KINEMATICS OF RIGID BODIES [PDF]

KINEMATICS OF RIGID BODIES

PROBLEMS 1. The center O of the disk has the velocity and acceleration shown. If the disk rolls without slipping on the horizontal surface, determine the velocity of A and the acceleration of B for the instant represented.

2. For an interval of its motion the hydraulic cylinder gives the piston rod A a velocity vA = 1.2 m/s which is increasing by 0.9 m/s each second for the instant when    = 60°. For this instant, determine the angular acceleration of link BC.

3. The center O of the disk rolling without slipping on the horizontal surface has the velocity and acceleration shown. Radius of the disk is 4.5 cm. Calculate the velocity and acceleration of point B.

vo=45 cm/s

O 37o

A

y

x

x=2 cm

ao=90 cm/s2

4.5 cm

6 cm

B 1 y  x2 4

4 cm

Disk: d 

45  10 rad / s 4.5

d 

90  20 rad / s 2 4.5

10 cm

VELOCITY: 





Disk v A  vO  v A / O

     v A  45i  10k   4 cos 37i  4 sin 37 j        

   v A  21i  32 j

 3.2i  2.4 j

Link AB

   vB  v A  vB / A

      v B   21i  32 j    AB k   10i  6 j 

     v B  21i  32 j  10 AB j  6 AB i 1

Particle B

dy 2 x  dx 4

1 = 2

x2 

 i  j

dy 1  dx

  45

  45 o

o

 vB



21  6 AB  0.707v B



32  10 AB  0.707v B vo=45 cm/s 37o

v B  35.5 cm / s

4 cm

ao=90 cm/s2

4.5 cm

6 cm

  d  10k 

B 1 y  x2 4

 AB  0.688 rad / s

O

A

y

     v B  v B  cos 45i  sin 45 j   0.707v B i  0.707v B j

x  v B x=2 cm

10 cm

  d  20k 

2

ACCELERATION: 





Disk a A  aO  a A / O Link AB

   aB  a A  aB / A



 



Particle B

a B n

2 3 / 2

  dy  1     2   dx   d y 1 dy 1 ,    dx dx 2 2 d2y

  45 o

a B t

 



         a A  90i  10k  10k   3.2i  2.4 j    20k   3.2i  2.4 j     a A  362i  304 j           a B  362i  304 j   0.688k   0.688k   10i  6 j   AB k   10i  6 j

 



 



     a B  366.73i  301.16 j  10 AB j  6 AB i  5.657 cm

dx 2

3

v B2

35.5 2 a B n    222.78 cm / s 2  5.657

     a B  a B t  cos 45i  sin 45 j   a B n  cos 45i  sin 45 j       a B  0.707a B t i  0.707a B t j  157.53i  157.53 j 4

3 = 4 vo=45 cm/s 37o A

y

4 cm 4.5 cm

ao=90 cm/s2

 i   j

366.73  6 AB  157.53  0.707a B t 301.16  10 AB  157.53  0.707a B t

6 cm

 AB  23.79 rad / s 2

B

1 y  x2 4

O

x x=2 cm

10 cm

a B t

 539.63 cm / s 2



4. The elements of a power hacksaw are shown in the figure. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counterclockwise speed of 60 rev/min,

determine the

acceleration of the blade for the position where  = 90°, and find the corresponding angular acceleration of the link AB.

 f  60

f

2  6.28 rad / s 60

  90o

B 100 mm O A

100 mm

O

438 mm 



















Velocity Analysis VB  VO  VB / O   f  rB / O  6.28k   0.1i  0.628 j

























Link AB V A  VB  V A / B  0.628 j   AB   0.1 j  0.438i  0.628 j  0.1 AB i  0.438 AB j Blade

  VA  VAi 2

Acceleration Analysis



1 = 2

1

VA 0.143m/ s , AB 1.43rad/ s

           a A  a B  a A / B   f   AB  rB / O    AB   AB  rA / B    AB  rA / B       

 



a



aA/ B

   B        a A  6.28k  6.28k   0.1i    1.43k   1.43k   0.1 j  0.438i    AB k   0.1 j  0.438i 

      a A i  3.943i  0.2045 j  0.895i  0.1 AB i  0.438 AB j  i  j

 

a A  3.943  0.895  0.1 AB    0  0.2045  0.438 AB

a A  4.885 m / s 2

 AB  0.467 rad / s 2

5. At a given instant, the gear has the angular motion shown. Determine the acceleration of points A and B on the link and the link’s angular acceleration at this instant.

VELOCITY Gear

     v  6  3   i    18 i Center O O      v A  18i  6k   2 j   6i

   v A  vO  v A / O Member AB

        v B  6i   AB k  8 cos 60i  8 sin 60 j   6i  4 AB j  6.93 AB i 1   

   vB  v A  vB / A   Collar B v B  v B i  i 1 = 2  j

4i  6.93 j

2

 

6  6.93 AB  v B 4 AB  0

 AB  0 v B  6 cm / s

ACCELERATION Gear



 





Center O aO  12  3  i  36i



Member AB



3 = 4

 i  j



          a B  12i  72 j   AB k   AB k  4i  6.93 j    AB k  4i  6.93 j 

   aB  a A  aB / A   Collar B a B  a B i



         a A  36i  6k  6k   2 j   12k   2 j   12i  72 j

   a A  aO  a A / O

     a B  12i  72 j  4 AB j  6.93 AB i

3

4

 

12  6.93 AB  a B 4 AB  72  0

 AB  18 rad / s

2

a B  112.74 cm / s 2

  AB  18k   a B  112.74i 

6. The rotation of link AB creates an oscillating movement of gear F. If AB has a constant angular velocity of AB=6 rad/s, determine the angular velocity and angular acceleration of gear F at the instant shown. Gear E is rigidly attached to arm CD and pinned to a fixed point.

Link AB

VELOCITY:     v B  6k  75 j  450i

   vB  v A  vB / A Link BC

   v C  v B  vC / B

        vC  450i   BC k  100 cos 30i  100 sin 30 j   450i  86.6 BC j  50 BC i 1      86.6i  50 j

Member CDG

    vC   E k  150 j  150 E i

   vC  v D  vC / D

1 = 2

 i  j



450  50 BC  150 E



50 BC  0

2

 BC  0  E  3 rad / s Gear F

vG   E 100   F 25

 F  12 rad / s

G

  F  12k 

  E  3k 

Link AB

ACCELERATION:





     a B  6k  6k  75 j  2700 j

   aB  a A  aB / A Link BC

   aC  a B  aC / B

          aC  2700 j   BC k  86.6i  50 j  2700 j  86.6 BC j  50 BC i

Member CDG



3 = 4

 i  j



        aC  3k  3k  150 j   E k  150 j  1350 j  150 E i

   aC  a D  aC / D 

50 BC  150 E



2700  86.6 BC  1350

4

 BC  15.58 rad / s 2  E  5.196 rad / s 2 Gear F

aG t   E 100   F 25  F  20.785 rad / s

G

2

  F  20.785k 

3