oldHWK27 - oldhomewk 27 VALENCIA DANIEL Due Apr 1 2008 4:00...

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

View Full Document Right Arrow Icon
oldhomewk 27 – VALENCIA, DANIEL – Due: Apr 1 2008, 4:00 am 1 Question 1, chap 13, sect 4. part 1 of 3 10 points A 3 kg bicycle wheel rotating at a 2221 rev / min angular velocity has its shaft supported on one side, as shown in the fig- ure. The wheel is a hoop of radius 0 . 6 m, and its shaft is horizontal. The distance from the center of the wheel to the pivot point is 0 . 5 m. When viewing from the left (from the posi- tive x -axes), one sees that the wheel is rotat- ing in a clockwise manner. Assume: All of the mass of the system is located at the rim of the bicycle wheel. The acceleration of gravity is 9 . 8 m / s 2 . y z x b m g ω R The magnitude of the angular momentum of the wheel is given by 1. bardbl vector L bardbl = 1 2 m R 2 ω 2. bardbl vector L bardbl = m R 2 ω 2 3. bardbl vector L bardbl = 1 4 m R 2 ω 2 4. bardbl vector L bardbl = m R 2 ω correct 5. bardbl vector L bardbl = 1 2 m R 2 ω 2 6. bardbl vector L bardbl = 1 4 m R 2 ω Explanation: Basic Concepts: vector τ = d vector L dt Solution: The magnitude of the angular momentum I of the wheel is L = I ω = m R 2 ω , and is along the negative x -axis. Question 2, chap 13, sect 4. part 2 of 3 10 points Let : m = 3 kg , ω = 2221 rev / min b = 0 . 5 m , and R = 0 . 6 m . Find the change in the precession angle after a 1 . 3 s time interval. Correct answer: 4 . 35896 (tolerance ± 1 %). Explanation: Let : ω = 2221 rev / min = 2 π (2221 rev / min) (60 s / min) = 232 . 583 rad / s . Solution: The magnitude of the angular momentum I of the wheel is L = I ω = m R 2 ω , and is along the negative x -axis. From the figure in Part 2, we get Δ φ = Δ L L . Using the relation, Δ L = τ Δ t , where τ is the torque, vector τ = vector b × mvectorg . The precession angle Δ φ is Δ φ = Δ L L = τ Δ t L = m g b Δ t m R 2 ω = g b Δ t R 2 ω = (9 . 8 m / s 2 ) (0 . 5 m) (1 . 3 s) (0 . 6 m) 2 (232 . 583 rad / s) = 0 . 0760781 rad = 4 . 35896 .
Image of page 1

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

View Full Document Right Arrow Icon
oldhomewk 27 – VALENCIA, DANIEL – Due: Apr 1 2008, 4:00 am 2 Question 3, chap 13, sect 4.
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