fisica1-youn-e-freedman-exercicios-resolvidos

9.19: a) Before the circuit breaker trips, the angle through which the wheel turned was
2
2
(24.0 rad s) (2.00 s) + (30.0 rad s ) (2.00 s) 2 = 108 rad, so the total angle is
108 rad + 432 rad = 540 rad. b) The angular velocity when the circuit breaker trips is
2
( 24.0 rad s) + ( 30.0 rad s )( 2.00 s) = 84 rad s, so the average angular velocity while the
432 rad
wheel is slowing is 42 .0 rad s, and the time to slow to a stop is = 10.3 s, so the
42.0 rad s
time when the wheel stops is 12 .3s
. c) Of the many ways to find the angular
acceleration, the most direct is to use the intermediate calculation of part (b) to find that
−84 rad s
2
while slowing down ∆ω
z
= − 84 rad s so α
z
= = − 8.17 rad s .
10.3s
α z
9.20: a) Equation (9.7) is solved for ω0 z
= ωz
− αzt,
which gives ωz
= ωz
− t,
−
or
1 2
θ − θ = ω t − α . b) 2( ) ω z θ
2
= 0.125 rad s . c) ω − α t = 5.5 rad s.
0 z
t
2 z
t
− ∆ 2
−
t
z
z
ave
2
9.21: The horizontal component of velocity is rω , so the magnitude of the velocity is
a) 47.1 m/s
⎛
⎛ π rad/s ⎞⎞
2
b) ⎜(5.0 m)(90 rev/ min) ⎜ ⎟⎟
+ (4.0 m/s) = 47.3 m/s.
⎝
⎝ 30 rev/min ⎠⎠
2
1.25 m s
1.25 m
9.22: a) −3 = 50.0 rad s , −3
= 21. 55
25.0 × 10 m
58.0×
10
b) (1.25 m s ) (74.0 min) (60 s min) = 5.55 km.
rad s , or 21.6 rad s to three figures.
50.0 rad s−21.55 rad s
c) α
z
= =
(74.0 min) (60 s min)
6.41× 10 −3 rad s
2 .
9.23: a) ω 2 r =
2
( 6.00 rad s) 0.500 m)
( = 18 m s
2 .
b) v = ωr = ( 6.00 rad s)
( 0.500 m) = 3 .00 m s , and =
v 2 (3.00 m s)
=
r (0.500 m)
2
18 m
s
2
.