978-0077687342 Chapter 17 Part 4

subject Type Homework Help
subject Pages 14
subject Words 1339
subject Authors Brian Self, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM 17.38
A long ladder of length l, mass m, and centroidal mass moment of inertia
I
is placed against a house at an angle
. Knowing that the ladder is
released from rest, determine the angular velocity of the ladder when
2.
Assume the ladder can slide freely on the horizontal ground and
on the vertical wall.
SOLUTION
Kinematics:
Kinetic energy: 22
11
Tmv I

page-pf2
PROBLEM 17.38 (Continued)
Position 2. 2;?
 

11
LL

page-pf3
PROBLEM 17.39
The ends of a 9-lb rod AB are constrained to move along slots cut in a
vertical plate as shown. A spring of constant
3k
lb/in. is attached to
end A in such a way that its tension is zero when
0.
If the rod is
released from rest when
50 ,

determine the angular velocity of the
rod and the velocity of end B when
0.
SOLUTION
22
2
2
B
L
v
vL
22
22 2
2
22
22
2
22 2 2
222
11
22
111
22 212
119lb25in.
0.2022
6632.212
ge
Tmv I
L
mmL
mL



 

 
 
 
 
 
 
page-pf4
PROBLEM 17.39 (Continued)
page-pf5
PROBLEM 17.40
The mechanism shown is one of two identical mechanisms
attached to the two sides of a 200-lb uniform rectangular door.
Edge ABC of the door is guided by wheels of negligible mass
that roll in horizontal and vertical tracks. A spring of constant k
40 lb/ft is attached to wheel B. Knowing that the door is
released from rest in the position 30

with the spring
unstretched, determine the velocity of wheel A just as the door
reaches the vertical position.
SOLUTION
Kinematics. Locate the instantaneous center at point I.

5sin 5cos
AB
vvv


Moment of inertia. 2
1:
12
W
I
l
g

22
120010 51.760 lb s ft
12 32.2
I

Kinetic energy. 22
11
22
W
TvI
g

 
222
1200 1
5cos 51.760
2 32.2 2
T


Potential energy.
277.640cos 90 25.880 25.880T
page-pf6
PROBLEM 17.40 (Continued)
page-pf7
PROBLEM 17.41
The mechanism shown is one of two identical mechanisms
attached to the two sides of a 200-lb uniform rectangular door.
Edge ABC of the door is guided by wheels of negligible mass
that roll in horizontal and vertical tracks. A spring of constant k
is attached to wheel B in such a way that its tension is zero
when 30 .
Knowing that the door is released from rest in
the position 45
and reaches the vertical position with an
angular velocity of 0.6 rad/s, determine the spring constant k.
SOLUTION
Kinematics. Locate the instantaneous center at point I.

5sin 5cos
AB
vvv


Moment of inertia. 2
1
12
W
I
l
g

22
120010 51.760 lb s ft
12 32.2
I

Kinetic energy. 22
11
22
W
TvI
g

 
222
1200 1
5cos 51.760
2 32.2 2
T


Potential energy.
Gravity. Datum at level A.
page-pf8
PROBLEM 17.41 (Continued)

277.640 cos 90 25.880 0.6 9.3168 ft lbT



2
22225 sin 90 0.5 1000sin 90
eg
VV V k
  
6.25 1000k
Conservation of energy. 11 2 2
:TV T V
0 1.07233 707.11 9.3168 6.25 1000kk

54.8 lb/ftk
page-pf9
PROBLEM 17.42
Each of the two rods shown is of length L 1 m and has a mass of 5 kg.
Point D is connected to a spring of constant k 20 N/m and is constrained
to move along a vertical slot. Knowing that the system is released from rest
when rod BD is horizontal and the spring connected to Point D is initially
unstretched, determine the velocity of Point D when it is directly to the
right of Point A.
2
2
31
22
mgL kx

page-pfa
PROBLEM 17.42 (Continued)
Kinematics.
AB AB
BABBAB
vL L

v
30°
3
(sin60) 2
GBDAB
vL L


222
111
DBDAB
vL L


(1)
21
()
12 2 2 2
AB
mL mgL k x x
  


(2)
2
12
20 N m, 0, 1 m
kxxL
  
22 2
21
22
11
22
mgL

 



page-pfb
PROBLEM 17.42 (Continued)
By Eq. (2), 22 2 2
735
kg m 21.096 J
mL


 
page-pfc
PROBLEM 17.43
The 4-kg rod AB is attached to a collar of negligible mass at A and to a flywheel
at B. The flywheel has a mass of 16 kg and a radius of gyration of 180 mm.
Knowing that in the position shown the angular velocity of the flywheel is 60 rpm
clockwise, determine the velocity of the flywheel when Point B is directly below C.
0.24
sin 19.471
0.72


11
2
1
22
0.3744
page-pfd
PROBLEM 17.43 (Continued)
Position 2. Point B is directly below C.
2
1
2
1(0.72) 0.24
AB
hLr

page-pfe
PROBLEM 17.44
If in Problem 17.43 the angular velocity of the flywheel is to be the same in the
position shown and when Point B is directly above C, determine the required value
of its angular velocity in the position shown.
PROBLEM 17.43
The 4-kg rod AB is attached to a collar of negligible mass at A
and to a flywheel at B. The flywheel has a mass of 16 kg and a radius of gyration of
180 mm. Knowing that in the position shown the angular velocity of the flywheel is
60 rpm clockwise, determine the velocity of the flywheel when Point B is directly
below C.
SOLUTION
Moments of inertia.
2
1
11
2
1
22
0.3744
page-pff
PROBLEM 17.44 (Continued)
Position 2. Point B is directly above C.
1
page-pf10
PROBLEM 17.45
The uniform rods AB and BC of masses 2.4 kg and 4 kg, respectively,
and the small wheel at C is of negligible mass. If the wheel is moved
slightly to the right and then released, determine the velocity of pin B
after rod AB has rotated through 90 .
Rod BC is at rest. 0
BC
00
B
v
vv v v L
  
page-pf11
PROBLEM 17.46
The uniform rods AB and BC of masses 2.4 kg and 4 kg, respectively,
and the small wheel at C is of negligible mass. Knowing that in the
position shown the velocity of wheel C is 2 m/s to the right,
determine the velocity of pin B after rod AB has rotated through 90.
2
page-pf12
PROBLEM 17.46 (Continued)
BB
vv
page-pf13
PROBLEM 17.47
The 80-mm-radius gear shown has a mass of 5 kg and a centroidal radius
of gyration of 60 mm. The 4-kg rod AB is attached to the center of the
gear and to a pin at B that slides freely in a vertical slot. Knowing that the
system is released from rest when
60 ,

determine the velocity of the
center of the gear when
20 .

22 2
11
(4)(0.320) 0.03413 kg m
12 12
AB AB
ImL
  
page-pf14
PROBLEM 17.47 (Continued)
Conservation of energy: 11 2 2
TV T V
1
1(4)(9.81)(0.320) cos 60
2
3.1392 J
V
 

Position 2: 20 ?
A
v
 
22
2
2
4.66124
1(4)(9.81)(0.320) cos 20
2
A
A
v
V
 
0.59225 m /s
A
v

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