Phys 141_HW Sol_Ch13_W08

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

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online