16P Kinetics Rigid Bodies

8/18/2019 16P Kinetics Rigid Bodies

1/12


2/12

1. In the mechanism shown, the flywheel has a mass of 50 kg and radius of gyration

about its center of 160 mm. Uniform connecting rod AB has a mass of 10 kg. Mass of the

piston B is 15 kg. Flywheel is rotating by the couple T ccw at a constant rate 50 rad/s.When θ=53o determine the angular velocity and angular acceleration of the connecting

rod AB (ωAB ve αAB). What are the forces transmitted by the pins at A and B? Neglect the

friction. Take sin 53=0.8, cos 53=0.6.


3/12

2. Crank AB rotates with an angular velocity of ωAB = 6 rad/s and angular acceleration of

αAB=2 rad/s2 , both in counterclockwise direction. Roller C can slide along the circular

slot within the fixed plate. For the position shown, angular velocity and angular

acceleration of rod BC are ωBC =2.9 rad/s (counterclockwise) and αBC=37.5 rad/s2

(counterclockwise). The masses of uniform bars AB and BC are mAB=2 kg and mBC=5 kg.

Mass of the roller C and friction can be neglected. Determine the reactions supported by

the pins B and C ?

ωAB = 6 rad/s

αAB=2 rad/s

2

mAB=2 kg

ωBC =2.9 rad/s (ccw) , αBC=37.5 rad/s2 (ccw)

mBC=5 kg

A

B

C

37o 45o

LBC

=500 mm L AB=300 mm

r =150 mm


4/12

3. Member AO is rotating at a constant angular velocity of ω AO =5 rad/s in ccw direction by a torque

T =10 N.m. The mass of the uniform slender bar AO is 2 kg. Bar AB has a mass of 5 kg and a radius of

gyration with respect to its mass center G of 400 mm. For the position shown, angular velocity and

angular acceleration of rod BC are ω AB

=1.19 rad/s (counterclockwise) and α AB

=11.79 rad/s2

(counterclockwise). Gear D can be assumed as a uniform thin disk with the radius of R=0.3 m. Its mass

is 8 kg and it rotates with an angular velocity of 7.93 rad/s in counterclockwise direction and an

angular acceleration of 5.67 rad/s2 in clockwise direction. Determine the reactions supported by the

pins A and B and contact point C .


5/12

4. The masses of uniform bars AB and BC are mAB = 2 kg and mBC = 1 kg, respectively. Bar BC

is pin connected to a fixed support at C . Bar AB is pin connected at A to a uniform wheel of

radius R = 0.50 m and mass mw

= 5 kg. At the instant shown, A is vertically aligned with O,

bar AB is horizontal and bar BC is vertical. For the given instant, bar BC is rotating with an

angular velocity of 2 rad/s and an angular acceleration of 1.2 rad/s2, both in clockwise

direction. Assuming that the wheel rolls without slipping, determine the force P that is

applied to the wheel. Also determine the coefficient of friction between the wheel and the

surface.

C

0.95 m

1.25 m

A

B

OP

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2


6/12

C

0.95 m

1.25 m

A

B

O

P

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2

D

( ) i. j.k r vvv C / B BC C / BC B

919502 =×−=×=+= ω

( ) j.i.i.k i.r vvvv AB AB B / A AB B B / A B A

ω ω ω 2519125191 −=−×+=×+=+=

( ) i. j.k r vvv w D D / Aw D / A D A

ω ω ω 950950 −=×=×=+=

(1)

(2)

(1) = (2) 02 =−= ABw s / rad ω ω


7/12

C

0.95 m

1.25 m

A

B

O

P

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2

D

( ) ( ) ( ) ( ) j.i. j.k . j.k k r r aaa C / B BC C / B BC BC C / BC B

831419502195022 −=×−+×−×−=×+××=+= α ω ω

( ) j. j.i.i.k j.i.aaa AB AB B / A B A

α α 2518314125183141 −−=−×+−=+=

( ) ( ) ( ) ( ) ( ) i. j.i. j.k j.k k i.aaa wwwwO / AO A

α α α α 45081504504502250 +−=×−+×−×−+=+=

(3)

(4)

(3) = (4)22

6121 s / rad .s / rad . ABw −== α α

{ }02 =−= ABw ,s / rad ω ω


8/12

C

0.95 m

1.25 m

A

B

O

P

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2

D

( ) ( )

y x aa

G j.i. j.k . j.k k a 915704750214750221 −=×−+×−×−=

B

C

G1

C x

B x

B y

C y

mBC g

≡

B

C

G1 x BC am

y BC am

BC BC I α

( )( ) 2

22

07520950112

1

12

1

mkg..lm I BC BC BC ⋅===

( ) ( )( )( ) ( )( )

N . B

..... B

I d am M

x

x

BC BC x BC c

380

210752047505701950

=

+=

+=∑ α

+

m AB = 2 kg

mBC = 1 kg

mw = 5 kg

G1


9/12

C

0.95 m

1.25 m

A

B

O

P

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2

D

( )

y x aa

B / G BG j.i.ik . j.i.aaa 821410625618314122 −=−×−+−=+=

A G2 A x B x =0.38 N

B y A y m AB g

≡ B x ABam

y ABam

AB AB I α

( )( ) 222 260251212

1

12

1mkg..lm I

AB AB AB

⋅===

( ) ( )( )( ) ( )( )( ) ( )( )

N . A

....... A I d am M

y

y AB AB AB B

347

61260625082262508192251

=

+−=−⇒+=∑ α +

m AB = 2 kg

mBC = 1 kg

A G2 B

( )( ) N . A. B AamF x x x x BC x 6621412 =⇒=−=∑

( )( )

( )( ) ( )( )

N . B

.. B.

.gm B A

amF

y

y

AB y y

y BC y

686

8228192347

822

−=⇒

−−−

−=−−

=

=

∑

mw = 5 kg

G2

{ }0= ABω

B y


10/12

C

0.95 m

1.25 m

A

B

O

P

R = 0.50 m

r

ωBC=2 rad/s

αBC=1.2 rad/s2

D

( )( ) i.i..aao

605021 ===

A

A x

A y

mw g

≡ ww I α

( )( ) 222 62505052

1

2

1mkg..r m I www ⋅===

( ) ( ) ( )( )( ) ( )( )

N .P

.....P. A

I d am M

x

www D

5549

2162505060550950

=

+=+−

⇒+=∑ α

+

m AB = 2 kg

mBC = 1 kg

( )( )

N .F

.F P AamF

f

f x xw x

8943

605

=⇒

=−+−=∑

N . N

gm N AF w y y

3956

00

=⇒

=−+−⇒=∑

mw = 5 kg 06903956

8943.

.

.

N

F f ===µ

O

P

N

F f

O amw

D


11/12

5. The unbalanced 20 kg wheel with the mass center at G has a radius of gyration

about G of 202 mm. The wheel rolls down the 20o incline without slipping. In the

position shown. The wheel has an angular velocity of 3 rad/s. Calculate the friction

force F acting on the wheel at this position.


12/12

SOLUTION

“General Motion”

FBD

N

mg

Ff

KD

=

am

α I

222 816.0)202.0(20 kgmk m I ===

α α 25.0== r ao x x

y

( ) ( )( ) jia

ik k ik iaaa

y xaa

G

OGOG

075.0675.025.0

075.033075.025.0/

−+−=

−××+−×+−=+=

α

α α

( ) ( )604805

67502502020

.F ..F sinmgamF

f

f xef x

=+

+−=+−⇒=∑

α

α

( )

α

α

51367184

07502020

.. N

.cosmg N amF yef y

−=

−=−⇒=∑

( ) α α 81602500750 .).(F ).( N I M f ef G =+⇒=∑ N . N

N .F

s / rad .

f

971160

6172

59715 2

=

=

=α

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16P Kinetics Rigid Bodies