center of gear 4 is also pinned to the ground at 1O and gear 5 is pinned to the ground at .5O Gears 3, 4 and 5 are all in rolling contact at point B. The radii of the fixed gear 1 and the moving gears 3, 4 and 5, are 1 10 cm,ρ = 3 10 cm,ρ = 4 30 cm,ρ = and 5 50 cm,ρ = respectively. Determine:

(i) The first-order kinematic coefficients for gears 3, 4, and 5.

(ii) The angular velocities of gears 3, 4, and 5. Specify the magnitudes and directions.

(iii) The angular accelerations of gears 3, 4, and 5. Specify the magnitudes and directions.

(iv) The velocity of point B fixed in gear 3.

Figure 4. A Gear Mechanism.

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Solution to Problem 1. Part I. (i) 4 points. There are five links in this mechanism and the joint types connecting these five links are as shown in Figure 1(a).

Figure 1(a). Joint Types of the Mechanism.

For this mechanism, the number of links, number of lower pairs (or 1J joints), and number of higher pairs (or 2J joints), respectively, are

1 2n 5, J 5, and J 1= = = (1)

The Kutzbach mobility criterion for a planar mechanism can be written as

1 2M 3(n 1) 2J 1J= − − − (2)

Substituting Equation (1) into Equation (2), the mobility of the mechanism is

M 3(5 1) 2(5) 1 1= − − − = (3)

This is the correct answer for this mechanism, that is, for a single input there is a unique output.

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(ii) 5 points. Suitable vectors for a kinematic analysis of the mechanism are shown in Figure 1(b).

Figure 1(b). Vectors for the Mechanism.

(iii) 4 points. There are 5 unknown variables, therefore, 2 independent vector loop equations are required and one rolling contact equation. If the input link is chosen to be the slider, that is, link 2 then the two independent vector loops can be written as

Loop 1: 2 13 23 33 30C C

R R R R RΙ√ √ ?√ √ √?+ + − − = (4a)

Loop 2: 3 333 4 7 9 1

? ? ? 0CR R R R R R√ √ √ √√ √ √√+ + + − − = (4b)

(a) Since the input link is the slider, that is, link 2, then the input variable is the length 2R . (b) The four unknown variables in Equations (4) are the angular displacements 3θ , 4θ , and the linear distances 23R and 9R . (c) There are four constraint equations, namely:

13 2 90θ θ= + ° (5a)

33 3 90θ θ= + ° (5b) and

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333 3 90θ θ= − ° (5c)

Note that the angular displacement of the wheel 5θ is constrained to the linear distance 9R by rolling contact. The rolling contact equation between link 5 and the ground link 1 can be written as

( )9 5 5 9 5 5R ρ θ θ ρ θ± Δ = Δ −Δ = Δ (6a)

Note that this equation can also be written in terms of the first-order kinematic coefficients as

9 5 5R ρ θ′ ′± = (6b)

The correct sign in Equations (6) is positive because for a positive rotation of the wheel (that is, counterclockwise) the length of the vector 9R is increasing or for a negative rotation of the wheel (that

is, clockwise) the length of the vector 9R is decreasing. Part II. 13 points. The vectors for the four-bar linkage are shown in Figure 1(b).

Figure 1(b). The vector loop for the four-bar linkage.

The vector loop equation (VLE) can be written as

√ I √ ? √ ? √ √ 2 3 4 1R R R R 0+ − − = (1a)

The X and Y components of Eq. (1) are

2 2 3 3 4 4 1 1cos cos cos cos 0R R R Rθ θ θ θ+ − − = (2a)

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and 2 2 3 3 4 4 1 1sin sin sin sin 0R R R Rθ θ θ θ+ − − = (2b)

Freudenstein's Equation can be written as

A cos B sin C4 4θ θ+ = (3) where

A 2 R R cos 2 R R cos1 4 1 2 4 2θ θ= − (4a)

1 4 2 4B 2 R R sin 2 R R sin1 2= −θ θ (4b) and

2 2 2 2C R R R R 2 R R cos ( )3 1 2 4 1 2 1 2= − − − + −θ θ (4c)

Substituting the known data into Equations (4) gives

2A 2 7 10 cos 0 2 8 10 cos 60 60 cm= × × − × × = + (5a)

2B 2 7 10 sin 0 2 8 10 sin 60 138.56 cm= × × × − × × × = − (5b) and

2 22 2 2C 5 7 8 10 2 x 7 x 8 cos (0 60 ) 132 cm= − − − + − = − (5c)

