fisica1-youn-e-freedman-exercicios-resolvidos

10.32:
I
= mL
1 2 1
2
2
= ( 117 kg)(2.08 m) = 42.2 kg ⋅ m
2 2
τ 1950 N ⋅ m
2
a) α = =
46.2 rad/s .
2
42.2 kg ⋅ m
=
I
2
b) ω = 2αθ
= 2(46.2 rad/s )(5.0 rev×
2π
rev) = 53.9 rad/s.
1 2
c) From either W = K = ω or Eq. (10.24),
2
4
W = τθ = (1950 N.m)(5.00 rev×
2π
rad/rev) = 6.13×
10 J.
d), e) The time may be found from the angular acceleration and the total angle, but the
instantaneous power is also found from P = τω = 105 kW(141hp).
The average power is
half of this, or 52 .6 kW.
3 ⎛
⎛ π rad/s ⎞⎞
10.33: a) τ = P / ω = (150×
10 W) ⎜(400 rev/min) ⎜ ⎟⎟
= 358 N ⋅ m.
⎝
⎝ 30 rev/min ⎠⎠
3
b) If the tension in the rope is F , F = w and so w = τ/
R = 1.79×
10 N.
c) Assuming ideal efficiency, the rate at which the weight gains potential energy is the
power output of the motor, or wv = P, so v = P w = 83.8 m/s. Equivalently, v = ωR.
10.34: As a point, the woman’s moment of inertia with respect to the disk axis is
and so the total angular momentum is
⎛ 1 ⎞ 2
L = Ldisk
+ Lwoman
= ( Idisk
+ Iwoman)
ω = ⎜ M + m⎟R
ω
⎝ 2 ⎠
⎛ 1
⎞
2
= ⎜ 110 kg + 50.0 kg⎟(4.00 m) (0.500 rev/s×
2π
rad/rev)
⎝ 2
⎠
= 5.28×
10
3
kg ⋅ m
2
/s.
2
mR ,
2
10.35: a) mvr sinφ = 115 kg ⋅ m /s,
with a direction from the right hand rule of into the
page.
2
b) dL dt = τ = ( 2 kg)( 9.8 N kg) ⋅ ( 8 m) ⋅ sin( 90° − 36.9°
) = 125 N ⋅ m = 125 kg ⋅ m s ,
out of the page.
2