This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: oldhomewk 30 VALENCIA, DANIEL Due: Apr 28 2008, 4:00 am 1 Question 1, chap 15, sect 1. part 1 of 2 10 points A 1 . 82 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4 . 74 N / m. The object is displaced 5 . 43 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3 . 93 s after it is released? Correct answer: 25 . 6933 N (tolerance 1 %). Explanation: Given : A = 5 . 43 m , k = 4 . 74 N / m , and m = 1 . 82 kg . The block exhibits simple harmonic motion, so the displacement is x = A cos( t ) where the angular frequency is = radicalbigg k m . Thus the force is F = k x = k A cos parenleftBigg radicalbigg k m t parenrightBigg = (4 . 74 N / m) (5 . 43 m) cos bracketleftBigg radicalBigg 4 . 74 N / m 1 . 82 kg (3 . 93 s) bracketrightBigg = 25 . 6933 N a force of 25 . 6933 N directed to the left. Question 2, chap 15, sect 1. part 2 of 2 10 points How many times does the object oscillate in 3 . 93 s? Correct answer: 1 . 00941 turn (tolerance 1 %). Explanation: The angular frequency is w = radicalbigg k m and the period of oscillation is T = 2 = 2 radicalbigg m k , so the number of oscillations made in 3 . 93 s is N = t T = t 2 radicalbigg k m = 3 . 93 s 2 radicalBigg 4 . 74 N / m 1 . 82 kg = 1 . 00941 turn . Question 3, chap 15, sect 1. part 1 of 3 10 points In an experiment conducted on the space shuttle ( i.e. , in free fall), a horizontal rod of mass 53 kg and length 89 m is pivoted about a point 32 m from one end, while the opposite end is attached to a spring of force constant 20 N / m and negligible mass. 3 2 m 5 3 k g 8 9 m 20N / m Calculate the moment of inertia I about the pivot point. Correct answer: 43265 . 7 kg m 2 (tolerance 1 %). Explanation: Let : L = 89 m , = 32 m , m = 53 kg , and k = 20 N / m . Using the Parallel Axis Theorem, the mo ment of inertia about a point midway between the center of mass and the end is oldhomewk 30 VALENCIA, DANIEL Due: Apr 28 2008, 4:00 am 2 I = I cm + M d 2 = 1 12 M L 2 + M bracketleftbigg L 2 bracketrightbigg 2 = 1 12 (53 kg) (89 m) 2 + (53 kg) bracketleftbigg (89 m) 2 (32 m) bracketrightbigg...
View
Full
Document
 Spring '08
 Turner
 Friction

Click to edit the document details