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Unformatted text preview: Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Homework Solutions Chapter 13 Assigned Problems: Chapter 13: 6, 10, 14, 18, 21, 30, 33, 47, 57, 68, 69, 72, 78 13.6. Model: The magnetic computer disk is a rigid rotating body. Visualize: Solve: Using the rotational kinematic equation f i , t ϖ ϖ α = + ∆ we get ϖ 1 = 0 rad + (600 rad/s 2 )(0.5 s  0 s) = 300 rad/s ϖ 2 = (300 rad/s) + (0 rad/s 2 )(1.0 s – 0.5 s) = 300 rad/s The speed of the painted dot v 2 = r ϖ 2 = (0.04 m)(300 rad/s) = 12 m/s. The number of revolutions during the time interval t to t 2 is 2 2 2 1 1 1 2 2 1 1 2 1 1 2 1 1 1 ( ) ( ) 0 rad 0 rad (600 rad/s )(0.5 s 0 s) 75 rad 2 2 1 ( ) ( ) 2 1 rev 75 rad (300 rad/s)(1.0 s 0.5 s) 0 rad 225 rad (225 rad) 35.8 rev 2 rad t t t t t t t t θ θ ϖ α θ θ ϖ α π = + + = + + = = + + = + + = = = 13.10. Visualize: Please refer to Figure Ex13.10. The coordinates of the three masses m A , m B , and m C are (0 cm, 0 cm), (10 cm, 10 cm), and (10 cm, 0 cm), respectively. Solve: The coordinates of the center of mass are A A B B C C cm A B C A A B B C C cm A B C (100 g)(0 cm) (200 g)(10 cm) (300 g)(10 cm) 8.33 cm (100 g 200 g 300 g) (100 g)(0 cm) (200 g)(10 cm) (300 g)(0 cm) 3.33 cm (100 g 200 g 300 g) m x m x m x x m m m m y m y m y y m m m + + + + = = = + + + + + + + + = = = + + + + 13.14. Model: The disk is a rotating rigid body. Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Visualize: The radius of the disk is 10 cm and the disk rotates on an axle through its center. Solve: The net torque on the axle is τ = F A r A sin φ A + F B r B sin φ B + F C r C sin φ C + F D r D sin φ D = (30 N)(0.10 m) sin ( 90 ° ) + (20 N)(0.05 m) sin 90 ° + (30 N)(0.05 m) sin 135 ° + (20 N)(0.10 m) sin ° = 3 N m + 1 N m + 1.0607 N m = 0.939 N m Assess: A negative torque means a clockwise rotation of the disk. 13.16. Model: Model the arm as a uniform rigid rod. Its mass acts at the center of mass. Visualize: Solve: (a) The torque is due both to the weight of the ball and the weight of the arm: ball arm b b a a 2 2 ( ) sin90 ( ) sin90 (3.0 kg)(9.8 m/s )(0.70 m)+(4.0 kg)(9.8 m/s )(0.35 m) 34.4 N m m g r m g r τ τ τ = + = ° + ° = = (b) The torque is reduced because the moment arms are reduced. Both forces act at φ = 45° from the radial line, so ball arm b b a a 2 2 ( ) sin45 ( ) sin45 (3.0 kg)(9.8 m/s )(0.70 m)(0.707) (4.0 kg)(9.8 m/s )(0.35 m)(0.707) 24.3 N m m g r m g r τ τ τ = + = ° + ° = + = 13.18. Model: The moment of inertia of any object depends on the axis of rotation. Cal Poly State University, SLO D. Niebuhr Physics Department Winter 2008 Phys 141 Visualize: Solve: (a) A A B B C C D D cm A B C D A A B B C C D D cm A B C D (100 g)(0 m) (200 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 m)(200 g)(0....
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This note was uploaded on 04/02/2008 for the course PHYS 141 taught by Professor Staff during the Winter '06 term at Cal Poly.
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