p11_022.pdf

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Chapter 11 - 11.22
22. (a) We obtain
ω = (200 rev / min)(2 π rad / rev)
60 s / min
=20 . 9rad / s .
(b) With r =1 . 20 / 2=0 . 60 m, Eq. 11-18 leads to
v = rω =(0 . 60)(20 . 9) = 12 . 6m / s .
(c) With t =1min, ω = 1000 rev/min and ω o = 200 rev/min, Eq. 11-12 gives
α = ω
−
ω o
= 800 rev / min 2
.
t
(d) With the same values used in part (c), Eq. 11-15 becomes
2 ( ω o + ω ) t = 1
2 (200 + 1000)(1) = 600 rev .
θ = 1

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