ch11-p078

# Ch11-p078 - 78(a The acceleration is given by Eq 11-13 acom = g 1 I com MR02 where upward is the positive translational direction Taking the

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() ( ) 2 com,0 2 com,0 0 2 2 52 2 11 22 1 1 1.3 m/s 0.12 kg 1.3 m/s 9.5 10 kg m 0.12 kg 9.8 m/s 1.2 m 2 2 0.0032 m 9.4 J. fi i v KK U m v I M g h R §· =+= + + ¨¸ ©¹ =+ × + = (c) As it reaches the end of the string, its center of mass velocity is given by Eq. 2-11: vv a t v gt IM R com com com com com =+=− + ,, . 00 0 2 1 78. (a) The acceleration is given by Eq. 11-13: a g R com com = + 1 0 2 where upward is the positive translational direction. Taking the coordinate origin at the initial position, Eq. 2-15 leads to yv t a t v t gt R com com,0 com com,0 com =− + 1 21 2 1 2 2 0 2 where y com = – 1.2 m and v com,0 = – 1.3 m/s. Substituting I com kg m =⋅ 0 000095 2 ., M = 0.12 kg, R 0 = 0.0032 m and g = 9.8 m/s 2 , we use the quadratic formula and find ( ) ( ) ( ) ( ) com 2 0 2 com 2 com 0 2 2 com,0 com,0 29 .8 1 .2 2 1 0.000095 0.12 0.0032 2 1 0.000095 0.12 0.0032 1 1 1.3 (1.3) 9.8 21.7 or 0.885 I MR gy R t g + + +− = = B B where we choose t = 0.89 s as the answer. (b) We note that the initial potential energy is U i = Mgh and h = 1.2 m (using the bottom as the reference level for computing

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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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Ch11-p078 - 78(a The acceleration is given by Eq 11-13 acom = g 1 I com MR02 where upward is the positive translational direction Taking the

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