17–61.
If a horizontal force of P = 100 N is applied to the 300-kg
reel of cable, determine its initial angular acceleration.
The reel rests on rollers at A and B and has a radius of
gyration of
kO=0.6 m.
SOLUTION
2020
O
AB
0.75 m
1 m
P
17–62.
SOLUTION
Ans.v=10.9rad/s
48.3
2(p
2)2=1
2v2
–Lo
p
2
48.3 udu=Lv
0
vdv
adu=vdv
–48.3 u=a
The 10-lb bar is pinned at its center Oand connected to a
torsional spring.The spring has a stiffness so
that the torque developed is where is in
radians.If the bar is released from rest when it is vertical at
determine its angular velocity at the instant u=0°.u=90°,
uM=15u2lb #ft,
k=5lb#ft>rad,
1ft
1ft
O
u
17–63.
The 10-lb bar is pinned at its center Oand connected to a
torsional spring.The spring has a stiffness so
that the torque developed is where is in
radians.If the bar is released from rest when it is vertical at
determine its angular velocity at the instant u=45°.u=90°,
uM=15u2lb #ft,
k=5lb#ft>rad,
SOLUTION
Ans.v=9.45 rad s
–24.15a(p
4)2–(p
2)2b=1
2v2
–Lp
4
p
2
48.3udu=Lv
0
vdv
adu=vdv
a=-48.3u
1ft
1ft
O
u
*17–64.
A cord is wrapped around the outer surface of the 8-kg disk.
If a force of F =
(
¼
u
2
)
N, where
u
is in radians,
is applied to the cord, determine the disk’s angular
acceleration when it has turned 5 revolutions. The disk has
an initial angular velocity of v
0=1 rad>s.
SOLUTION
300 mm
O
F
v
17–65.
Disk Ahas a weight of 5 lb and disk Bhas a weight of 10 lb.
If no slipping occurs between them, determine the couple
moment Mwhich must be applied to disk Ato give it an
angular acceleration of 4 rad>s2.
SOLUTION
Disk B:
Solving:
Ans.M=0.233 lb #ft
aB=2.67 rad>s2;FD=0.311 lb
0.5(4) =0.75aB
rAaA=rBaB
+©MB=IBaB;FD(0.75) =c1
2a10
32.2 b(0.75)2daB
0.75ft
B
M
A
a4rad/s
2
0.5 ft
17–66.
The kinetic diagram representing the general rotational
motion of a rigid body about a fixed axis passing through O
is shown in the figure.Show that may be eliminated by
moving the vectors and to point P,located a
distance from the center of mass Gof the
body.Here represents the radius of gyration of the body
about an axis passing through G.The point Pis called the
center of percussion of the body.
kG
rGP =k2
G>rOG
m(aG)n
m(aG)t
IGA
SOLUTION
However,
and
Q.E.D.=m(aG)t(rOG +rGP)
m(aG)trOG +IGa=m(aG)trOG +(mrOG rGP)c(aG)t
rOG d
a=(aG)t
rOG
k2
G=rOG rGP
m(aG)trOG +IGa=m(aG)trOG +
A
mk2
G
B
a
rGP
rOG
m(aG)n
G
IG
m(aG)t
O
P
a
a
17–67.
If the cord at B suddenly fails, determine the horizontal
and vertical components of the initial reaction at the
pin A, and the angular acceleration of the 120-kg beam.
Treat the beam as a uniform slender rod.
SOLUTION
B
A
2 m 2 m
800 N
*17–68.
SOLUTION
Ans.
Equation of Motion: Applying Eq. 17–16, we have
a
Ans.a1.25 rad s2
–200(9.81) cos 45°(0.75) =-174.75a
+©MA=IAa; 100(9.81)(0.625) +200(9.81) sin 45°(0.15)
=174.75 kg #m2=175 kg #m2
+1
12(200)
A
0.52+0.32
B
+200
A
20.752+0.152
B
2
12 (100)
AC
1.25 m
0.5 m
0.5 m
0.3 m
B
The device acts as a pop-up barrier to prevent the passage
of a vehicle. It consists of a 100-kg steel plate AC and a
200-kg counterweight solid concrete block located as
shown. Determine the moment of inertia of the plate and
block about the hinged axis through A. Neglect the mass of
the supporting arms AB. Also, determine the initial angular
acceleration of the assembly when it is released from rest at
u = 45°.
859
17–69.
