fisica1-youn-e-freedman-exercicios-resolvidos

10.36: For both parts, L = Iω.
Also, ω = v r,
so L = I( v r).
2
a) L = ( mr )( v r)
= mvr
L = (5.97 × 10
24
2
b) L = (2 5mr
)( ω)
L = (2 5)(5.97 × 10
= 7.07 × 10
33
kg)(2.98×
10
24
kg ⋅ m
s
4
kg)(6.38×
10
2
m s)(1.50×
10
6
11
m) = 2.67 × 10
40
kg ⋅ m
2
m) (2π
rad (24.0 hr × 3600 s hr))
2
s
10.37: The period of a second hand is one minute, so the angular momentum is
M 2
L = Iω = l
3
−
⎛ 6.0 × 10
=
⎜
⎝ 3
3
2π
T
kg ⎞
⎟(15.0
× 10
⎠
−2
m)
2
2π
= 4.71×
10
60 s
−6
kg ⋅ m
2
s.
10.38: The moment of inertia is proportional to the square of the radius, and so the
angular velocity will be proportional to the inverse of the square of the radius, and the
final angular velocity is
ω
2
2
⎛ R1
⎞ 2 3
2
= ω1
⎜
⎟
= 4.6×
10
R2
⎝
⎠
5
⎛ π rad ⎞⎛ 7.0 × 10 km ⎞
=
⎜
(30 d)(86,400 s d
⎟
⎜
16 km
⎟
⎝
⎠⎝
⎠
rad
s.
10.39: a) The net force is due to the tension in the rope, which always acts in the radial
direction, so the angular momentum with respect to the hole is constant.
b) 2
2
2
L = mω r 1,
L = mω
, and with L = L , ω = ω ( r r ) = 7.00 rad .
1 1 2 2r2
2
K = (1 2) m((
ω2r2
) − ( ω1r
1
2
c) ∆
) ) = 1.03×
10 J.
1 2 2 1 1 2
s
−2
−2
d) No other force does work, so 1.03× 10 J of work were done in pulling the cord.