fisica1-youn-e-freedman-exercicios-resolvidos

5.49: a) Setting arad
= g in Eq. (5.16) and solving for the period T gives
= R 400 m
T 2π
= 2
40.1s,
2
g
π 9.80 m s
=
so the number of revolutions per minute is ( 60 s min) (40.1s) = 1.5 rev min .
b) The lower acceleration corresponds to a longer period, and hence a lower rotation
rate, by a factor of the square root of the ratio of the accelerations,
T ′ = ( 1.5 rev min) × 3.70 9.8 = 0.92 rev min. .
5.50: a) 2 π R T = 2π(50.0 m) (60.0 s) = 5.24 m s.
b) The magnitude of the radial force
2
2 2
2 2
is mv R = m4π R T = w(4π
R gT ) = 49 N (to the nearest Newton), so the apparent
weight at the top is 882 N − 49 N = 833 N, and at the bottom is 882 N + 49 N = 931 N .
c) For apparent weightlessness, the radial acceleration at the top is equal to g in
magnitude. Using this in Eq. (5.16) and solving for T gives
= 50.0 m
2 π R
T = 2 π
14 s.
2
g 9.80 m s
=
d) At the bottom, the apparent weight is twice the weight, or 1760 N.
2
5.51: a) If the pilot feels weightless, he is in free fall, and a = g = v R , so
2
v = Rg = (150 m)(9.80 m s ) = 38.3 m s , or 138 km h . b) The apparent weight is
the sum of the net inward (upward) force and the pilot’s weight, or
⎛ a ⎞
w + ma = w ⎜1
+ ⎟
⎝ g ⎠
2
⎛ (280 km h)
⎞
= ( 700 N)
⎜1
⎟
+
2
2
(3.6(km h) (m s)) (9.80 m s )(150 m)
⎝
⎠
= 3581 N,
or 3580 N to three places.
5.52: a) Solving Eq. (5.14) for R,
2 2
2
2
R = v a = v 4g
= (95.0 m s) (4 × 9.80 m s ) = 230 m.
b) The apparent weight will be five times the actual weight,
2
5 mg = 5 (50.0 kg) (9.80 m s ) = 2450 N
to three figures.