74. (a) We use Table 10-2(e) and the parallel-axis theorem to obtain the rod’s rotational inertia about an axis through one end: IIMhML MLML=+=+FHGIKJ=com2222112213whereL= 6.00 m and M= 10.0/9.8 = 1.02 kg. Thus, the inertia is212.2 kg mI=⋅.(b) Using ω= (240)(2π/60) = 25.1 rad/s, Eq. 11-31 gives the magnitude of the angular
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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.