PROBLEM 16.107
A 12–in.–radius cylinder of weight 16 lb rests on a 6-lb carriage. The system
is at rest when a force P of magnitude 4 lb is applied. Knowing that the
cylinder rolls without sliding on the carriage and neglecting the mass of the
wheels of the carriage, determine (a) the acceleration of the carriage, (b) the
acceleration of Point A, (c) the distance the cylinder has rolled with respect to
the carriage after 0.5 s.
SOLUTION
Masses and moments of inertia.
22
16 lb 6 lb
AB
Cylinder alone. Point C is contact point.
PROBLEM 16.107 (Continued)
2
/
AB
PROBLEM 16.108
Gear C has a mass of 5 kg and a centroidal radius of gyration of 75 mm.
The uniform bar AB has a mass of 3 kg and gear D is stationary. If the
system is released from rest in the position shown, determine (a) the
angular acceleration of gear C, (b) the acceleration of Point B.
SOLUTION
Kinematics:
Et
But
/E B EB
= +aaa
/
+( ) ( ) ( )
Et Bt EBt
aa= + a
0 0.2 0.1
aa
= −
AB C
1
2
AB C
aa
=
(1)
eff
BB
(0.1m)
a
=
CC
QI
2
C
PROBLEM 16.108 (Continued)
Bar AB and gear C
(a)
eff
( ):
AA
MMΣ=Σ
(0.1) (0.2) (0.3) ( ) 0.1 ( )0.2
aa
+ −= ++ −
AB C AB AB AB C B C C
W W Q m a I ma I
2
1
AB C
(1.3) 0.3 0.24 0.028125
aa
−= −
AB C
gQ
AB
1
1.3 0.3(0.28125 ) 0.24 0.028125
2
aa a
−=−
CC C
g
1.3 0.17625
a
=
C
g
7.3759(9.81)
a
=
C
2
2
B AB C C
2
7.24 m/s=
B
a
PROBLEM 16.109
Two uniform disks A and B, each of mass 2 kg, are connected
by a 2.5-kg rod CD as shown. A counterclockwise couple M of
moment
2.25 N m⋅
is applied to disk A. Knowing that the
disks roll without sliding, determine (a) the acceleration of the
center of each disk, (b) the horizontal component of the force
exerted on disk B by pin D.
2
( )
( )
2
2.25 N m 0.08 m 0.0432 k mC
a
⋅− = ⋅
(1)
Disk B:
2
1
2
PROBLEM 16.109 (Continued)
( ) ( ) ( )
22
1
: 0.08 m 2
M M D ma r I mr mr
aa a
Σ=Σ = += +
PROBLEM 16.110
A single axis personal transport device starts from rest with the rider leaning
slightly forward. Together, the two wheels weigh 25 lbs, and each has a
radius of 10 in. The mass moment of inertia of the wheels about the axle is
0.15 slug ft2. The combined weight of the rest of the device and the rider
(excluding the wheels) is 200 lbs, and the center of gravity G of this weight
is located at x = 4 in. in front of the axle A and y = 36 in. above the ground.
An initial clockwise torque M is applied by the motor to the wheels.
Knowing that the coefficients of static and kinetic friction are 0.7 and 0.6
respectively, determine (a) the torque M that will keep the rider in the same
angular position, (b) the corresponding linear acceleration of the rider.
SOLUTION
200 lb 25 lb
,,
rx wx
PROBLEM 16.110 (Continued)
Assume no slip:
Substitute into (3)→(1):
( ) ( )
0.15 12 200 12 200
+ 0.833
0.833 26 6.211 3 26 26 6.211 3
MM
,
,
(8)
wx
wx w w w
w
a
ar r
aa
=− ⇒=−
,+ (9)
rx
a
,
26 3 w rx
,
12 200
(11)
26 3
rx
r
aM
m
= −
12
12 200 (12)
w
m
( ) ( )
( ) ( )
26 3 26 26 3
12 0.7764
0.15 12 200 12 200
+ 0.833
0.833 26 6.211 3 26 26 6.211 3
w
wr R
rm m
MM
= −+ −
1.4461 29.738
20.565 ft-lb
M
M
=
=
12
12 200 20.5646
26 26 3
23.94 lb
w
r
m
=
( )
0.7 225
157.5 lb
s
N
µ
=
=
x=a
PROBLEM 16.111
A hemisphere of weight W and radius r is released from rest in the position
shown. Determine (a) the minimum value of
µ
s for which the hemisphere starts to
roll without sliding, (b) the corresponding acceleration of Point B. [Hint: Note that
OG =
3
8
r and that, by the parallel-axis theorem,
22
2
5
( ) .]I mr m OG= −
2
() ( )mg x mr r mx x mk
aa a
=++
222
gx
rxk
a
=++
(2)
PROBLEM 16.111 (Continued)
PROBLEM 16.112
Solve Problem 16.111, considering a half cylinder instead of a hemisphere.
