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.

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/s2

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 LAB=300 mm

r =150 mm

3. Member AO is rotating at a constant angular velocity of ωAO =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 ωAB =1.19 rad/s (counterclockwise) and αAB=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.

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

O P

R = 0.50 m

r

ωBC=2 rad/s αBC=1.2 rad/s2

µ

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

µ

D

( ) i.j.krvvv C/BBCC/BCB 919502 =×−=×=+= ω

( ) j.i.i.ki.rvvvv ABABB/AABBB/ABA ωωω 2519125191 −=−×+=×+=+=

( ) i.j.krvvv wDD/AwD/ADA ωωω 950950 −=×=×=+=

(1)

(2)

(1) = (2) 02 =−= ABw s/rad ωω

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

µ

D

( ) ( )[ ] ( ) ( ) j.i.j.k.j.kkrraaa C/BBCC/BBCBCC/BCB 831419502195022 −=×−+×−×−=×+××=+= αωω

( ) j.j.i.i.kj.i.aaa ABABB/ABA αα 2518314125183141 −−=−×+−=+=

( ) ( ) ( )[ ] ( ) ( ) i.j.i.j.kj.kki.aaa wwwwO/AOA αααα 45081504504502250 +−=×−+×−×−+=+=

(3)

(4) (3) = (4) 22 6121 s/rad.s/rad. ABw −== αα

{ }02 =−= ABw ,s/rad ωω

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

µ

D

( )[ ] ( ) ( )

yx aaG j.i.j.k.j.kka 915704750214750221 −=×−+×−×−=

B

C

G1

Cx

Bx

By

Cy

mBCg

≡

B

C

G1 xBC am

yBC am

BCBCI α

( )( ) 222 075209501121

121

mkg..lmI BCBCBC ⋅===

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

N.B

.....B

IdamM

x

x

BCBCxBCc

380

210752047505701950

=

+=

+=∑ α

+

mAB = 2 kg

mBC = 1 kg

mw = 5 kg

G1

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

µ

D

( ) ( )

yx aaB/GBG j.i.ik.j.i.aaa 821410625618314122 −=−×−+−=+=

A G2 Ax Bx=0.38 N

By Ay mABg

≡ B xAB am

yAB am

ABABI α

( )( ) 222 2602512121

121

mkg..lmI ABABAB ⋅===

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

N.A

.......AIdamM

y

yABABABB

347

61260625082262508192251

=

+−=−⇒+=∑ α+

mAB = 2 kg

mBC = 1 kg

A G2 B

( )( ) N.A.BAamF xxxxBCx 6621412 =⇒=−=∑

( )( )

( )( ) ( )( )

N.B

..B.

.gmBA

amF

y

y

AByy

yBCy

686

8228192347

822

−=⇒

−−−

−=−−

=

=

∑

mw = 5 kg

G2

{ }0=ABω

By

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

µ

D

( )( ) i.i..aao 605021 ===

A Ax

Ay

mwg

≡ wwI α

( )( ) 222 625050521

21

mkg..rmI www ⋅===

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

N.P

.....P.A

IdamM

x

wwwD

5549

2162505060550950

=

+=+−

⇒+=∑ α

+

mAB = 2 kg

mBC = 1 kg

( )( )

N.F

.FPAamF

f

fxxwx

8943

605

=⇒

=−+−=∑

N.N

gmNAF wyy

3956

00

=⇒

=−+−⇒=∑

mw = 5 kg 06903956

8943 ..

.N

F f ===µ

O

P

N

Ff

O amw

D

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.

SOLUTION “General Motion”

FBD

N

mg

Ff

KD

=

am

αI

222816.0)202.0(20 kgmkmI ===

αα 25.0== raox x

y

( ) ( )[ ]( ) jia

ikkikiaaa

yxaa

G

OGOG

075.0675.025.0

075.033075.025.0/

−+−=

−××+−×+−=+=

α

αα

( ) ( )604805

67502502020

.F

..FsinmgamF

f

fxefx

=+

+−=+−⇒=∑α

α

( ) ( )α

α

51367184

07502020

..N

.cosmgNamF yefy

−=

−=−⇒=∑

( ) αα 81602500750 .).(F).(NIM fefG =+⇒=∑ N.N

N.Fs/rad.

f

971160

617259715 2

=

==α