fisica1-youn-e-freedman-exercicios-resolvidos

10.29: a)
b)
∆ω
τ = Iα = I
∆t
=
2
( 1 2)( 1.50 kg)( 0.100 m)
)( 1200 rev min)
= 0.377 N ⋅ m.
ω
c) τ ∆θ = 59.2 J.
d)
1 K = Iω
2
2
( 600 rev/min)( 2.5 s)
60 s/min
2.5 s
25.0 rev
ave
∆t =
=
1
=
2
the same as in part (c).
2
((1/
2)(1.5 kg)(0.100 m) )
⎛ π rad s ⎞
⎜ ⎟
⎝ 30 rev min ⎠
157 rad.
⎛
⎛ π
⎜(1200 rev/min) ⎜
⎝
⎝ 30
= 59.2 J,
rad/s
rev/min
⎞⎞
⎟⎟
⎠⎠
2
10.30: From Eq. (10.26), the power output is
⎛
2π
rad/s ⎞
P = τω = ( 4.30 N ⋅ m) ⎜4800 rev/min × ⎟ = 2161 W,
⎝
60 rev/min ⎠
which is 2.9 hp.
10.31: a) With no load, the only torque to be overcome is friction in the bearings
(neglecting air friction), and the bearing radius is small compared to the blade radius, so
any frictional torque could be neglected.
τ P / ω (1.9 hp)(746 W/hp)
b) F = = =
= 65.6 N.
R R
⎛ π rad/s ⎞
(2400 rev/min) ⎜ ⎟(0.086 m)
⎝ 30 rev/min ⎠