Substituting Equations (5) into Equation (3) gives

260 cos 138.56 sin 132 cm4 4θ θ+ − = − (6)

To determine the output angle we can write this transcendental equation as an algebraic equation, (namely, a quadratic equation). The procedure is to use the tangent of the half-angle relationship; i.e.,

4Z tan ( )2θ= (7a)

which gives

22 Zsin 4 1 Z

θ =+

and 2

21 Zcos 4 1 Z

θ −=+

(7b)

Substituting Equations (7b) into Equation (3), and rearranging, gives

2(A C) Z (2 B) Z (C A) 0+ − + − = (8)

The solution to this equation can be written as

2B B (A C)(C A)ZA C

+ ± − + −=+

(9)

Substituting Equations (5) into Equation (9) gives

2138.56 ( 138.56) (60 132)( 132 60)Z60 132

− ± − − − − −=−

(10a)

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The two roots to this quadratic equation are

1 2Z 0.90620 and Z 2.94269= + = + (10b)

Equation (7a) can be written as

I 14 12 tan Zθ −= and II 1

4 22 tan Zθ −= (11)

Substituting Equations (10b) into Equations (11) gives

I 14 2 tan 0.90620θ −= and II 1

4 2 tan 2.94269θ −= (12)

Therefore, the two possible answers for the angular position of link 4 are

84.374θ = and 142.464θ = (13)

The answer for the angular position of link 4, for the given open configuration shown in Figure 1(b), is

84.374θ =

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Solution to Problem 2. (i) 8 Points. A suitable set of vectors for a kinematic analysis of the mechanism are shown in Fig. 2(a).

Figure 2(a). Suitable Vectors for the Mechanism.

The vector loop equation (VLE) can be written as

√ I √ ? ? C √ √ 2 3 34 1R R R R 0+ − − = (1a)

where the constraint is 34 4 3 90θ θ θ= = + ° (1b)

The X and Y components of the VLE, see Eq. (1), are

2 2 3 3 34 34 1 1R cos R cos R cos R cos 0θ + θ − θ − θ = (2a) and

2 2 3 3 34 34 1 1R sin R sin R sin R sin 0θ + θ − θ − θ = (2b)

(ii) 7 Points. Differentiating Equations (2) with respect to the input position 2θ gives

2 2 3 3 3 34 34 34 34 3R sin R sin R cos R sin 0′ ′ ′− θ − θ θ − θ + θ θ = (3a) and

2 2 3 3 3 34 34 34 34 3R cos R cos R sin R cos 0′ ′ ′θ + θ θ − θ − θ θ = (3b)

Then writing Equations (3) in matrix form gives

3 3 34 34 34 3 2 2

3 3 34 34 34 34 2 2

R sin R sin cos R sinR cos R cos sin R R cos

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

(4)

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Substituting the known data into Equation (4) gives

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12.99 cm 0 rad 12.99 cmR22.5 cm 1 rad 7.5 cm′θ+ − +⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥′+ − −⎣ ⎦ ⎣ ⎦⎣ ⎦ (5a)

The determinant of the coefficient matrix in Equation (5a) is

DET ( 12.99 cm)( 1 rad) ( 22.5 cm)(0 rad) 12.99 cm= + − − + = − (5b)

Using Cramer’s rule, then from Eq. (5a), the first-order kinematic coefficient for link 3 is

312.99 cm 1cm / cm12.99 cm

−′θ = = +−

(6a)

The positive sign indicates that link 3 is rotating in the same direction as the input link, i.e., counterclockwise. Also this answer indicates that link 3 is rotating with same angular velocity as the input link; i.e., by definition

3 2 1rad / rad′ ′θ = θ = (6b)

Also, using Cramer’s rule, the first-order kinematic coefficient for links 3 and 4 is

234

( 12.99 cm) ( 7.5 cm) ( 22.5 cm)( 12.99 cm)R 30 cm / cm12.99 cm

+ − − + +′ = = +−

(7a)

The positive sign indicates that the vector 34R is increasing in length for a positive input, that is, point B on link 3 is moving away from the ground pivot of link 4. (iii) 7 Points. The angular velocity of link 4 can be written as

4 4 2′ω = θ ω (8)

Note that the first-order kinematic coefficient for link 4 is

4 34 3 1 rad / rad′ ′ ′θ = θ = θ = + (9)

Substituting Eq. (9) and the input angular velocity into Eq. (8), the angular velocity of link 4 is

