{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw11 - al-ameri(aoa434 Hwk10 Stokes(19102 This print-out...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
al-ameri (aoa434) – Hwk10 – Stokes – (19102) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No 10. It is due before 6.00 am on Tuesday 21 April, Abu Dhabi Time. (Monday 9.00 pm in Texas) 001 10.0 points A homogeneous cylinder of radius 20 cm and mass 50 kg is rolling without slipping along a horizontal floor at 2 m / s. How much work was required to give it this motion? Correct answer: 0 . 15 kJ. Explanation: Let : r = 20 cm , m = 50 kg , and v = 2 m / s . Because the cylinder rolls without slipping, the translational speed is v = r ω and the total kinetic energy of the cylinder is K total = 1 2 m v 2 + 1 2 I ω 2 = 1 2 m v 2 + 1 2 parenleftbigg 1 2 m r 2 parenrightbigg parenleftBig v r parenrightBig 2 = 3 4 m v 2 = 3 4 (50 kg) (2 m / s) 2 · 1 kJ 1000 J = 0 . 15 kJ , which is also the work required to give the cylinder this motion. keywords: 002 (part 1 of 2) 10.0 points A skater spins with an angular speed of 13 . 8 rad / s with his arms outstretched. He lowers his arms, decreasing his moment of in- ertia from 47 kg · m 2 to 43 kg · m 2 . a) Calculate his initial rotational kinetic energy. Correct answer: 4475 . 34 J. Explanation: Basic Concepts: L = KE = 1 2 2 Given: I i = 47 kg · m 2 I f = 43 kg · m 2 ω i = 13 . 8 rad / s Solution: KE i = 1 2 I i ω 2 i = 1 2 (47 kg · m 2 )(13 . 8 rad / s) 2 = 4475 . 34 J 003 (part 2 of 2) 10.0 points b) Calculate his final rotational kinetic en- ergy. Correct answer: 4891 . 65 J. Explanation: Solution: Angular momentum of the system is con- served, so I f ω f = I i ω i ω f = I i ω i I f KE f = 1 2 I f parenleftbigg I i ω i I f parenrightbigg 2 = 1 2 I 2 i ω 2 i I f = 1 2 (47 kg · m 2 ) 2 (13 . 8 rad / s) 2 43 kg · m 2 = 4891 . 65 J 004 (part 1 of 2) 10.0 points A cylinder with moment of inertia 26 . 3 kg m 2 rotates with angular velocity 5 . 93 rad / s on a
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
al-ameri (aoa434) – Hwk10 – Stokes – (19102) 2 frictionless vertical axle. A second cylinder, with moment of inertia 25 . 2 kg m 2 , initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same an- gular velocity. I 2 I 1 ω 0 Before ω After Calculate the final angular velocity. Correct answer: 3 . 02833 rad / s. Explanation: From conservation of angular momentum ( I 1 + I 2 ) ω = I 1 ω 0 , or ω = I 1 I 1 + I 2 ω 0 = (26 . 3 kg m 2 ) (26 . 3 kg m 2 ) + (25 . 2 kg m 2 ) (5 . 93 rad / s) = 3 . 02833 rad / s . 005 (part 2 of 2) 10.0 points Show that energy is lost in this situation by calculating the ratio of the final to the initial kinetic energy. Correct answer: 0 . 51068. Explanation: K i = 1 2 I 1 ω 0 2 and K f = 1 2 ( I 1 + I 2 ) ω 2 , Therefore K f = 1 2 ( I 1 + I 2 ) ω 2 = 1 2 ( I 1 + I 2 ) parenleftbigg I 1 I 1 + I 2 parenrightbigg 2 ω 2 0 = parenleftbigg I 1 I 1 + I 2 parenrightbigg 1 2 I 1 ω 2 0 = parenleftbigg I 1 I 1 + I 2 parenrightbigg K i .
Image of page 2
Image of page 3
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