problem12_88

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

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12.88: As suggested in the problem, divide the disk into rings of radius r and thickness dr . Each ring has an area dr r dA 2 π = and mass . 2 2 2 dr r dA dM a M a M = = π The magnitude of the force that this small ring exerts on the mass m is then ), ) ( )( ( 2 3 2 2 x r x dM m G + the expression found in Problem 12.82, with dM instead of M and the variable r instead of a . Thus, the contribution dF to the force is . ) ( 2 2 3 2 2 2 r x rdr a GMmx dF + = The total force F is then the integral over the range of r ; . ) ( 2 0 2 3 2 2 2 dr r x r a GMmx dF F a + = = The integral (either by looking in a table or making the substitution 2 2 a r u + = ) is . 1 1 1 1 ) ( 2 2 2 2 0 2 3 2 2
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Unformatted text preview: +-= +-= + ∫ x a x x x a x dr r x r a Substitution yields the result . 1 2 2 2 2 +-= x a x a GMm F The second term in brackets can be written as 2 2 1 2 2 2 1 1 ) ) ( 1 ( ) ( 1 1 -≈ + = +-x a x a x a if x >> a , where the binomial approximation (or first-order Taylor series expansion) has been used. Substitution of this into the above form gives , 2 x GMm F ≈ as it should....
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