Solution 6.107
22 2 2 2
1111 1
2mg sin 2mg sin
22 2 2 2 2 2
TOOACABD T
LL
kII I k
ω
θ
φ
ωωω θ
φ
↑ ↑
′
+=+++ +
AOAABOAAB
LL
υω ωωωω
==→==
and
12 2 6
CC
ImLmd dL ImL
=+ =−→=
ωω
BB
TV V TV V AB
gegC11 1 12 2 2 2
+++
′=+ +
−
At instant shown, Bar is
rotatting about point .C
⎛
⎝
⎜⎞
⎠
⎟
b
B
b
G
θ
2
Solution 6.108
When
θ
= 0, the cord length between A′ and the upper pulley is LO = 0.25b.
1
bbk
TV TV
11 12 2 2
0++ = +
−
*
,.sec datum at
0
Solution 6.109
23
2
22 2
290
554cos
554cos
hL L bb
b
θ
θ
θ
=°
=−=−−
=−−
()
12
3
g1sin g554cos
2
b
mmb
θθ
+−− −−
22.23rad/s at
Solution 6.110
()
Roll w o slip, Raw
υ
=+
2
CG
Velocities:
υθ
=++ −
2
22
22
1
Ring : 2 mg sin
R
R
OmaaRR x
aaRR
()
()
3
diff diff
If 0.2 , 1.080 g sin and 1.078 g sin
2.18 10 g sin Torus wins
TR
TR
aR x x
x
υθυ
θ
υ
υυ υ θ
−
== =
=−→ =
C
x
m
θ
4 lb
Solution 6.111
Solution 6.112
υ
−
+=
112 2
2
TT
=
or 0.211
x
()
ω
ω
=
−
==
max 0.211 2
22
g 0.211
0.211
0.211
x
Solution 6.113
rr
υυυυ
−
ω
OV
g
=O
m, k
r
i
r
O
D
P
C
P
D’
O’
θ
Solution 6.114
22
υ
υ
′=Δ +Δ
=
=
g
22
222
F
or system,
5.22 m s
2.29 m s
UTV
Solution 6.115
()( )
Δ= +× − =
2
2
11500 0.1 2 0.05 0.1 22.5
2
e
VJ
BB
A
G
0.4 m
A
O
O
h
1
1
h
2
0.4 m
ω
υ
υ
Solution 6.116
()
υ
Δ
=−= −
2
2
translational
1172
010,000 0
Tm
Δ=Δ+Δ
g
;
ETV
()
ωω
=× =
2662 66
22
2
262
22
2
168.75 25.65 10 , 152,000 rad s
2
Δ
Solution 6.117
()
υ
== ++
g
dU d
PTVR
υB
C
B
θ
Solution 6.118
Δ
θ
′
=
UM
Bar BO is rotating about O so
2
22
111
B
BO O
TIOmb
υ
ω
Δ= −=
6424
BO A
Also AB is rotating about C so
()
22
22
22
g
111 3
71
;mg1cos
24 24
A
AA
b
UTVM m m b
υ
θυ υ θ
′=Δ +Δ = + + −
P
A
CD
m
O
Solution 6.119
()
()
g
22
g
11
mg g sin mg sin
2
gcos
Ot
OD O
O
dU dT dV
dT d I mb d
b
dV d h m d b d
m
mbd
ωωω
θθ
θθ
′=+
==
== +
=+
()
gcos
2
OO
O
mmb d m b d
αθ θθ
=+ + +
A
b
b
Sb
b cos θ
O
θ
θ
Solution 6.120
For the equilibrium position, the work done by the spring is equal and opposite to the work
done by the weight, so that
()
θ
′
=+ +
′=
′=−
g
becomes
,
e
d
UdtdVdV
d
UdT
d
UMd
() ( )
()
2
1
Thus cos cos 2
12
Md mb b d mb d
θθθθθ θθ
−= +
Solution 6.121
dU = dT
C = instantaneous center of zero velocity for AO
2
b
BA
υ
A
θθ
Solution 6.122
918
θ
=°
7.08
A
18″
Solution 6.123
() ( ) ( )
θθ
′
== + +
=+
g
g
0
2 6 8 cos 10 18 cos
e
dU dT dV dV
dV d d
() ()
2 18 cos30 31.2 in.; stretch 2 18 cos 18,
2
CA x
θ
=°= =−
Solution 6.124
θ
=−
2tan g
52
P
am
P
V
g
= 0 P
b
b
θ
x = b cos θ
2
Solution 6.125
g
;
d
UdTdVdUMd
θ
′′
=+ =
Solution 6.127
7.85 sin
J
θδθ
=
() ()
9.81
2 2 19.62 cos
Tas y b
δδ δδθ
=⋅= =
()
2
Thus 7.85 sin 3.92 sin 520 1 cos sin
b
θ δθ θ δθ θ θ δθ
=− + −
b
bb
a = g/2
θθ
y = b cos θ
C
Solution 6.128
Replace P by force P at B and couple M = Pb
dU dT
=
2 cos since 0
b
θθ θ
==
()
()()()
2
1
so 2 2 cos 2 sin 2 2
AC
dT m b d b m b d
θθ θ θθ
=+
()
2
8cos 1
mb
Solution 6.129
θθ
θ
==
22 22
2sin cos
Fd b Fb d
b
F
m
b
2
b
2
sin θ
2
3b
2
y = 4b sin θ