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Exercicios Resolvidos do Halliday sobre Rotação- Cap11- Exercicio 32

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32. (a) The angular speed in rad/s is
ω =
(
33
1
3
rev/min
)(
2π rad/rev
60 s/min
)
= 3.49 rad/s .
Consequently, the radial (centripetal) acceleration is (using Eq. 11-23)
a = ω2r = (3.49 rad/s)2
(
6.0× 10−2m) = 0.73m/s2 .
(b) Using Ch. 6 methods, we have ma = fs ≤ fs,max = µsmg, which is used to obtain the (minimum
allowable) coefficient of friction:
µs,min =
a
g
=
0.73
9.8
= 0.075 .
(c) The radial acceleration of the object is ar = ω2r, while the tangential acceleration is at = αr. Thus
|�a| =
√
a2r + a2t =
√
(ω2r)2 + (αr)2 = r
√
ω4 + α2 .
If the object is not to slip at any time, we require
fs,max = µsmg = mamax = mr
√
ω4max + α2 .
Thus, since α = ω/t (from Eq. 11-12), we find
µs,min =
r
√
ω4max + α2
g
=
r
√
ω4max + (ωmax/t)2
g
=
(0.060)
√
3.494 + (3.49/0.25)2
9.8
= 0.11 .
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	Chapter 1 Measurement
	Chapter 2 Motion Along a Straight Line
	Chapter 3 Vectors
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