fisica1-youn-e-freedman-exercicios-resolvidos

9.1: a)
1.50 m
= 0.60 rad = 34.4°
.
2.
50 m
b)
(14.0 cm)
= 6.27 cm.
(128°)(
π rad 180°
)
c) ( 1.50 m)(0.70 rad) = 1.05 m.
9.2: a)
⎛ rev ⎞ ⎛ 2π
rad ⎞ ⎛1min
⎞
⎜1900 ⎟ × ⎜ ⎟ ⎜ ⎟ = 199 rad s.
⎝ min ⎠ ⎝ rev ⎠ ⎝ 60 s ⎠
b)
−3
(35° × π rad 180°
) (199 rad s) = 3.07 × 10 s.
dωz
3
2
9.3: a) αz
= = (12.0 rad s ) t,
so at t = 3.5 s, α = 42 rad s . The angular acceleration
dt
is proportional to the time, so the average angular acceleration between any two times is
the arithmetic average of the angular accelerations. b) ω (6.0 rad s
3 ) t 2
z
= , so at
t = 3.5 s, ωz
= 73.5 rad s. The angular velocity is not linear function of time, so the
average angular velocity is not the arithmetic average or the angular velocity at the
midpoint of the interval.
dω
dt
z
9.4: a) α (t) = = −2βt
= ( −1.60 rad s
3 ) t.
z
3
2
b) α (3.0 s) = ( −1.60 rad s )(3.0 s) = −4.80 rad s .
z
ω( 3.0 s) − ω(0)
− 2.20 rad s − 5.00 rad s
2.
αav−
z
=
=
= −2.40 rad s ,
3.0 s
3.0 s
which is half as large (in magnitude) as the acceleration at t = 3.0 s.
2
3 2
9.5: a) ω z
= γ + 3βt
= (0.400 rad s) + (0.036 rad s ) t b) At t = 0, ωz
= γ =
3.50rad
.400rad s. c)At t 5.00 s, ω = 1.3 rad s, θ = 3.50 rad, so ω = 0.70rad s.
0 =
z
av − z
=
5.00s
The acceleration is not constant, but increasing, so the angular velocity is larger than the
average angular velocity.