Tutorial 8 Solutions

8/2/2019 Tutorial 8 Solutions

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SOLUTION 8

MP2002 Kinematics and Dynamics of Machinery

8.1 ( Homework)

A uniform slender rod, of length L = 750 mm and mass m = 2 kg and

2

121 mL I G = , hangs freely from a hinge at A. If a force P of magnitude

12 N is applied at B horizontally to the left ( h = L), determine

(a) the angular acceleration of the rod,

(b) the components of the constraint force at A.

Solution

Kinematic Analysis:i j k r a GAGˆ375.0) ˆ375.0( ˆ

α−

Rod AB rotates about a fixed point A, so we have

22

22

3

1

212

1 mL

L m mL mr I I

GAG A =

+ = 0.375 (kg m

2)

Dynamic Analysis:

• For Rod A

α

A A I M =

k ri GAˆ375.0) ˆ12( α

⇒ α375.0)75.0)(12( −

⇒ 24 (rad/s)

⇒ i aGˆ9

(rad/s)

G a m F

=

⇒ ) ˆ9(2 ˆ ˆ ˆ12 ˆ2 i j Fi Fi j g y A

x A

−

⇒ 62.19

6

=

−

y A

x A

F

F(N)

B

I Aα

G a G

=

y A

F

A

L

12N

B

mg

G

i ˆ

j ˆ

x A F



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8.2 ( Homework)

The 2-kg collar C is free to slide in the vertical plane along the smooth

rod AB, which is welded onto collar A as shown. The collar A is sliding

along the vertical shaft with acceleration 4 m/s2

upward.

Determine:

(a) the acceleration of collar C (b) the force acting on collar C by rod AB

Solution

Kinematics Analysis:

Acceleration of collar C

we can find t AC AC a a a /

+ t AC e a j ˆ ˆ4 /

Dynamic Analysis:• For collar C:

C n a me N j mg

= ˆ ˆ = ( ) t AC e a j m ˆ ˆ4 /

Changing the coordinates by

jie tˆ45sin ˆ45cos ˆ

+ , jie nˆ45sin ˆ45cos ˆ

+

Then

i ˆ : 45cos45cos / AC ma N = (1)

j ˆ : )45sin4(45sin /

AC a m N mg + (2)

Solving (1) and (2)

⇒ 765.9 / = AC a (m/s2), 53.19 N (N),

Therefore,

49.7 / = t AC AC a a a

(m/s2) ∠-157.2

o.

C 45°

A

B ne ˆ te ˆ

mg

i ˆ

j ˆ



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8.3 (Wilson & Sadler, Q10.8, p.759) (Tutorial )

The crank of the slider-crank mechanism of the figure has an

instantaneous angular velocity of 10 rad/s clockwise and an angular

acceleration of 200 rad/s2

clockwise. Acceleration information is given

in the figure.

The connecting rod has a mass of 15 kg and a mass moment of inertiaof 7500 kg-mm

2about its mass center G2. The slider has a mass of 8

kg. The crank has a moment of inertia of 4000 kg-mm2

about its

stationary center of mass G1.

Determine all bearing forces and the input torque T 1 for the position

shown.

82

36

44.186

90

OB =30 mm

BC =90 mm

aG2 = 3810 ∠325o mm/s2

α2 = 66.3 rad/s2

(ccw)

aG3 = 5800 mm/s2



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Additional Questions

8.4 A pipe (thin ring) with mass 500 kg and radius of 0.5 m rolls down

a ramp without slipping as shown. If the ramp is fixed horizontally,

determine the angular acceleration of the pipe and forces between the

pipe and ramp.

Solution:

Kinematic Analysis:

t n p e re r k r a ˆ) ˆ( ˆα−

1252= mr I p (kg m

2)

Dynamic Analysis:• For pipe:

α

P P I M =

⇒ ) ˆ( ˆ ˆ 2 k mre Fe r t n α

⇒ α125 Fr (1)

p a m F

=

⇒ ) ˆ( ˆ ˆ ˆ t t n e r m j mge Fe N α

Changing the coordinates by

n t eei ˆ30sin ˆ30cos ˆ − , n t ee j ˆ30cos ˆ30sin ˆ

+

Then

ne ˆ : 0)30(cos = mg N (2)

te ˆ : r m mg F α)30(sin (3)

Solving (1), (2) and (3)⇒ 9.4 (rad/s)

⇒

=

=

1225)30(sin

5.4243)30(cos

r m mg F

mg N

α

(N)

P a P

A30°

B te ˆ ne ˆ

Ι pα

=

P

N 30°

B

te ˆ ne ˆ mg

F

i ˆ

j ˆ

Tutorial 8 Solutions

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