Download MOTION RELATIVE TO ROTATING AXES

PROBLEMS
The gear has the angular motion shown. Determine the angular velocity and angular
acceleration of the slotted link BC at this instant. The pin at A is fixed to the
gear.
C
A
w=2 rad/s
2m
0.5 m
0.7 m
B
O
a=4 rad/s2
PROBLEMS
Link 1, of the plane mechanism shown, rotates about the fixed point O
with a constant angular speed of 5 rad/s in the cw direction while
slider A, at the end of link 2, moves in the circular slot of link 1.
Determine the angular velocity and the angular acceleration of link 2 at
the instant represented where BO is perpendicular to OA. The radius
of the slot is 10 cm.
Take sin 37=06, cos 37=0.8
1
A
10 cm
2
20 cm
37o
C
37o
w1=5 rad/s
B
O
16 cm
BO  OA
PROBLEMS
For the instant shown, particle A has a velocity of 12.5 m/s towards point C
relative to the disk and this velocity is decreasing at the rate of 7.5 m/s each
second. The disk rotates about B with angular velocity w=9 rad/s and angular
acceleration a=60 rad/s2 in the directions shown in the figure. The angle b
remains constant during the motion. Telescopic link has a velocity of 5 m/s and an
acceleration of -2.5 m/s. Determine the absolute velocity and acceleration of
point A for the position shown.
Problem 7
Velocity Analysis
  


v A  v B  w  r  v rel

 24  7  
v B  5 i 
j
25 
 25



v B  4.8 i  1.4 j
vB
b 25
24
7
Velocity Analysis
2
yx 
9
2

2
2
2
x 
3
9

  2  2 
 
w  r  9k   i  j 
3 
3
 
 
w  r  -6 i  6 j
2
x m
3
Velocity Analysis
2
2
2
2
yx 

x 
9
3
9
dy
4
tan  
 2x 
dx x 2 / 3
3
2
vrel

3
5
4

 3  4 
v rel  12.5 i  j 
5 
5



v rel  7.5 i  10 j



v A  6.3 i  17.4 j

2
x m
3
Acceleration Analysis


    
 

a A  a B  w  w  r   a  r  2w  v rel  a rel

aB

aB
 24  7  
 -2.5 i 
j
25 
 25


 -2.4 i - 0.7 j
aB
b 25
24
7
Acceleration Analysis
2
yx 
9
2

2
2
2
x 
3
9

2
x m
3


    2  2  

 


  
w  w  r   9k  9k   i  j   9k  - 6 i  6 j  -54 i - 54 j
3 
3

  2  2 


 
a  r  -60k   i  j   40 i - 40 j
3 
3
Acceleration Analysis
 




2w  v rel  2 9k  7.5 i  10 j



2w  v rel  -180 i  135 j

Acceleration Analysis
+n
+t
(arel)n
vrel

3
dy
4
 2x 
dx x  2 / 3
3
d2y
2
dx 2

  dy 
1   
  dx 

d2y
dx 2
2 3/ 2



v 2rel 12.52
a rel n  
 67.49 m / s 2

2.315



 4  3 
a rel n  67.49 - i  j   -53.992 i  40.494 j
5 
 5
 

 3  4 
a rel t  -7.5 i  j   -4.5 i - 6 j
5 
5
(arel)t
 2.315 m



a A  -254.892 i  74.794 j
5
4
PROBLEMS
The pin A in the bell crank AOD is guided by the flanges of the collar
B, which slides with a constant velocity vB of 0.9 m/s along the fixed
shaft for an interval of motion. For the position =30o determine the
acceleration of the plunger CE, whose upper end is positioned by the
radial slot in the bell crank. .
Problem 8
Velocity Analysis
vA
vrel
vA
30o
vB=(vA)x
129.9 mm
60o
vA 

30o
vB
0.9

 1.039 m / s
cos 30 cos 30
1.039
wAOD  w 
 6.928 rad / s
0.15


vC  -v C j
(1)








 



v C  v O  w  rC / O  v rel  -6.928k  0.225 i  0.13 j  v rel cos 30 i  sin 30 j

0





v C  0.9 i - 1.56 j  0.866v rel i  0.5v rel j
(2)
vrel
(1)=(2)
vrel=-1.039 m/s
vc=2.079 m/s

Acceleration Analysis
aA
vrel
VB=constant
(aA)n
o
30
aA
So aA must be vertical.
60o
(aA)t
30o
129.9 mm
a A n  w2 OA  6.9282  0.15  7.195 m / s 2
a A n
2
aA 
 8.308 m / s
cos 30
a A t  a A sin 30  4.154 m / s 2
a A t  a AOD AO

a AOD  a  27.695 rad / s 2


aC  aC j
(3)