The 20-kg roll of paper has a radius of gyration kA = 90 mm
about an axis passing through point A. It is pin supported at
both ends by two brackets AB. If the roll rests against a wall
for which the coefficient of kinetic friction is μk = 0.2 and a
vertical force F = 30 N is applied to the end of the paper,
determine the angular acceleration of the roll as the paper
unrolls.
Ans:
a
=
7.28 rad
>
s
2
Solving:
NC=103 N
TAB =267 N
a
=7.28 rad>s2
Ans.
300 mm
B
A
860
17–70.
The 20-kg roll of paper has a radius of gyration kA = 90 mm
about an axis passing through point A. It is pin supported at
both ends by two brackets AB. If the roll rests against a wall
for which the coefficient of kinetic friction is μk = 0.2,
determine the constant vertical force F that must be applied
to the roll to pull off 1 m of paper in t = 3 s starting from rest.
Neglect the mass of paper that is removed.
Ans:
F=22.1 N
SOLUTION
(
+ T
)
s=s0+v0t+
1
2
aC
t2
1=0+0+
1
2
aC
(3)2
a
C=
0.222 m
>
s
2
a=
a
C
0.125
=1.778 rad
>
s2
+
S
ΣFx=m(aG x);
NC–TAB cos 67.38°=0
+
c
ΣF
y=
m(a
G
)
y
;
TAB sin 67.38°–0.2NC–20(9.81) –F=0
c
+ΣMA=IA
a
;
–0.2NC(0.125) +F(0.125) =20(0.09)2(1.778)
Solving:
NC=99.3 N
TAB =258 N
F=22.1 N
Ans.
300 mm
B
A
C
125 mm
861
17–71.
The reel of cable has a mass of 400 kg and a radius of
gyration of kA
=
0.75 m. Determine its angular velocity
when t
=
2 s, starting from rest, if the force P
=
(20t2
+
80) N,
when t is in seconds. Neglect the mass of the unwound cable,
and assume it is always at a radius of 0.5 m.
Ans:
v
=0.474 rad>s
A
1 m
0.5 m P
*17–72.
The 30-kg disk is originally spinning at
v
=
125 rad
>
s. If it
is placed on the ground, for which the coefficient of kinetic
friction is μC
=
0.5, determine the time required for the
motion to stop. What are the horizontal and vertical
components of force which the member AB exerts on the
pin at A during this time? Neglect the mass of AB.
SOLUTION
a
B
0.3 m
0.5 m
0.5 m v
125 rad/s
C
863
17–73.
Cable is unwound from a spool supported on small rollers
at A and B by exerting a force T
=
300 N on the cable.
Compute the time needed to unravel 5 m of cable from the
spool if the spool and cable have a total mass of 600 kg and
a radius of gyration of kO
=
1.2 m. For the calculation,
neglect the mass of the cable being unwound and the mass
of the rollers at A and B. The rollers turn with no friction.
Ans:
t=6.71 s
30
O
T 300 N
0.8 m
AB
1.5 m
864
17–74.
The 5-kg cylinder is initially at rest when it is placed in
contact with the wall Band the rotor at A. If the rotor
always maintains a constant clockwise angular velocity
, determine the initial angular acceleration of
the cylinder.The coefficient of kinetic friction at the
contacting surfaces Band Cis .mk=0.2
v=6 rad>s
SOLUTION
Equations of Motion: The mass moment of inertia of the cylinder about point Ois
given by .Applying Eq. 17–16,
we have
(1)
(2)
a(3)
Solving Eqs. (1), (2), and (3) yields;
Ans.a=14.2 rad>s2
N
A=51.01 N N
B=28.85 N
+©MO=IOa; 0.2N
A(0.125) –0.2N
B(0.125) =0.0390625a
+c©F
y=m(aG)y; 0.2N
B+0.2N
Asin 45° +N
Acos 45° –5(9.81) =0
:
+©Fx=m(aG)x;NB+0.2NAcos 45° –NAsin 45° =0
IO=1
2mr2=1
2(5)(0.1252)=0.0390625 kg #m2
C
A
125 mm
B
v
Ans:
a
=
14.2 rad
>
s
2
865
17–75.
(1)
(2)
a(3)
Solvings Eqs. (1),(2) and (3) yields:
Ans.
Ans.
