Dynamics pulley porblems chpter 5

Questiosns about dynamics

1. Asst. Prof. Murat Saribay
Bilgi University
ME 232 Dynamics (Chapter V - Examples)

2. Sample Problem
Cable C has a constant acceleration of 225
mm/s
2
and an initial velocity of 300 mm/s,
both directed to the right.
Determine (a) the number of revolutions
executed by the pulley in 2 s, (b) the
velocity and change in position of the load B
after 2 s, and (c) the acceleration of the
point D on the rim of the inner pulley at t=0.
SOLUTION:
•Due to the action of the cable, the
tangential velocity and acceleration of
D are equal to the velocity and
acceleration of C. Calculate the initial
angular velocity and acceleration.
•Apply the relations for uniformly
accelerated rotation to determine the
velocity and angular position of the
pulley after 2 s.
•Evaluate the initial tangential and
normal acceleration components of D.

3. Sample Problem
SOLUTION:
•The tangential velocity and acceleration of D are equal to
the velocity and acceleration of C.



srad
r
v
rv
smm.
D
D
CD
4
75
300
300
0
0
00
00





vv 


2
2
3
75
225
225
srad
r
a
ra
smm.
tD
tD
CtD





aa
•Apply the relations for uniformly accelerated rotation to determine velocity
and angular position of pulley after 2s.
  srad ssradsradαtωω 10234
2
0

   rad ssrad ssradtt 142324
22
2
12
2
1
0  
  revs ofnumber
rad 2
rev 1
rad 14 







N rev23.2N
  
   rad mmrθΔy
srad mmrωv
B
B
14125
10125


mmΔy
smm
B
B
1750
1250

v

4. Sample Problem
•Evaluate the initial tangential and normal
acceleration components of D.
  smm
CtD
225aa
   
222
0 1200475 smmsrad mmωra
DnD 
  
22
1200225 smmsmm
nDtD aa

22
22
1200225

nDtDD aaa
2
91220 smm.a
D


225
1200


tD
nD
a
a
tan
4.79

5. Sample Problem
The pinion A of the hoist motor drives gear
B, which is attached to the hoisting drum.
The load L is lifted from its rest position and
acquires an upward velocity of 2 m/s in a
vertical rise of 0.8 m with constant
acceleration. As the load passes this
position, compute (a) the acceleration of
point C on the cable in contact with the
drum and (b) the angular velocity and
angular acceleration of the pinion A.
(a) The load is raised by a constant
acceleration. Initially at rest and reaches a
velocity of 2 m/s after travelling a distance
of 0.8m. Thus the acceleration is:
The point C has a tangential acceleration
that is equal to the acceleration of the load
Similarly, the point C has a velocity equal
to the velocity of the load,
savv
LLoL
2
22

80202
2
.a
L
L
as/m. 
2
52

2
52 s/m.a
tC

s/mvv
LC 2

6. Sample Problem
Thus, the normal component of the acceleration of point
C is,
The total acceleration is then,
Angular motion of A is determined from the angular
motion of gear B. The point of contact between these two
gears has the common velocity v
1 and acceleration a
1.
Angular velocity of B is determined from the velocity of
point C;
Las/m. 
2
52

2
52 s/m.a
tC

s/mvv
LC 2

2
22
10
40
2
s/m
.
v
a
C
nC



22222
31101052 s/m..aaa
nCtCC

BCC
rv


B
.

402 CCWs/rad
B
5


7. Sample Problem
The tangential component of the acceleration of point C is,
Thus, the velocity and tangential acceleration of point 1 on
gear B are,
The velocity and tangential acceleration of point 1 on gear
A are,
L
as/m. 
2
52

2
52 s/m.a
tC
s/mvv
LC
2
CCWs/rad
B 5


CBtC ra


B
..

4052
CCWs/rad.
B

2
256


11 ra
Bt



Brv

11 s/m..v 51530
1


2
1 875125630 s/m...a
t

AArv

11 CWs/rad..
AA
151051  


AAtA ra
11


CWs/rad...
AA
2
7518108751  

◄
◄

8. Sample Problem
The center of the double gear has a
velocity and acceleration to the right
of 1.2 m/s and 3 m/s
2
, respectively.
The lower rack is stationary.
Determine (a) the angular acceleration
of the gear, and (b) the acceleration of
points B, C, and D.
SOLUTION:
•The expression of the gear position as a
function of q is differentiated twice to
define the relationship between the
translational and angular accelerations.
•The acceleration of each point on the
gear is obtained by adding the
acceleration of the gear center and the
relative accelerations with respect to
the center. The latter includes normal
and tangential acceleration
components.

