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HW_12_Solutions_F05

# HW_12_Solutions_F05 - Physics 112 HW#12 Solutions...

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- 1 - Physics 112 - HW #12 Solutions Fall 2005 11-78 [Dragging A Hay Bale] (a) Since the bale is moving at constant velocity, the net force on it is zero. F x = F – f k = 0 F = f k (1) F y = N – w = 0 N = mg (2) Since block is sliding, f k = µ k N = µ k mg Substituting for f k into eq. (1), we have F = µ k mg = (0.35)(30.0 kg)(9.81 m/s 2 ) = 103 N (b) When block is about to tip about edge A (see diagram), the normal force N is acting at that edge, and the net torque about A is zero. τ A = w(0.125 m) – F h = 0 h = w(0.125 m) F = (30.0 kg)(9.81 m/s 2 )(0.125 m) 103 N = 0.357 m 10-39 [Block on String] (a) Yes, angular momentum of the block is conserved, because there are no forces that can exert a torque on the block. (b) L f = L i so m R f 2 ω f = m R i 2 ω i ω f = R i 2 R f 2 ω i = (0.300 m) 2 (0.150 m) 2 1.75 rad/s = 7.00 rad/s (c) Change in KE of block is K f - K i = ½ mR f 2 ω f 2 - ½ mR i 2 ω i 2 = ½ (0.0250 kg)[ (0.150 m) 2 (7.00 rad/s) 2 - (0.300 m) 2 (1.75 rad/s) 2 ] = 1.03 x 10 -2 J (d) The work done in pulling on the cord is equal to the change in KE of the block, so W = 1.03 x 10 -2 J

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- 2 - 10-40 [Spinning Figure Skater] Angular momentum is conserved, so L f = L i I f ω f = I i ω i [ I body + I arms wrapped ] ω f = [I body + I arms out ] ω i ω f = [ 0.40 kg m 2 + ( 1 / 12 ) (8.0 kg)(1.8 m) 2 ] [ 0.40 kg m 2 + (8.0 kg)(0.25 m) 2 ] (0.40 rev/s) = 1.14 rev/s or 7.15 rad/s 10-43 [Parachuting onto a Turntable] (a) Vertical landing Cons. of Angular Momentum about turntable axis: L before = L after I i ω i = I f ω f I i = 1 2 MR 2 = 1 2 (120 kg)(2.00 m) 2 = 240 kg-m 2 I f = 1 2 MR 2 + mR 2 = 240 kg-m 2 + (70 kg)(2.00 m) 2 = 520 kg-m 2 ω f = I i I f ω i = 240 kg-m 2 520 kg-m 2 (3.00 rad/s) = 1.38 rad/s (b) K i = 1 2 I i ω i 2 = 1 2 (240 kg-m 2 )(3.00 rad/s) 2 = 1080 J K f = 1 2 I f ω f 2 = 1 2
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HW_12_Solutions_F05 - Physics 112 HW#12 Solutions...

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