P113_06_ps10solns

# P113_06_ps10solns - of the sample, we have m = water V ....

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Solutions to problem set 10 – P113 – Fall 2006

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12- 71. ( a ) The cylinder will roll about the contact point A. We write Σ τ = I α about the point A: F a (2 R h ) + F N1 [ R 2 – ( R h ) 2 ] 1/2 Mg [ R 2 – ( R h ) 2 ] 1/2 = I A . When the cylinder does roll over the curb, contact with the ground is lost and F N1 = 0. Thus we get F a = { I A + Mg [ R 2 – ( R h ) 2 ] 1/2 }/(2 R h ) = [ I α /(2 R h )] + [ Mg (2 Rh h 2 ) 1/2 /(2 R h )]. α = 0: F a min = Mg [ h (2 R h )] 1/2 /(2 R h ) = Mg [ h /(2 R h )] 1/2 . ( b ) The cylinder will roll about the contact point A. We write τ = α about the point A: F b ( R h ) + F N1 [ R 2 – ( R h ) 2 ] 1/2 Mg [ R 2 – ( R h ) 2 ] 1/2 = I α . When the cylinder does roll over the curb, contact with the ground is lost and F N1 = 0. Thus we get F b = { I α + Mg [ R 2 – ( R h ) 2 ] 1/2 }/( R h ) = [ A α /( R h )] + [ Mg (2 Rh h 2 ) 1/2 /( R h The minimum force occurs when α = 0: F b min = Mg [ h (2 R h )] 1/2 /( R h ) . h R F a F N1 M g F N2 A

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13-31. Because the mass of the displaced water is the apparent change in mass
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Unformatted text preview: of the sample, we have m = water V . For the density of the sample we have = m / V = ( m / m ) water = [(63.5 g)/(63.5 g 56.4 g)](1.00 10 3 kg/m 3 ) = 8.94 10 3 kg/m 3 . From the table of densities, the most likely metal is copper . 13-45. If we ignore viscosity, we can use Bernoullis equation. We choose the initial point at the top of the water, where the velocity is essentially zero, and the final point at the hole. Thus we have P 1 + ! v 1 2 + gy 1 = P 2 + ! v 2 2 + gy 2 ; P atm + 0 + gh = P atm + ! v 2 2 + 0, which gives v 2 = (2 gh ) 1/2 = [2(9.80 m/s 2 )(4.6 m)] 1/2 = 9.5 m/s ....
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## This note was uploaded on 04/07/2008 for the course PHY 113 taught by Professor Jordan during the Spring '08 term at Rochester.

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P113_06_ps10solns - of the sample, we have m = water V ....

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