fisica1-youn-e-freedman-exercicios-resolvidos

2 2 2 2
9.73: a) ∆ a = ω r − ω r = ( ω − ω ) r
rad
=
0
[ ω − ω ][ ω + ω ]
0
⎡ω − ω0
⎤
=
⎢
⎣ t ⎥
⎦
=
0
[(
ω + ω ) t]
[ α] [2 ( θ − θ ) r.
b) From the above,
2
2
∆arad
(85.0 m s − 25.0 m s )
2
αr = =
= 2.00 m s .
2∆θ
2(15.0 rad)
c) Similar to the derivation of part (a),
0
0
r
0
r
∆K
= ω
2
1 2
2
0
1
I − ω
2
I
1
= [ α][2∆θ]
I
2
= Iα∆θ.
d) Using the result of part (c),
I
∆K
45.0 J − 20.0 J)
= =
= 0.208 kg ⋅ m
2
α∆θ ((2.00 m s )/(0.250 m))(15.0 rad)
( 2
.
9.74: I = I wood
+ Ilead
m
w
2
= mwR
5
= ρ
w
V
w
2
+
= ρ
2
mLR
3
4 3
πR
3
w
2
m
L
= σ
L
A
2 ⎛
I = ⎜ ρ
5 ⎝
8
3
L
w
= σ
L
4
π R
3
4π
R
3
2
⎞
⎟R
⎠
⎛ ρwR
⎜ + σ
⎝ 5
4
= π R
L
2
⎞
⎟
⎠
2
+ ( σ
3
L
4 π R
2
) R
2
8
4 ⎡(800 kg
= π (0.20 m)
3
⎢
⎣
= 0.70 kgm
2
3
m )(0.20 m)
+ 20 kg
5
m
⎤
⎥
⎦
2