9. Sample Problem
SOLUTION:
•The expression of the gear position as a function of
q is differentiated twice to define the relationship
between the translational and angular
accelerations.


11
1
rrv
rx
A
A



srad 8
m 0.150
sm2.1
1

r
v
A


11
rra
A

m 150.0
sm3
2
1

r
a
A

 kkα
2
20 srad

10. Sample Problem
•The acceleration of each
point is obtained by
adding the acceleration of
the gear center and the
relative accelerations with
respect to the center.
The latter includes normal
and tangential acceleration
components.

     
   jii
jjki
rrka
aaaaaa
222
222
2
40623
100081000203
sm.smsm
m.sradm.sradsm
ωα
ABABA
n
AB
t
ABAABAB




  
222
1284065 sm.asm.sm
BB  jia

11. Sample Problem
     
   jii
jjki
rrkaaaa
222
222
2
60933
150081500203
sm.smsm
m.sradm.sradsm
ACACAACAC


 
 ja
2
609 sm.
c

     
   iji
iiki
rrkaaaa
222
222
2
60933
150081500203
sm.smsm
m.sradm.sradsm
ADADAADAD


 
  
222
95123612 sm.asmsm.
DD
 jia

12. Sample Problem
Crank AB of the engine system has a
constant clockwise angular velocity of
2000 rpm.
For the crank position shown,
determine the angular acceleration of
the connecting rod BD and the
acceleration of point D.
SOLUTION:
•The angular acceleration of the
connecting rod BD and the acceleration
of point D will be determined from
•The acceleration of B is determined from
the given rotation speed of AB.
•The directions of the accelerations a
D
,
(a
D/B)
t, and (a
D/B)
n are determined from
the geometry.
•Component equations for acceleration of
point D are solved simultaneously for
acceleration of D and angular
acceleration of the connecting rod.

n
BD
t
BDBBDBD aaaaaa 

13. Sample Problem
•The acceleration of B is determined from the given
rotation speed of AB.
SOLUTION:
•The angular acceleration of the connecting rod BD
and the acceleration of point D will be determined
from

n
BD
t
BDBBDBD aaaaaa 
  
222
6328842090750
0
42092000
sm.srad. m.rωa
constsrad.rpmω
ABB
AB
AB




  jia  40sin40cos6.3288
2
sm
B
 jia 9.21132.2519
B

14. Sample Problem
•The directions of the accelerations a
D, (a
D/B)
t, and (a
D/B)
n are determined from the
geometry.
From Sample Problem 15.3, w
BD
= 61.98 rad/s, b = 13.95
o
.
  
222
3.76898.612.0 smsradmBDa
BD
n
BD

  jia  95.13sin95.13cos3.768
2
sm
n
BD
 
BDBDBD
t
BD
α.αmαBDa 202.0 
The direction of (a
D/B
)
t
is known but the sense is not known,
  jia  0576057620 .sin.cos.
BD
t
BD 
ia
DDa

15. Sample Problem
•Component equations for acceleration of point D are solved
simultaneously.
x components:
 95.13sin2.095.13cos3.76840cos6.3288
BD
o
Da 
 95.13cos2.095.13sin3.76840sin6.32880
BD
o

y components:
 
 ia
kα
2
2
2778
9900
sm
srad
D
BD



n
BD
t
BDBBDBD aaaaaa 

16. Sample Problem
In the position shown, crank AB has a
constant angular velocity w
1
= 20
rad/s counterclockwise.
Determine the angular velocities and
angular accelerations of the
connecting rod BD and crank DE.
SOLUTION:
•The angular velocities are
determined by simultaneously
solving the component equations
for
BDBD
vvv 
•The angular accelerations are
determined by simultaneously
solving the component equations for
BDBD
aaa 

