HW5-soln - Physics 105 Mechanics Fall 2009 Homework 5...

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Unformatted text preview: Physics 105: Mechanics Fall 2009 Homework 5 Solutions 4.HW.78 I. Find the interatomic bondlength in Ti: The volume of the wire is V = ( L )( πR 2 ) = (3m)( π )(1 . × 10- 3 m) 2 = 9 . 43 × 10- 6 m 3 . The mass of the wire is M = ρV = (4 . 51g / cm 3 ) 1kg 1000g ( 100cm 1m ) 3 (9 . 43 × 10- 6m 3 ) = 0 . 425kg . The mass of a single Ti atom is m Ti = (48g / mol)(1kg / 1000g) / (6 . 02 × 10 23 atoms / mol) = 1 . 66 × 10- 21 kg / atom . Hence the number of atoms in the wire is N atoms = M/m Ti = 5 . 33 × 10 23 . Assuming the Ti atoms sit on a cubic lattice, the volume occupied by each atom is d 3 , where d is the interatomic spacing. So the number of atoms in the whole volume is V/d 3 . N atoms = V/d 3 = 5 . 33 × 10 23 Or d = 2 . 61 × 10- 10 m . II. Determine the number of chains of atoms (which run along the length of the wire), and the number of bonds (or atoms) per chain: N chain = A/d 2 = πR 2 /d 2 = 4 . 61 × 10 13 where A is the cross sectional area of the wire, and d 2 is the area/atom. N bonds/chain = L/d = 1 . 15 × 10 10 , where L is the length of the wire. III. Relate the spring stiffness of interatomic bonds, k i , to the stiffness of the wire, k w : For a single chain, consisting of N bonds/chain interatomic springs in series, k chain = k i N bonds/chain . For N chain chains in parallel, the stiffness is k w = k chain N chain = k i N chain N bonds/chain . Rearranging: k i = k w N bonds/chain N chain . IV. Determine the stiffness of the wire, and calculate k i : k w = mg/s = (5kg)(9 . 8m / s 2 ) / ( . 4035 × 10- 3 m) = 1 . 21 × 10 5 N / m ....
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HW5-soln - Physics 105 Mechanics Fall 2009 Homework 5...

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