fisica1-youn-e-freedman-exercicios-resolvidos

9.24: From a
rad
= ω
2 r,
4
which is (1.25 10 rad s)
2
a 400,000 × 9.80 m s
4
ω = =
= 1.25× 10 rad s,
r 2.50×
10 − 2
m
1rev
2π
rad
× ( ) = 1. 20×
1min 60 s
10 5
rev
min.
2
9.25: a) a = 0, a = αr = ( 0.600 rad s )( 0.300 m)
rad tan
=
0.180 m s
2
and so
π
2
2
2
b) θ = rad, so a 2( 0.600 rad s )( 3 rad)( 0.300 m) 0.377 m s .
3
rad
= ω r =
π
=
a =
0.180 m
2
s .
2
The tangential acceleration is still 0.180 m s , and so on
a =
2 2
2 2
2
( 0.180 m s ) + ( 0.377 m s ) = 0.418 m s .
For an angle of 120° , a = 0.754 m s , and a = 0.775 m s , since a is
2
2
c)
rad
tan
2
still 0 .180 m/s
ω = ω α t
2
9.26: a) + = 0.250 rev s + ( 0.900 rev s )( 0.200 s) 0.430 rev s
(note that since
z
0z z
=
ω 0 z
and
z
b) ω
av −z∆t
= (0.340 rev s) (0.2s) =
α are given in terms of revolutions, it’s not necessary to
convert to radians).
to radians must be made to use Eq. (9.13), and
⎛ 0.750 m ⎞
v = rω = ⎜ ⎟ 0.430 rev/s×
2π
⎝ 2 ⎠
d) Combining equations (9.14) and (9.15),
0.068 rev
( rad rev) = 1.01m s.
. c) Here, the conversion
a =
a
2
rad
= [((0.430 rev s × 2π
rad rev)
= 3.46 m
+ a
s
2
2
tan
.
=
2
( ω r)
2
+ ( αr)
2
2
(0.375 m))
2
+ ((0.900 rev
2
s × 2π
rad
rev)(0.375 m))
2
]
1
2
9.27:
a
r =
ω
(3000)(9.80 m
2
rad
=
=
2
2
π rad s
((5000 rev min) ( )
s
)
30 rev min
10.7 cm,
so the diameter is more than 12.7 cm, contrary to the claim.