fisica1-youn-e-freedman-exercicios-resolvidos

5.45: a) The magnitude of the force F is given to be equal to 3.8w. “Level flight” means
that the net vertical force is zero, so F cos β = (3.8) w cos β = w,
, and
β = arccos(1 3.8) = 75°
.
(b) The angle does not depend on speed.
5.46: a) The analysis of Example 5.22 may be used to obtain tanβ = ( v 2 gR),
but the
subsequent algebra expressing R in terms of L is not valid. Denoting the length of the
πR
horizontal arm as r and the length of the cable as l , R = r + l sin β.
The relation v = 2 T
is
2
2
4π
R 4π
( r + l sin β )
still valid, so tan β =
2
=
2
. Solving for the period T,
T =
4π
2
gT
gT
( r + l sinβ)
=
g tanβ
4π
2
(3.00 m + (5.00 m)sin 30°
)
2
(9.80 m s ) tan 30°
= 6.19 s.
Note that in the analysis of Example 5.22, β is the angle that the support (string or cable)
makes with the vertical (see Figure 5.30(b)). b) To the extent that the cable can be
considered massless, the angle will be independent of the rider’s weight. The tension in
the cable will depend on the rider’s mass.
5.47: This is the same situation as Example 5.22, with the lift force replacing the tension
in the string. As in that example, the angle β is related to the speed and the turning
2
v
radius by tan β = . Solving for β ,
gR
2
⎛ v ⎞ ⎛ (240 km h × ((1m s)
β = arctan ⎜
⎟ = arctan
⎜
2
⎝ gR ⎠ ⎝
(3.6 km
( 9.80 m s )( 1200 m)
h)))
2
⎞
⎟ = 20.7°
.
⎠
5.48: a) This situation is equivalent to that of Example 5.23 and Problem 5.44, so
v
µ 2
. Expressing v in terms of the period T, 2
2
πR
4π
R
v = , so µ = . A platform speed
s
=
Rg
of 40.0 rev/min corresponds to a period of 1.50 s, so
2
4π
(0.150 m)
µ
s
=
= 0.269.
2
2
(1.50 s) (9.80 m s )
b) For the same coefficient of static friction, the maximum radius is proportional to
the square of the period (longer periods mean slower speeds, so the button may be moved
further out) and so is inversely proportional to the square of the speed. Thus, at the higher
40.
0
speed, the maximum radius is (0.150 m) ( ) 2
60. 0
= 0. 067 m .
T
S
2
T g