University Physics with Modern Physics with Mastering Physics (11th Edition)

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12.39: a) Refer to the derivation of Eq. (12.26) and Fig. (12.22). In this case, the red ring in Fig. (12.22) has mass M and the common distance s is . 2 2 a x + Then, . 2 2 a x GMm U + - = b) When x >> a , the term in the square root approaches 2 x and x GMm U - , as expected. c) x F δx δU - = , 2 3 2 2 ) ( a x GMmx + - = , with the minus sign indicating an attractive force. d) when x >> a, the term inside the parentheses in the above expression approaches 2 x and 2 3 2 ) ( x GMmx F x - , 2 x GMm - = as expected.
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Unformatted text preview: e) The result of part (a) indicates that a GMm U-= when . = x This makes sense because the mass at the center is a constant distance a from the mass in the ring. The result of part (c) indicates that = x F when . = x At the center of the ring, all mass elements that comprise the ring attract the particle toward the respective parts of the ring, and the net force is zero....
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