ch11-p006

# ch11-p006 - 6. From I = 2 MR 2 (Table 10-2(g) we find 3 3 (...

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() 2 2 2 3 0.040 kg m 3 2.7 kg. 2 20 .15 m I M R == = It also follows from the rotational inertia expression that 1 2 2 1 3 22 IM R ωω = . Furthermore, it rolls without slipping, v com = R ω , and we find K KK MR mR MR rot com rot + = + 1 3 1 2 1 3 . (a) Simplifying the above ratio, we find K rot / K = 0.4. Thus, 40% of the kinetic energy is rotational, or K rot = (0.4)(20 J) = 8.0 J. (b) From 1 rot 3 8.0J KM R (and using the above result for M ) we find 1 015 380 27 20 . . . m J kg rad s b g which leads to v com = (0.15 m)(20 rad/s) = 3.0 m/s. (c) We note that the inclined distance of 1.0 m corresponds to a height h = 1.0 sin 30° =
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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