Solution 2.102
T = 2L
y
x
3 m
π
T
Radius of curvature
22
dydx
ρ
=
2
x
Solution 2.103
243; 4
0.814 0.581
=+
t
ij
n
n
ρ
or
ρ
==
8.60 31.8 m
CA
Solution 2.104
2
120 m/s 35 5 m/s==°=
va
θ
(Constant)
2
00
1
sin gt
yv t
θ
=−
x
ρ
θθ
x, m
(to C)
e
n
_
• Flight Time:
Evaluate x
v and y
v
, compute
θ
,,
tn
aa
, and
ρ
at each time increment from t = 0 to
14.03 s,=
f
t and plot the values.
d
–5
0
500
1000
1500
Radius of Curvature, ρ, m
ρ
min
Solution 2.105
r
0
y
x
θ = 60°
θ
υ
r
65 mi/hr
100 115.5 ft
sin 95.3sin60 0.715 rad/sec
or 0.715 41.0 deg/sec
v
r
v
θ
θ
θ
θπ
=
==
°
=− =− =−
=− =−
Solution 2.106
r
O
θ
υ
r
12.5deg/s
=
r
θ
Solution 2.107
Position r
r
r
θ
θ
θ
Solution 2.108
r
P
y
α
θ
θ
υ
θ
21.3
=°
θ
α
−
Solution 2.109
2
2.07 m/s
=
r
Solution 2.110
120 mm
r
O
θ
υ
r
υ = 25 mm/s
r = 200 mm
3
sin 5
=
θ
Solution 2.111
O
θ
r = 200 mm
Solution 2.112
()
θθ
π
θ
=+ = ++
1.5 24 7 5180
rr
vre re e e
Solution 2.113
≠≠≠ ≠, , , and because scalars are NEVER equal to vectors.rvrvra ra
Solution 2.114
π
θθ
==− = =
==
2
6 in./sec, 2 in./sec , 10 rad/s, 0
180
6 ft 72 in.
rr
r
θθ
θ
rr
r
Solution 2.115
A (Ar r
vrerevele
θθ
θ
=+ =+Ω
2
Br
θ
Solution 2.116
d
θ
a
θ
υ
θ
a
r
28 m/s
=
v
()()
22 2 2 2
sin sin 40 40 62.2 m
2 cos 62.2 160 2 62.2 160 cos 40 119.2 m
=→ °= →=
=+ − = + − °→=
rhr r
srd rd s
θ
θ
θ
Solution 2.117
2
0.8t rad 1.6 0.2t m
=− =−
tr
θ
For car B,
11
22.7
D
−−
D
x
θ
υ
A
‘
α
Solution 2.119
y
x
Oθ
θ
υ
r
r
350‘
175‘
a = g
y
max
= 275‘
22 1
175
175 350 391 ft, tan 26.7
r
θ
−
=+= = =°
cos 80.2 cos 26.7 391 0.1834 rad/sec
vr v
θ
θθ θθ
Solution 2.120
x
Oθ
θ
υ
350‘
175‘
391 ft and 35.9ft/sec
rr
==
θ
υθ
y
Solution 2.121
()
9000 0.02 180 m/s
== =
vr
θ
θ
Solution 2.122
0
2.5 m/s
=
v
0
cos (7)
vv r
θ
θθ
==
0
0
22
fyff f ff
υ
υ
r
a
a
r
max
max
21.9 rad/s @ 1.633s Both values
33.4 rad/s @ 1.373s
==
==
t
t
θ
θ
0.0
–2.0
⋅
, rad/s
|θ
max
–0.5
0.5 1.0 1.5
Time, t, s
0.0
–3
–2
–1
⋅⋅
, rad/s2
0
1
0.5 1.0 1.5
Time, t, s
⋅⋅
Solution 2.123
50 20 30
sin 161.7sin30
=°−°=°
°
v
θ
α
α
Solution 2.124
Values of , , andr
α
θ
from solution to previous problem.
()
2
2
2
14 1000 0.0808
=−
=− −
r
a
rr
θ
Then 2
7.47 8.62 ft/s
== =
r
a
a
r
θa
a
r
r
P
20°
α
Solution 2.125
22
1000 400
1077 m
=+
=
r
r
2
12.15 m/s
=
r
Solution 2.126
2
0.3m/s 0
=− =
r
θ
22
0.296 m/s
=+=
r
vvv
θ
2
e
θ
_
unit
Solution 2.127
,0
rd
θ
==
Solution 2.128
1
5280 ft
12,149 17,819
: 17,819 sin30
ft 4200
r
v
vr r
−
==
=°=
()
()
2
2
: 7.159 8400 5280 3.48 10
r
arr r
θ
=− − =− ×
υr
O
7275 mi
Solution 2.129
Solution 2.130:
2
22
360 100 m/s
100 3.33m/s
3000
==
==
v
a
cos 3.33 cos 43.2 2.43 m/s=− =− °=−
a
a
θ
52
−
P
θa
υ
a
r
y
r
⋅ = 80(0.4)
γ
= 32.0
°
β
θ
Solution 2.131
υ
β
υr
er
eθ
100 ft/sec
y
2
cos 93.0 cos10.54 91.4 ft/sec
sin 93.0sin10.54 17.02 ft/sec
17.02 51.0 , 0.334 rad/sec
32.2 ft/sec
r
vv r
vv
r
ay
θ
β
β
θθθ
== °= =
=− =− °=−
−
== =−
==−
Solution 2.132
β
θ
θ
θ
O
υ
υ
r = 90 m
· = 15.5 m/s
sin( ) sin(30 ) 15.5(m/s) (1)
== − = °− =
r
vrv v
θ
ββ
2
S
olving, 50.2 m/s, 6.01m/s , 50.5 m, 12.00
== ==°
p
vv p
β