17. Sample Problem
SOLUTION:
•Velocity of point D can be obtained from the
expressions,
BDBD
vvv 
 
ji
jikrωv
DEDE
DEE/DDED


4242
4242


 
ij
jikrωv
700400
352020


A/BABB
 
ji
jikrωv
BDBD
BDBDBDBD


307
730


BDDE  770042 
BDDE
 3040042 
   kωkω sradsrad
DEBD
71.1137.29 
  jijiji
BDBDDEDE  3074007004242 
E/DDED
rωv 

18. Sample Problem
Similar to the velocity case consider the two
expressions for the acceleration of point D,
  
jiji
jijik
rrαa
574057404242
424271114242
2
2



DEDE
DE
DDEDDED
.



 
ji
j0irrαa
140008000
352200
22


BABBABB 
  
jiji
jij0ik
rrαa
616026400307
730732973
2
2



DBDB
DB
DBBDDBBDBD
.



x components: 40140742 
BDDE
y components: 144203042 
BDDE
   kαkα
22
840695 sradsrad
DEBD 
BDBD
aaa  DDEDDED rrαa
2


19. Sample Problem
The crane rotates with a constant
angular velocity w
1
= 0.30 rad/s and
the boom is being raised with a
constant angular velocity w
2
= 0.50
rad/s relative to the cab. The length
of the boom is l
OP = 12 m.
Determine:
•angular velocity of the boom,
•angular acceleration of the boom,
•velocity of the boom tip, and
•acceleration of the boom tip.
•Angular acceleration of the boom,

21
22221
ωω
ωΩωωωωα




Oxyz
•Velocity of boom tip,
rωv
•Acceleration of boom tip,
 vωrαrωωrαa 
SOLUTION:
With
•Angular velocity of the boom,
21ωωω 
 
ji
jir
kωjω
63910
303012
500300
21



.
sincos
..

20. Sample Problem
jir
kωjω
63910
500300
21


.
..
SOLUTION:
•Angular velocity of the boom,
  kjω srad.srad. 500300 
•Angular acceleration of the boom,

  kjωω
ωΩωωωωα
srad.srad.
Oxyz
500300
21
22221

 
 iα
2
150 srad.
•Velocity of boom tip,
063910
50300
.
..
kji
rωv 
   kjiv sm.sm.sm. 123205543 
21
ωωω 

21. Sample Problem
•Acceleration of boom tip,

kjiik
kjikji
a
vωrαrωωrαa
900501602940900
1232053
5003000
063910
00150
.....
..
..
.
.




   kjia
222
801501543 sm.sm.sm. 
jir
kωjω
63910
500300
21


.
..

22. Sample Problem
For the disk mounted on the arm, the
indicated angular rotation rates are
constant. Determine:
•the velocity of the point P,
•the acceleration of P, and
•angular velocity and angular
acceleration of the disk.
SOLUTION:
•Define a fixed reference frame OXYZ at O
and a moving reference frame Axyz or F
attached to the arm at A.
•With P′ of the moving reference frame
coinciding with P, the velocity of the
point P is found from
FPPP
vvv 

•The acceleration of P is found from
cFPPP
aaaa 

•The angular velocity and angular
acceleration of the disk are
 ωΩωα
ωΩω
F
F



D

23. Sample Problem
SOLUTION:
•Define a fixed reference frame OXYZ at O
and a moving reference frame Axyz or F
attached to the arm at A.
jΩ
jir
1
 RL
kω
jr
F 2

D
AP
R
•With P’ of the moving reference frame
coinciding with P, the velocity of the point P
is found from
 
ijkrωv
kjijrΩv
vvv
FF
F
RR
LRL
APDP
P
PPP
22
11







kiv LR
P 12

24. Sample Problem
•The acceleration of P is found from
cPPP
aaaa
F


   ikjrΩΩa LL
P
2
111  

 
  jik
rωωa
FFF
RR
APDDP
2
222
 

  kij
vΩa
F
RR
Pc
2121
22
2
 

kjia RRL
P 21
2
2
2
1
2 
•Angular velocity and acceleration of the disk,
FωΩω
D kjω
21


 kjj
ωΩωα
F
211
 

iα
21

27. At a given instant, the gear racks have the
velocities and accelerations as shown.
Determine (a) the angular velocity and
angular acceleration of the gear A and (b)
acceleration of the center of the gear A and the
point B on the perimeter.