Phys 141_HW Sol_Ch13_W08

# Phys 141_HW Sol_Ch13_W08 - Cal Poly State University SLO...

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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 0 ° = - 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) 0.0571 m 100 g 200 g 200 g 200 g (100 g)(0 m) (200 g)(0.10 m) (200 g)(0.10 cm) (200 i i i m x m x m x m x m x x m m m m m m y m y m y m y y m m m m + + + = = + + + + + + = = + + + + + + = + + + + + + = g)(0 m) 0.0571 m 700 g = (b) The moment of inertia about a diagonal that passes through B and D is 2 2 BD A A C C I m r m r = + where r A = r C = (0.10 m) cos 45 ° = 7.07 cm and are the distances from the diagonal. Thus, 2 2 2 BD A C (0.100 kg) (0.200 kg) 0.0015 kg m

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