HW31 - homework 31 PAPAGEORGE MATT Due 4:00 am Question 1...

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

View Full Document Right Arrow Icon
homework 31 – PAPAGEORGE, MATT – Due: Apr 15 2008, 4:00 am 1 Question 1, chap 15, sect 5. part 1 of 3 10 points A uniform disk of radius 0 . 1 m and 9 . 1 kg mass has a small hole a distance from the disk’s center that can serve as a pivot point. 0 . 1 m 9 . 1 kg The acceleration of gravity is 9 . 81 m / s 2 . What should be the distance so that the period of this physical pendulum is 8 . 9 s? (If there are two possible answers, answer with the smaller distance.) Correct answer: 0 . 000254031 m (tolerance ± 1 %). Explanation: Let : R = 0 . 1 m , m = 9 . 1 kg , T = 8 . 9 s , and g = 9 . 81 m / s 2 . Using the parallel-axis theorem, I = I cm + m ℓ 2 = 1 2 m R 2 + m ℓ 2 . The period of a physical pendulum is T = 2 π radicalBigg I m g ℓ T = 2 π radicaltp radicalvertex radicalvertex radicalbt 1 2 m R 2 + m ℓ 2 m g ℓ T 2 π = radicaltp radicalvertex radicalvertex radicalbt 1 2 R 2 + 2 g ℓ T 2 g ℓ 4 π 2 = 1 2 R 2 + 2 2 - g T 2 4 π 2 + R 2 2 = 0 . Applying the quadratic formula, = g T 2 4 π 2 ± radicalbigg g 2 T 4 16 π 4 - 2 R 2 2 . Since g T 2 4 π 2 = (9 . 81 m / s 2 ) (8 . 9 s) 2 4 π 2 = 19 . 6829 m and g 2 T 4 16 π 4 - 2 R 2 = (9 . 81 m / s 2 ) 2 (8 . 9 s) 4 16 π 4 - 2 (0 . 1 m) 2 = 387 . 397 m 2 , then = 19 . 6829 m - 387 . 397 m 2 2 = 0 . 000254031 m . Question 2, chap 15, sect 5. part 2 of 3 10 points What should be the distance so that this physical pendulum will have the shortest pos- sible period? Correct answer: 0 . 0707107 s (tolerance ± 1 %). Explanation: The period is T = 2 π radicalBigg R 2 2 g 1 + g and its derivative is ℓ T dℓ = 2 π 1 2 parenleftbigg R 2 2 g ℓ + g parenrightbigg 1 / 2 × parenleftbigg - R 2 2 g 2 + 1 g parenrightbigg .
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
homework 31 – PAPAGEORGE, MATT – Due: Apr 15 2008, 4:00 am 2 For minimum period, ℓ T dℓ = 0, so - R 2 2 g 2 + 1 g = 0 2 = R 2 2 = R 2 = 0 . 1 m 2 = 0 . 0707107 m . Question 3, chap 15, sect 5. part 3 of 3 10 points What will be the period at this distance? Correct answer: 0 . 754402 s (tolerance ± 1 %). Explanation: T = 2 π radicaltp radicalvertex radicalvertex radicalvertex radicalvertex radicalbt 1 2 R 2 + 1 2 R 2 g R 2 = 2 π radicalBigg 2 R g = 2 π radicalBigg 2 (0 . 1 m) 9 . 81 m / s 2 = 0 . 754402 s .
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