fisica1-youn-e-freedman-exercicios-resolvidos

9.87: The moment of inertia of the hoop about the nail is 2MR 2 (see Exercise 9.52), and
the initial potential energy with respect to the center of the loop when its center is directly
below the nail is gR ( 1− cos β).
From the work-energy theorem,
1 2
2 2
K = Iω = Mω R = MgR(1
−
2
cos β),
from which ω = ( g R)(1
− cos β).
9.88: a)
1 Iω
2
2
K =
J.
1 ⎛ 1
2 ⎞⎛
= ⎜ (1000kg)(0.90m) ⎟ 3000rev
2 2
⎜
⎝
⎠⎝
= 2.00×
10
7
2π
rad s ⎞
min ×
60 rev min
⎟
⎠
2
7
K 2.00×
10 J
b) =
= 1075 s,
4
P 1.86 × 10 W
ave
which is about 18 min.
1 2 1 2 1
−2
2
−2
2
9.89: a) M
1R1
+ M
2R2
= ((0.80kg)(2.50×
10 m) + (1.60kg)(5.
00 × 10 m) )
2 2 2
= 2.25×
10
−3
kg ⋅ m
b) See Example 9.9. In this case, ω = v R 1
,and so the expression for v becomes
2
.
v =
=
2gh
1+
( I mR
2
)
2
2(9.80m s )(2.00m)
−3
2
(1 + ((2.25×
10 kg ⋅ m ) (1.50kg)(0.025m)
2
))
= 3.40m s.
c) The same calculation, with R
2
instead of R
1
gives v = 4.95 m s. This does make
sense, because for a given total energy, the disk combination will have a larger fraction of
the kinetic energy with the string of the larger radius, and with this larger fraction, the
disk combination must be moving faster.