978-0133915426 Chapter 17 Part 7

subject Type Homework Help
subject Pages 9
subject Words 1996
subject Authors Russell C. Hibbeler

Unlock document.

This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
page-pf1
902
17–111.
4 (0.4) m
3p
SOLUTION
For roll A.
For roll B
a(2)
(3)
Kinematics:
(4)
also,
(5)
Solving Eqs. (1)–(5) yields:
Ans.
Ans.
Ans.
aB=7.85 m>s2aO=3.92 m>s2
T=15.7 N
aB=43.6 rad>s2
aA=43.6 rad>s2
A
+T
B
aO=aA(0.09)
A
+T
B
aB=aO+0.09aB
caB
Td=caO
Td+caB(0.09)
Td+[0]
aB=aO+(aB>O)t+(aB>O)n
+c©Fy=m(aG)y;T-8(9.81) =-8aB
MO(Mk)O;8(9.81)(0.09) =1
2(8)(0.09)2aB+8aB(0.09)
The semicircular disk having a mass of 10 kg is rotating
at v = 4 rad>s at the instant u = 60º. If the coefficient of
static friction at A is m = 0.5, determine if the disk slips at
this instant.
VO
G
0.4 m
A
u
v
page-pf2
*17–112.
4 ft
0.5 ft
G
O
v
The circular concrete culvert rolls with an angular velocity
of when the man is at the position shown. At
this instant the center of gravity of the culvert and the man is
located at point G,and the radius of gyration about Gis
Determine the angular acceleration of the
culvert. The combined weight of the culvert and the man is
500 lb.Assume that the culvert rolls without slipping, and the
man does not move within the culvert.
kG=3.5 ft.
v=0.5 rad>s
SOLUTIONS
page-pf3
904
17–113.
v0
r
The uniform disk of mass mis rotating with an angular
velocity of when it is placed on the floor. Determine the
initial angular acceleration of the disk and the acceleration
of its mass center.The coefficient of kinetic friction between
the disk and the floor is .mk
v0
r
page-pf4
905
17–114.
v
0
r
The uniform disk of mass mis rotating with an angular
velocity of when it is placed on the floor. Determine the
time before it starts to roll without slipping.What is the
angular velocity of the disk at this instant? The coefficient
of kinetic friction between the disk and the floor is .mk
v0
page-pf5
17–115.
SOLUTION
c(1)
For B:
+
90 mm
A
D
C
A cord is wrapped around each of the two 10-kg disks.
If they are released from rest, determine the angular
acceleration of each disk and the tension in the cord C.
Neglect the mass of the cord.
MA=IAaA;T(0.09) =c1
2(10)(0.09)2daA
©
page-pf6
*17–116.
The disk of mass mand radius rrolls without slipping on the
circular path.Determine the normal force which the path
exerts on the disk and the disk’s angular acceleration if at
the instant shown the disk has an angular velocity of V.
SOLUTION
R
page-pf7
908
17–117.
The uniform beam has a weight W. If it is originally at rest
while being supported at Aand Bby cables, determine the
tension in cable Aif cable Bsuddenly fails. Assume the
beam is a slender rod.
SOLUTION
BA
7
page-pf8
17–118.
SOLUTION
a
Ans.
Ans.
means that the beam stays in contact with the roller support.By70
By=9.62 lb
a=23.4 rad>s2
(+T)(aG)y=a(3)
-aBi=-(aG)xi-(aG)yj+a(3)j
aB=aG+aB>G
+aMB=a(Mk)B; 500(3) +1000 a3
5b(8) =500
32.2(aG)y(3) +c1
12 a500
32.2 b(10)2da
+TaFy=m(aG)y; 1000 a3
5b+500 -By=500
32.2 (aG)y
;
+aFx=m(aG)x; 1000a4
5b=500
32.2 (aG)x
The 500-lb beam is supported at Aand Bwhen it is
subjected to a force of 1000 lb as shown. If the pin support
at Asuddenly fails, determine the beam’s initial angular
acceleration and the force of the roller support on the beam.
For the calculation, assume that the beam is a slender rod
so that its thickness can be neglected.
BA
8ft2ft
1000 lb
3
4
5
page-pf9
910
17–119.
SOLUTION
Ans.a=10g
1322r
©Ma-a(Mk)a-a;mg sin 45°ar
2b=c2
5mr2+mar
2b2da
d=rsin 30° =r
2
The solid ball of radius rand mass mrolls without slipping
down the trough. Determine its angular acceleration.60°
30°
Ans:
a=
10g
1322 r
page-pfa
*17–120.
By pressing down with the finger at B, a thin ring having a
mass m is given an initial velocity and a backspin when
the finger is released. If the coefficient of kinetic friction
between the table and the ring is determine the distance
the ring travels forward before backspinning stops.
mk,
V0
v0
SOLUTION
a
(c
Ans.s=v0r
mkgv0-1
2v0r
s=0+v0av0r
mkgb-a1
2b(mkg)av0
2r2
mk
2g2b
A
;
+
B
s=s
0+v0t+1
2act2
t=v0r
mkg
0=v0-amkg
rbt
+)v=v0+act
a=mkg
r
MG=IGa;mk(mg)r=mr2a
aG=mkg
:
+©F
x=m(aG)x;mk(mg)=m(aG)
N
A=mg
+c©F
y=0; N
A-mg =0
B
A
0
0
v
ω
r

Trusted by Thousands of
Students

Here are what students say about us.

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.