ch11-p079 - 79. (a) When the small sphere is released at...

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K K I IM v M I v rot com com = + = + 1 2 2 1 2 2 1 2 2 2 1 1 ω b g di . Substituting v com = R (Eq. 11-2) and R = 2 5 2 (Table 10-2(f)), we obtain () 2 rot 2 5 2 12 0.29. 7 1 R K K R == + (c) The small sphere is executing circular motion so that when it reaches the bottom, it experiences a radial acceleration upward (in the direction of the normal force which the “bowl” exerts on it). From Newton’s second law along the vertical axis, the normal force F N satisfies F N – mg = ma com where avR r com com =− 2 /( ). Therefore, 2 2 com com . N mg R r mv mv Fm g Rr −+ =+ = −− But from part (a), mg ( R – r ) = K , and from Eq. 11-5, 1 2 2 mv K K com rot . Thus, rot rot 2 32 . N KK K K K F R rR r R r +− §· ¨¸ ©¹ We now plug in R – r = K / mg and use the result of part (b): 42 2 21 7 1 7 3 2 (5.6 10 kg)(9.8 m/s ) 1.3 10 N. 77 7 N g m g m g = = × = × 79. (a) When the small sphere is released at the edge of the large “bowl” (the hemisphere
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