fisica1-youn-e-freedman-exercicios-resolvidos

10.102: Denoting the upward forces that the hands exert as F L
and F R
, the conditions that
F
L
and F R
must satisfy are
F + F = w
L
R
Iω
FL
− FR
= Ω ,
r
where the second equation is τ = ΩL,
divided by r. These two equations can be solved
for the forces by first adding and then subtracting, yielding
1 ⎛ Iω
⎞
FL
= ⎜ω
+ Ω ⎟
2 ⎝ r ⎠
1 ⎛ Iω
⎞
FR
= ⎜ω
− Ω ⎟.
2 ⎝ r ⎠
2
Using the values ω = mg = (8.00 kg)(9.80 m s ) = 78.4 N and
gives
a) Ω = 0, F
2
Iω
(8.00 kg)(0.325 m) (5.00 rev s × 2π
rad rev)
=
= 132.7 kg ⋅ m s
r
(0.200 m)
L
F
= F
39.2 N + Ω(66.4 N ⋅s),
F = 39.2 N − Ω(66.4 N ⋅s).
= 39.2 N.
L
=
R
R
b) Ω = 0.05 rev s = 0.314 rad s, F
c) Ω = 0.3 rev s = 1.89 rad s, F
d)
= 0
gives
L
= 60.0 N, F
= 165 N, F
with the minus sign indicating a downward force.
F
R
Ω =
39.2 N
=
66.4 N⋅s
L
R
R
= 18.4 N.
= −86.2 N,
0.575 rad s, which is 0.0916 rev
s.
2 2 3
10.103: a) See Problem 10.92; T = mv r .
W = −
c) v = v ( r ), 2
so
2 1 1
r
1 1
r
r r
b) T and d
are always antiparallel, so
2
r2
r1
2 2 dr mv 2
⎡ 1 1 ⎤
∫ T dr = mv r ∫ =
1 r1
⎢ −
1 3
2 ⎥ ⎦
r1
1
∆K
= m(
v
2
2
2
1
.
r 2 2
r 2
2
r1
− v
2
1
mv
) =
2
which is the same as the work found in part (b).
2
1
⎡⎛
r ⎞
1
⎢
⎜
⎟
⎢⎣
⎝ r2
⎠
2
⎣r
⎤
−1⎥
'
⎥⎦