Ans.t
=
1.26 s
0=40 +(–31.71) t
v=v0+act
Ay=3
5FAB =0.6(111.48) =66.9 N
Ax=4
5FAB =0.8(111.48) =89.2 N
a=-31.71 rad>s2
FAB =111.48 N
NC=178.4 N
+©MB=IBa; 0.5NC(0.2) =0.5625(–a)
:
+©Fx=m(aG)x; 0.5NC–
A
4
5
B
FAB =0
+c©Fy=m(aG)y;
A
3
5
B
FAB +NC–25(9.81) =0
The wheel has a mass of 25 kg and a radius of gyration
It is originally spinning at If it is
placed on the ground, for which the coefficient of kinetic
friction is determine the time required for the
motion to stop.What are the horizontal and vertical
components of reaction which the pin at Aexerts on AB
during this time? Neglect the mass of AB.
mC=0.5,
v=40 rad>s.kB=0.15 m.
0.3 m
B
0.4 m
A
0.2 m
V
Ans:
Ax=89.2
N
A
y
=66.9
N
t=1.25
s
866
*17–76.
The 20-kg roll of paper has a radius of gyration
kA = 120 mm about an axis passing through point A. It is pin
supported at both ends by two brackets AB. The roll rests
on the floor, for which the coefficient of kinetic friction is
μk = 0.2. If a horizontal force F = 60 N is applied to the end
of the paper, determine the initial angular acceleration of
the roll as the paper unrolls.
SOLUTION
300 mm
C
400 mm
B
A
F
Ans:
a
=
3.38 rad
>
s
2
867
17–77.
Disk Dturns with a constant clockwise angular velocity of
30 .Disk Ehas a weight of 60 lb and is initially at rest
when it is brought into contact with D.Determine the time
required for disk Eto attain the same angular velocity as
disk D.The coefficient of kinetic friction between the two
disks is .Neglect the weight of bar BC.mk=0.3
rad>s
SOLUTION
(1)
(2)
a(3)
Solving Eqs. (1), (2) and (3) yields:
Kinematics: Applying equation , we have
(a
Ans.t=1.09 s
+)30=0+27.60t
v=v0+at
F
BC =36.37 lb
N=85.71 lb
a=27.60 rad>s2
+©MO=IOa; 0.3N(1) =0.9317a
+c©F
y=m(aG)y;N–F
BC sin 45° –60 =0
:
+©F
x=m(aG)x; 0.3N–F
BC cos 45° =0
A
B
1ft
2ft
2ft
1ft
30 rad
/
s
C
E
D
v
Ans:
t=1.09
s
868
17–78.
SOLUTION
. Applying Eq. 17–16, we have
a
Kinematic: Applying equation , we have
Ans.y=0+9.758 (0.5) =4.88 ft s
y=y0+at
a=9.758 ft>s2
–ca 5
32.2 bad(0.75) –ca 10
32.2 bad(0.75)
+©MO=IOa; 5(0.75) –10(0.75) =-0.02620aa
0.75 b
a=a
r=a
0.75
2mr2=1
2a3
32.2 b
Two cylinders Aand B, having a weight of 10 lb and 5 lb,
respectively, are attached to the ends of a cord which passes
over a 3-lb pulley (disk). If the cylinders are released from
rest, determine their speed in The cord does not
slip on the pulley. Neglect the mass of the cord. Suggestion:
Analyze the “system” consisting of both the cylinders and
the pulley.
t=0.5 s.
B
O
0.75 ft
Ans:
v=
4.88 ft
>
s
869
17–79.
rO
The two blocks Aand Bhave a mass of 5 kg and 10 kg,
respectively.If the pulley can be treated as a disk of mass 3kg
and radius 0.15 m, determine the acceleration of block A.
Neglect the mass of the cord and any slipping on the pulley.
SOLUTION
The mass moment of inertia of the pulley about point Ois
Equation of Motion: Write the moment equation of motion about point Oby
referring to the free-body and kinetic diagram of the system shown in Fig. a,
a;
Ans. a=2.973 m>s2=2.97 m>s2
=-0.03375(6.6667a)–5a(0.15) –10a(0.15)
5(9.81)(0.15) –10(9.81)(0.15)+©Mo=©(Mk)o
1
2 (3)(0.152)=0.03375 kg #m2
Io=1
2Mr2 =
Ans:
a
=
2.97
m
>
s
2
*17–80.
SOLUTION
c
Ans.a=g(mB–mA)
a1
2M+mB+mAb
a=g(mB–mA)
ra1
2M+mB+mAb
+©MC=©
(Mk)C;mBg(r)–mAg(r)=a1
2Mr2ba+mBr2a+mAr2a
a=ar
The two blocks Aand Bhave a mass and
respectively, where If the pulley can be treated
as a disk of mass M, determine the acceleration of block A.
Neglect the mass of the cord and any slipping on the pulley.
mB7mA.
m
B
,m
A
A
B
rO