    
 

a C  a O  w  w  r   a  r  2w  v rel  a rel

0








a C  -6.928k  - 6.928k  0.225 i  0.13 j  27.695k  0.225 i  0.13 j





 2 - 6.928k  - 1.039 cos 30 i - 1.039 sin 30 j  arel cos 30 i  sin 30 j





a C  -21.58 i  12.464 j  0.866a rel i  0.5a rel j
(4)










(3)=(4)
arel=24.92 m/s2
aC=27.92 m/s2
PROBLEMS
1. The uniform 30-kg bar OB is secured to the accelerating frame in the
30o position from the horizontal by the hinge at O and roller at A. If the
horizontal acceleration of the frame is a=20 m/s2, compute the force FA on
the roller and the x- and y-components of the force supported by the pin
at O.
PROBLEMS
2. The block A and attached rod have a combined mass of 60 kg and are
confined to move along the 60o guide under the action of the 800 N applied
force. The uniform horizontal rod has a mass of 20 kg and is welded to the
block at B. Friction in the guide is negligible. Compute the bending moment M
exerted by the weld on the rod at B.
SOLUTION
FBD
Kinetic Diagram
mTax=60ax
x
x

N
60o
W=60(9.81) N
 Fx ext. forces  max 
800 - 60(9.81) sin 60  60a x
a x  4.84 m / s 2
By
FBD of rod
KD of rod
m1ax=20ax
Bx
M
W1=20(9.81) N
M
B
 ma x d 
M  196 m / s 2
M - 20 (9.81)0.7  (20 )( 4.94 )( 0.7 sin 60 )
PROBLEMS
3. The parallelogram linkage shown moves in the vertical plane with the
uniform 8 kg bar EF attached to the plate at E by a pin which is welded both to
the plate and to the bar. A torque (not shown) is applied to link AB through its
lower pin to drive the links in a clockwise direction. When  reaches 60o, the
links have an angular acceleration an angular velocity of 6 rad/s2 and 3 rad/s,
respectively. For this instant calculate the magnitudes of the force F and
torque M supported by the pin at E.
PROBLEMS
4. The uniform 100 kg log is supported by the two cables and used as a
battering ram. If the log is released from rest in the position shown, calculate
the initial tension induced in each cable immediately after release and the
corresponding angular acceleration a of the cables.
SOLUTION
+n
FBD
KD
+n
TA
TB
ma n

+t
mat
+t
W=100(9.81) N
When it starts to move, v=0, w=0 but a≠0
 Fn ext. forces  0

 Ft ext. forces  mat
at  ar 

a
an  w 2 r  0
TA  TB - mg cos 30  0
mg sin 30  mat

TA  TB  849.57

at  4.905 m / s
2

3TA  TB
4.905
 2.45 rad / s 2
2
Length of the cables
The motion of the log is curvilinear translation.
 M 
G d .k .
0 
TA  212 .39 N
TA sin 60 (1.5) - TB sin 60 (0.5)  0
TB  637 .17 N
*
*
PROBLEMS
5. An 18 kg triangular plate is supported by cables AB and CD. When the plate
is in the position shown, the angular velocity of the cables is 4 rad/s ccw. At
this instant, calculate the acceleration of the mass center of the plate and the
tension in each of the cables.
C
A
60°
24 cm
B
10 cm
60°
D
G
Answer:
20 cm
20 cm
a  6.23 m / s 2
TAB  143 .11 N
TCD  78 .93 N
PROBLEMS
6. The uniform 8 kg slender bar is hinged about a horizontal axis through O
and released from rest in the horizontal position. Determine the distance b
from the mass center to O which will result in an initial angular acceleration of
16 rad/s2, and find the force R on the bar at O just after release.
PROBLEMS
7. The spring is uncompressed when the uniform slender bar is in the vertical
position shown. Determine the initial angular acceleration a of the bar when it
is released from rest in a position where the bar has been rotated 30o
clockwise from the position shown. Neglect any sag of the spring, whose mass
is negligible.
SOLUTION
Unstrecthed length of the spring:
When =30o , length of the spring:
When
=30o
lo  (2l / 4) 2  l 2 
l spring 
5
l
2
3
l
2
 5
 5
3 
3 


, spring force: Fspring  k
ll  kl
 2

 2
2
2 



(in compression)
 M O  Ia  mat
30o
W
+t
l
l 1
l

- mg cos 60  Fspring   ml 2 a  m at
 4
4
2  12

l
.
60o
Ot +n
O
lspring
G
60o
30o
l
On
a
Fspring

+t
mat
+n
G
Ia
man  mw 2
a  0.864
l
0
4
k
g
- 0.857
m
l
4