()2223 0.040 kg m32.7 kg.220.15mIMR⋅===It also follows from the rotational inertia expression that 1221322IMRωω=. Furthermore,it rolls without slipping, vcom= Rω, and we find KKKMRmRMRrotcomrot+=+131213.(a) Simplifying the above ratio, we find Krot/K= 0.4. Thus, 40% of the kinetic energy is rotational, or Krot= (0.4)(20 J) = 8.0 J. (b) From 1rot38.0JKMR(and using the above result for M) we find 10153802720...mJkgrad sbgwhich leads to vcom= (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.