ch11-p017 - 17(a The derivation of the acceleration is...

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17. (a) The derivation of the acceleration is found in §11-4; Eq. 11-13 gives a g IM R com com =− + 1 0 2 where the positive direction is upward. We use I com gcm =⋅ 950 2 , M =120g, R 0 = 0.320 cm and g = 980 cm/s 2 and obtain () ( ) 2 22 com 2 2 980 cm/s | | 12.5 cm/s 13 cm/s . 1 950 g cm 120 g 0.32 cm a == +⋅ (b) Taking the coordinate origin at the initial position, Eq. 2-15 leads to ya t com com = 1 2 2 . Thus, we set y com = – 120 cm, and find com 2 com 2 120cm 2 4.38 s 4.4 s. 12.5 cm s y t a = (c) As it reaches the end of the string, its center of mass velocity is given by Eq. 2-11: 2 com com 12.5 cm s 4.38s 54.8 cm s va t = , so its linear speed then is approximately
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