fisica1-youn-e-freedman-exercicios-resolvidos

9.28: a) Combining Equations (9.13) and (9.15),
2 2 ⎛ v ⎞
a
rad
= ω r = ω ⎜ ⎟ = ωv.
⎝ ω ⎠
arad 0.500 m s
b) From the result of part (a), ω = = 0.250 rad s.
= v
2.00 m s
ωr
rad s
−3
π
12.7×
10 m
9.29: a) = 1250 rev min) ( )( ) = 0.831m s.
(
30 rev min 2
2
v (0.831m s)
2
b) = 109 m s .
=
r
−3
(12.7×
10 m) 2
2
a
r
tan −10.0 m s
2
9.30: a) α = = = − 50.0 rad s b) At t = 3 .00 s, v = 50.0 m s and
0.200 m
v 50.0 m s
ω = r
= = 250 rad s and at t = 0,
0.200
2
v =
50.0 m s + ( −10.0 m
( 0 − 3.00 s) = 80.0 m s, so ω = 400 rad s.
c)
av
t = (325 rad s)(3.00 s)
2
= 975 rad = 155 rev. d) v = a r = (9.80 m s )(0.200 m) 1.40 m s.
This speed will
be reached at time
50.0 m s−1.40 m s
10.0 m s
equivalent ways to do this calculation.)
ω e
rad
=
= 4.86 s after t = 3.00 s, or at t = 7.86 s . (There are many
s
2
)
9.31: (a) For a given radius and mass, the force is proportional to the square of the
640 rev min
angular velocity; ( ) 2. 29
423 rev min
2
= (note that conversion to rad s is not necessary for this
part). b) For a given radius, the tangential speed is proportional to the angular velocity;
640
= 1.51 (again conversion of the units of angular speed is not necessary).
423
a
rad s 0.470 m
c) (640 min) ( )( ) 15.75 m s, or 15.7 m s
rad
π
rev =
30 rev min 2
to three figures, and
2
2 (15.75 m s)
3 2
=
v = = 1.06×
10 m s = 108 g.
r
(0.470 m 2)
9.32: (a) v = T
Rω
⎛ 7.5 rev ⎞ ⎛1min
⎞ ⎛ 2π rad ⎞
2.00 cm s = R ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ min ⎠ ⎝ 60 s ⎠ ⎝ 1rev ⎠
R = 2.55 cm
D = 2R
= 5.09 cm
b) a Rα
= T
2
T
0.400 m s
2
a
α = R
=
0.0255 m
= 15.7 rad
s