17. (a) The derivation of the acceleration is found in §11-4; Eq. 11-13 gives agIMRcomcom=−+102where the positive direction is upward. We use Icomgcm=⋅9502,M=120g, R0= 0.320 cm and g= 980 cm/s2and obtain ()()222com22980 cm/s||12.5 cm/s13 cm/s .1950 g cm120 g 0.32 cma==≈+⋅(b) Taking the coordinate origin at the initial position, Eq. 2-15 leads to yatcomcom=122.Thus, we set ycom= – 120 cm, and find com2com2120cm24.38 s 4.4 s.12.5 cm syta−=≈−(c) As it reaches the end of the string, its center of mass velocity is given by Eq. 2-11: 2comcom12.5 cm s4.38s54.8 cm svat−=−,so its linear speed then is approximately
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