4 ( 1 rad / rad)( 30 rad / s) 30 rad / sω = + + = + (10)

The positive sign means that link 4 is, indeed, rotating counterclockwise. (iv) 5 Points. The velocity of point B fixed in link 3 relative to the velocity of point B fixed in link 4 can be written as

B 34 2R R′= ω (11)

Substituting Equation (7a) and the input angular velocity into Equation (11), the velocity is

2BR ( 30 cm / cm)( 30 rad / s) 900 cm / s= + + = + (12)

The magnitude and the direction of the velocity of point B fixed in link 3 relative to the velocity of point B fixed in link 4 is

B BV R j 900 j cm / s= + = + (13)

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The velocity of point B fixed in link 3 relative to the velocity of point B fixed in link 4 is directed vertically upwards. This can be verified by finding the instantaneous centers of velocity, see Figure 2(c).

Figure 2(c). The location of the instant centers.

Check. Using instant centers, the velocity of point B fixed in link 3 relative to the velocity of the coincident point B fixed in link 4 can be written as

34 3 4B B BV V V= − (14)

Using instant centers, the magnitude of the velocity of point B fixed in link 3 can be written as

3B 13 3V (I B)= ω (15a)

The distance 13I B is measured as 13I B 32.692 cm= . Therefore, the velocity of point B fixed in link 3 is

3BV (32.692)(30 rad / s) 980.76 cm / s= = (15b)

Using instant centers, the velocity of point B fixed in link 4 can be written as

4B 14 4V (I B) i= − ω (16a)

The distance 14I B is measured as 14I B 12.990 cm= . Therefore, the velocity of point B fixed in link 4 is

4BV (12.990)(30 rad / s) 389.7 i cm / s= − = − (16b)

Substituting Eqs. (15b) and (16b) into Eq. (14), the velocity of point B fixed in link 3 relative to the velocity of the coincident point B fixed in link 4 is

34 3 4B B BV V V (389.7 i 900 j) 389.7 i 900 j cm / s= − = + − = (17)

The direction of this vector is indeed vertically upward as shown in Figure 2(c).

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Solution to Problem 3. (i) 10 Points. The number of links in the mechanism is four, therefore, the total number of instant centers for this mechanism is six; i.e.,

( 1) 4 x 3 62 2

n nN −= = = (1)

There are three primary instant centers; namely, I12, I13, and I14. The three secondary instant centers; namely, I23, I24, and I34 can be obtained as follows: (i) It is important to note that the secondary instant center I23 is not located at point C because there is slip between links 2 and 3. However, the instant center I23 must lie on the line that is perpendicular to the line BC through the point of contact C. The point of intersection of this line with the line connecting the instant centers I12I13 is the instant center I23. (ii) The point of intersection of the line through I13I14 and the perpendicular to the slot in link 3 that passes through point E is the instant center I34. (iii) The point of intersection of the line through I12I14 and the line through I23I34 is the instant center I24. The procedure to locate the three secondary instant centers are marked on the Kennedy circle.

Figure 3(a). The Kennedy Circle.

The location of the six instant centers for this mechanism are shown on Figure 3(b). (ii) 5 Points. The first-order kinematic coefficient of link 3 can be written as

12 233

13 23

I II I

θ ′ = (4a)

The distance 12 23I I is measured as 12 23 1.875 inI I = and the distance 13 23I I is measured as

13 23 3.125 in.I I = Therefore, the first-order kinematic coefficient of link 3 is

31.875 in 0.6 in/in3.125 in

θ ′ = = (4b)

Note that the correct sign is negative because the relative instant center 23I lies between the absolute instant centers, that is

3 0.6 rad/radθ ′ = − (4c)

The first-order kinematic coefficient of link 4 can be written as

12 244

14 24

I II I

θ ′ = (5a)

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The distance 12 24I I is measured as 12 24 2.785 inI I = and the distance 14 24I I is measured as

14 24 2.307 in.I I = Therefore, the first-order kinematic coefficient of link 4 is

42.785 in 1.207 in / in2.307 in

θ ′ = = (5b)

Note that the correct sign is negative because the relative instant center 24I lies between the absolute instant

centers 12I and 14I , that is

4 1.207 rad/radθ ′ = − (5c)

Figure 3(b). The location of the instant centers.

