Chap8 - 2 Strategy Use Eq(8-2 to find the rotational...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
2. Strategy Use Eq. (8-2) to find the rotational inertia. Use Eq. (7-9) for the center of mass. Solution (a) 2 2 2 2 2 (200 g)(5.0 cm) (300 g)(0 cm) (500 g)(4.0 cm) 13,000 g cm C x i i i A I m r = = = + + = (b) 2 2 2 2 2 (200 g)(3.0 cm) (300 g)(6.0 cm) (500 g)(5.0 cm) 25,000 g cm C y i i i A I m r = = = + + = (c) 2 2 2 2 2 2 2 (200 g)[(3.0 cm) (5.0 cm) ] (300 g)(6.0 cm) (500 g)[(5.0 cm) (4.0 cm) ] 38,000 g cm C z i i i A I m r = = = + + + + = (d) CM (200 g)( 3.0 cm) (300 g)(6.0 cm) (500 g)( 5.0 cm) 1.3 cm 200 g 300 g 500 g x + + = = + + CM (200 g)(5.0 cm) (300 g)(0 cm) (500 g)( 4.0 cm) 1.0 cm 200 g 300 g 500 g y + + = = + + 13. Strategy Use Eq. (8-3) to compute the torque in each case. Solution (a) The force is applied perpendicularly to the door, so (1.26 m)(46.4 N) 58.5 N m . rF τ = = = (b) The force is applied at 43.0 ° from the door’s surface, so sin (1.26 m)(46.4 N)sin 43.0 39.9 N m . rF rF τ θ = = = ° = (c) Since the force is applied such that its line of action passes through the axis of the door hinges— the axis of rotation—there is no perpendicular component of the force and the torque is 0 . 17. Strategy Use Eq. (8-3).
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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