Phys 141_HW Sol_Ch13_W08

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

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

View Full Document Right Arrow Icon
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 0 to t 2 is 2 2 2 1 0 0 1 0 0 1 0 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.
Image of page 1

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

View Full Document Right Arrow Icon
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.
Image of page 2
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
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern