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PROBLEM 16.112
Solve Problem 16.111, considering a half cylinder instead of a hemisphere.
[Hint. Note that
4 /3
OG r
π
=
and that, by the parallel-axis theorem,
22
1
2
( ) .]I mr m OG= −
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 − =− = −
a
µa
= = −
r
a
µa
=−
(3)
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.
PROBLEM 16.113 (Continued)
Substitute:
2
1.5 kg
0.018 m
0.06 m
0.044 mm and 9.81 m/s in Eq. (1)
m
r
R
kg
=
=
=
= =
22 2
22
2
0.018 0.018 0.044 0.018
1.5(9.81) 1.5( ) 1 1.5
0.06 0.06 0.06
BB
Pa v
+
= ++ +
4.4145 2.9300 0.9188
8.263 N
=++
= +
8.26 N=P
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
.
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.
PROBLEM 16.116 (Continued)
Using the bar alone as a free body,
eff
():
xx
FFΣ=Σ
( ) cos20 ( ) sin 20
x BPt BPn
P ma ma= °− °
44
(10.779)cos 20 (2.6667)sin 20
1.1450 lb
x
P
= °− °
=
±
eff
( ) : ( ) sin 20 ( ) cos20
y y y B BPt BPn
F F P mg ma maΣ =Σ − =− °− °
22
44
(4) (10.779)sin 20 (2.6667)cos20
32.2 32.2
3.2307 lb
1.145 3.2307 3.4276 lb
y
y
P
P
P
= − °− °
=
=+=
3.2307
tan 1.145
70.5
β
β
=
= °
3.43 lb=P
70.5°
Recognizing that P is the CG of the bar.
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
=+
