fisica1-youn-e-freedman-exercicios-resolvidos

β 2
β 2
9.101: a) ds = r dθ
= r0
dθ
+ βθ dθ
, so s(
θ ) = r0θ
+
2
θ . b) Setting s = vt = r0θ
+
2
θ
gives a quadratic in θ . The positive solution is
1 ⎡ 2
θ(t) = = r + − ⎤
0
2βvt
r0
.
β ⎢⎣
⎥⎦
(The negative solution would be going backwards, to values of r smaller than r 0
.)
c) Differentiating,
dθ
ωz
( t)
=
dt
=
r
2
0
v
,
+ 2βvt
α =
z
dω
= −
dt
2
( r = 2βvt)
0
βv
2
3 2
.
The angular acceleration α
z
is not constant. d) r = 25.0
0
mm; It is crucial that θ is
measured in radians, so β = ( 1 .55 µ m rev)( 1rev 2π
rad) = 0.247 µ m rad.
The total angle
turned in 74.0 min = 4440 s is
1
θ =
2.47 × 10
−7
= 1.337 × 10
5
⎡
⎢
m/rad ⎢
⎣
rad
2
+
−7
( 2.47 × 10 m/rad)( 1.25 m/s)( 4440 s)
−3
( 25.0×
10 m)
2
− 25.0 × 10
−3
m
⎤
⎥
⎥
⎦
which is 2.13 ×
10 4 rev.
e)