(iii) 5 Points. The magnitude of the angular velocity of link 3 can be written as

3 3 2 ( 0.6 rad rad)( 15 rad sec) 9 rad secω θ ω′= = − − = + (6)

The positive sign indicates that the direction of the angular velocity of link 3 is counterclockwise, as shown in Figure 3(c). The magnitude of the angular velocity of link 4 can be written as

4 4 2 ( 1.207 rad rad)( 15 rad sec) 18.105 rad secω θ ω′= = − − = + (7)

The positive sign indicates that the direction of the angular velocity of link 4 is counterclockwise, as shown in Figure 3(c).

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Figure 3(c). Angular velocities of links 3 and 4.

(iv) 5 Points. The velocity of point B can be written as

( )13 3BV I B ω= (8a)

The distance 13I B is measured as 13 4.940 in,I B = therefore, the velocity of point B is

( ) ( )13 3 (4.94 in) 9 rad sec 44.46 in secBV I B ω= = = (8b)

The direction of the velocity of point B is perpendicular to the line connecting the instant center 13I to

point B as shown in Figure 3(d). The velocity of point B is directed 52 degrees below the X-axis.

The slipping velocity at point E can be written as

34 43 34 4 3( ) ( )( )SlipV I E I Eω ω ω= = − (9)

where the slipping velocity at point E is defined here as

4 3 4 3E /E E ESlipV V V V= = − (10)

The direction of the slipping velocity. The angular velocity of link 4 is counterclockwise and the angular

velocity of link 3 is counterclockwise. Since the angular velocity of link 4 is greater than the angular

velocity of link 3 then the angular velocity of link 4 relative to the angular velocity of link 3 is

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counterclockwise about the instant center I34. This means that the slipping velocity of point E along the

slot is pointed downward as shown in Figure 3(d).

The distance 34I E is measured as

34 1.510 inI E = (11)

Substituting Eqwuation (11) into Equation (10), the slipping velocity at point E is

( ) ( )34 43 1.510 18.105 9 13.75 in secSlipV I E ω= = × − = (12)

If the slipping velocity at point E is defined as

3 4 3 4E /E E ESlipV V V V= = − (13)

then Equation (11) would be written as

( ) ( )34 34 1.510 9 18.105 13.75 in secSlipV I E ω= = × − = − (14)

Since the angular velocity of link 3 relative to the angular velocity of link 4 is clockwise about the

instant center I34. This means that the slipping velocity of point E along the slot (based on this

definition) is pointed upward as shown in Figure 3(d).

Figure 3(d). Velocity of point B and the slipping velocity at point E.

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Solution to Problem 4.

(i) 6 Points. The rolling contact equation for gear 3 rolling on the ground link 1 can be written as

3 21

3 1 2

θ θρρ θ θ

′ ′⎛ ⎞−± = ⎜ ⎟′ ′−⎝ ⎠

(1a)

The correct sign on the left hand side is negative since there is external contact, that is

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3

10 1θρ

ρ′⎛ ⎞−

− = ⎜ ⎟−⎝ ⎠ (1b)

and

31

3

00 0θρ

ρ′′⎛ ⎞−

− = ⎜ ⎟−⎝ ⎠ (1c)

The rolling contact equation between gear 3 and gear 4 can be written as

3 4 2

4 3 2

ρ θ θρ θ θ

⎛ ⎞′ ′−± = ⎜ ⎟′ ′−⎝ ⎠

(2a)

The correct sign on the left hand side is positive since there is internal contact, that is

3 4

4 3

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ρ θρ θ

⎛ ⎞′ −+ = ⎜ ⎟′ −⎝ ⎠

(2b)

and

3 4

4 3

00

ρ θρ θ

⎛ ⎞′′−+ = ⎜ ⎟′′−⎝ ⎠

(2c)

The rolling contact equation between gear 4 and gear 5 can be written as

5 14

5 4 1

θ θρρ θ θ

′ ′⎛ ⎞−± = ⎜ ⎟′ ′−⎝ ⎠

(3a)

The correct sign on the left hand side is positive since there is internal contact, that is

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5 4

00

θρρ θ

′⎛ ⎞−+ = ⎜ ⎟′ −⎝ ⎠

(3b)

and

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5 4

00

θρρ θ

′′⎛ ⎞−+ = ⎜ ⎟′′−⎝ ⎠

(3c)

(ii) 7 Points. Substituting the known radius for gear 1 and the radius of gear 3into Equation (1b), and

rearranging, gives

310 110

θ ′+ = − (4a)

Therefore, the first-order kinematic coefficient for gear 3 is

3 2 rad/radθ ′ = + (4b)

