fisica1-youn-e-freedman-exercicios-resolvidos

2π
2 2
9.49: a) ω =
T
, so Eq. (9.17) becomes K = 2π
I T .
b) Differentiating the expression found in part (a) with respect to T,
dK
dt
2
= ( −4π
I
T
3
)
dT
dt
.
2
2
2
c) 2π
(8.0 kg ⋅ m ) (1.5 s) = 70.2 J, or 70 to two figures.
2
2
3
d) ( −4π
(8.0 kg ⋅ m ) (1.5 s) )(0.0060) = −0.56
W.
9.50: The center of mass has fallen half of the length of the rope, so the change in
gravitational potential energy is
1
− 1 mgL = − (3.00 kg)(9.80 m s
2 )(10.0 m) = −147 J.
2 2
2
9.51: (120 kg)(9.80 m s )(0.700 m) = 823 J.
2
2
2
9.52: In Eq; (9.19), I = MR and d = R , so I = P
2MR
.
cm
2 2 2 2 2 2 4 2
9.53: MR = MR + Md , so d = R , and the axis comes nearest to the center of
3 5
15
the sphere at a distance d = ( 2 15) R = (0.516) R.
9.54: Using the parallel-axis theorem to find the moment of inertia of a thin rod about an
axis through its end and perpendicular to the rod,
2 M 2 ⎛ L ⎞
I p
= Icm
+ Md = L + M ⎜ ⎟ =
12 ⎝ 2 ⎠
2
M
3
2
L .
2
1 2 2
a 2 b 2
9.55: Ιp = Ιcm
+ md , so Ι = Μ ( a + b ) + Μ
( ) + ( ) ),
which gives
12
2 2
1 2 2 1 2 2 1 2 2
Ι = Μ ( a + b ) + Μ( a + b ),
or Ι = Μ ( a + b ).
12
4
3