[Hint. Note that
4 /3OG r
π
=
and that, by the parallel–axis theorem,
22
1
2
( ) .]I mr m OG= −
2
() ( )mg x mr r mx x mk
aa a
=++
222
gx
rxk
a
=++
(2)
PROBLEM 16.112 (Continued)
eff
():
xx x
F F F ma F mr
a
Σ=Σ = =
eff
():
yy y
F F N W ma N mg mx
a
Σ=Σ − =− = −
F mr
a
µa
= = −
r
a
µa
=−
(3)
3
π
23
O
π
2
22
14
23
Ir
kr
m
π
= = −
PROBLEM 16.113
The center of gravity G of a 1.5-kg unbalanced tracking wheel is
located at a distance r = 18 mm from its geometric center B. The
radius of the wheel is R = 60 mm and its centroidal radius of
gyration is 44 mm. At the instant shown the center B of the wheel
has a velocity of 0.35 m/s and an acceleration of 1.2 m/s2, both
directed to the left. Knowing that the wheel rolls without sliding
and neglecting the mass of the driving yoke AB, determine the
horizontal force P applied to the yoke.
22
BB
RRR
PROBLEM 16.113 (Continued)
PROBLEM 16.114
A small clamp of mass
B
m
is attached at B to a hoop of mass
.
h
m
The system is
released from rest when
90
θ
= °
and rolls without sliding. Knowing that
3,
hB
mm=
determine (a) the angular acceleration of the hoop, (b) the horizontal
and vertical components of the acceleration of B.
PROBLEM 16.115
A small clamp of mass mB is attached at B to a hoop of mass mh. Knowing that the
system is released from rest and rolls without sliding, derive an expression for the
angular acceleration of the hoop in terms of mB, mh, r, and
θ
.
B hB
2 (1 cos )
hB
rm m
PROBLEM 16.116
A 4–lb bar is attached to a 10–lb uniform cylinder by a square pin, P, as
shown. Knowing that r = 16 in., h = 8 in.,
θ
= 20°, L = 20 in. and
ω
= 2 rad/s at the instant shown, determine the reactions at P at this instant
assuming that the cylinder rolls without sliding down the incline.
SOLUTION
Masses and moments of inertia.
4lb 0.12422 slug
B
W
( )( ) ( )
( )
( )
22
10 16 4 24
32.2 12 32.2 12
2
24
12
0.27605 (0.028755)
5.3896 rad/s
( ) (5.3896) 10.7791 f
Pt
a
=
= =
( )
2
22
8
12
t/s
( ) (2) 2.6667 ft/s
Pn
a= =
PROBLEM 16.116 (Continued)
Using the bar alone as a free body,
P
PROBLEM 16.117
The uniform rod AB of mass m and length 2L is attached to collars of
negligible mass that slide without friction along fixed rods. If the rod is
released from rest in the position shown, derive an expression for
(a) the angular acceleration of the rod, (b) the reaction at A.
2
1
3
sin
L
θ
+
PROBLEM 16.117 (Continued)
( )
eff
: sin
yy
F F A mg ma mL
aθ
Σ=Σ − =− =−
2
1
3
sin sin
sin
g
A mg mL L
θθ
θ
= −
+
2
1
3
sin
A mg
θ
+
=
2
sin
θ
−
2
1
3
sin
θ
+
2
1 3sin
mg
A
θ
=+