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Equation (1a) can also be written as

1 1 2 3 3 2( ) ( )ρ θ θ ρ θ θ′ ′ ′ ′± − = − (5a) or as

1 1 2 3 3 2( ) ( )ρ θ θ ρ θ θ′′ ′′ ′′ ′′± − = − (5b) Substituting the known second-order kinematic coefficients into Equation (5b) gives

1 3 3(0 0) ( 0)ρ ρ θ ′′± − = − (6a)

Therefore, the second-order kinematic coefficient for gear 3 is

3 0 rad/radθ ′′ = + (6b)

Substituting the known radius for gears 3 and 4 into Equation (2b) gives

4

3

11030 1

θθ⎛ ⎞′ −

+ = ⎜ ⎟′ −⎝ ⎠ (7a)

Then rearranging this equation gives

( )4 310 1 130

θ θ′ ′= − + (7b)

Substituting Equation (4b) into this equation gives

( )410 402 1 1 rad/rad30 30

θ ′ = − + = + (8a)

Therefore, the first-order kinematic coefficient for gear 4 is

4 1.333 rad/radθ ′ = + (8b)

Substituting the known radius for gears 3 and 4 into Equation (2c) the second-order kinematic

coefficient for gear 4 is 2

4 0 rad/radθ ′′ = + (8c)

Substituting the known radius for gears 5 and 4 into Equation (3b) gives

5

4

3050

θθ′

+ =′

(9a)

Substituting Equation (6b) into this equation gives

530 40 40 rad/rad50 30 50

θ ⎛ ⎞⎛ ⎞′ = + + = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(9b)

Therefore, the first-order kinematic coefficient for gear 5 is

5 0.8 rad/radθ ′ = + (10a)

Substituting the known radius for gears 5 and 4 into Equation (3c), the second-order kinematic

coefficient for gear 5 is

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25 0 rad/radθ ′′ = + (10b)

(iii) 8 Points. The angular velocity of gear 3 can be written as

3 3 2ω θ ω′= (11a)

Substituting Equation (4a) into Equation (11a), the angular velocity of gear 3 is

3 ( 2)( 50) 100 rad/sω = + − = − (11b)

The negative sign indicates that gear 3 is indeed rotating clockwise.

The angular velocity of gear 4 can be written as

4 4 2ω θ ω′= (12a)

Substituting Equation (6b) into Equation (10a), the angular velocity of gear 4 is

4 ( 1.333)( 50) 66.67 rad/sω = + − = − (12b)

The negative sign indicates that gear 4 is indeed rotating clockwise.

The angular velocity of gear 5 can be written as

5 5 2ω θ ω′= (13a)

Substituting Equation (8a) into Equation (11a), the angular velocity of gear 5 is

5 ( 0.8)( 50) 40 rad/sω = + − = − (13b)

The negative sign indicates that gear 5 is indeed rotating clockwise.

The angular acceleration of gear 3 can be written as 2

3 3 2 2α θ α θ ω′ ′′= + (14a)

Substituting Equation (4c) into Equation (14a), the angular acceleration of gear 3 is 2 2

3 ( 2)( 15) (0)( 50) 30 rad/sα = + − + − = − (14b)

The negative sign indicates that gear 3 is accelerating clockwise.

The angular acceleration of gear 4 can be written as 2

4 4 2 4 2α θ α θ ω′ ′′= + (15a)

Substituting Equation (6c) into Equation (15a), the angular acceleration of gear 4 is 2 2

4 ( 1.333)( 15) (0)( 50) 20 rad/sα = + − + − = − (15b)

The negative sign indicates that gear 4 is accelerating clockwise.

The angular acceleration of gear 5 can be written as 2

5 5 2 5 2α θ α θ ω′ ′′= + (16a)

Substituting Equation (8b) into Equation (16a), the angular acceleration of gear 5 is 2 2

5 ( 0.8)( 15) (0)( 50) 12 rad/sα = + − + − = − (16b)

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The negative sign indicates that gear 5 is accelerating clockwise.

(iv) 5 Points. The velocity of point B fixed in gear 3 can be written as

( )14 43 4

V V I BB B ω= = (17a)

Since point B is coincident with the instant center 34I . Substituting the values into Equation (17a) gives

3 4

200 30 2000 cm/s3

V VB B⎛ ⎞= = =⎜ ⎟⎝ ⎠

(17b)

The velocity of point B fixed in gear 3 is directed vertically downward.