OLDHW30 - oldhomewk 30 PAPAGEORGE MATT Due 4:00 am and the...

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oldhomewk 30 – PAPAGEORGE, MATT – Due: Apr 28 2008, 4:00 am 1 Question 1, chap 15, sect 1. part 1 of 2 10 points A 1 . 83 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 1 . 72 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 . 97 s after it is released? Correct answer: 8 . 10693 N (tolerance ± 1 %). Explanation: Given : A = 1 . 72 m , k = 4 . 74 N / m , and m = 1 . 83 kg . The block exhibits simple harmonic motion, so the displacement is x = A cos( ωt ) where the angular frequency is ω = radicalbigg k m . Thus the force is F = - k x = - k A cos parenleftBigg radicalbigg k m t parenrightBigg = - (4 . 74 N / m) (1 . 72 m) cos bracketleftBigg radicalBigg 4 . 74 N / m 1 . 83 kg (3 . 97 s) bracketrightBigg = - 8 . 10693 N a force of 8 . 10693 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 . 97 s? Correct answer: 1 . 01689 turn (tolerance ± 1 %). Explanation: The angular frequency is w = radicalbigg k m and the period of oscillation is T = 2 π ω = 2 π radicalbigg m k , so the number of oscillations made in 3 . 97 s is N = Δ t T = Δ t 2 π radicalbigg k m = 3 . 97 s 2 π radicalBigg 4 . 74 N / m 1 . 83 kg = 1 . 01689 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 24 m is pivoted about a point 9 . 1 m from one end, while the opposite end is attached to a spring of force constant 26 N / m and negligible mass. 9 . 1 m 53 kg 24 m θ 26 N / m Calculate the moment of inertia I about the pivot point. Correct answer: 2989 . 73 kg m 2 (tolerance ± 1 %). Explanation: Let : L = 24 m , = 9 . 1 m , m = 53 kg , and k = 26 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
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oldhomewk 30 – PAPAGEORGE, MATT – 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) (24 m) 2 + (53 kg) × bracketleftbigg (24 m) 2 - (9 . 1 m) bracketrightbigg 2 = 1 12 (53 kg) (24 m) 2 + (53 kg) (2 . 9 m) 2 = (2544 kg m 2 ) + (445 . 73 kg m 2 ) = 2989 . 73 kg m 2 .
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