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hw23 - Muraj Hamza Homework 23 Due 4:00 am Inst Florin This...

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Muraj, Hamza – Homework 23 – Due: Mar 27 2006, 4:00 am – Inst: Florin 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Consider the diatomic molecule oxygen, O 2 , which is rotating in the xy plane about the z axis passing through its center, perpendicular to its length. The mass of each oxygen atom is 2 . 66 × 10 - 26 kg, and at room temperature, the average separation distance between the two oxygen atoms is 2 . 76 × 10 - 10 m(the atoms are treated as point masses). If the angular speed of the molecule about the z axis is 5 . 18 × 10 12 rad / s, what is its rotational kinetic energy? Correct answer: 1 . 35925 × 10 - 20 J. Explanation: Since the distance of each atom form the z axis is d/ 2 (where d = 2 . 76 × 10 - 10 m is the separation distance between the two oxygen atoms), the moment of inertial about the z axis is I = 2 X n =1 m n r 2 n = m d 2 2 + m d 2 2 = 1 2 m d 2 , where m = 2 . 66 × 10 - 26 kg is the mass of an oxygen atom. Using the formula for the rotational kinetic energy, we obtain K R = 1 2 I ω 2 = 1 4 m d 2 ω 2 = 1 . 35925 × 10 - 20 J . 002 (part 1 of 3) 10 points Three particles of mass 8 kg, 3 kg, and 2 kg are connected by rigid rods of negligible mass lying along the y axis and are placed at 8 m, - 3 m, and - 4 m , respectively as in the figure. The system rotates about the x axis with an angular speed of 2 . 87 rad / s . Contrary to what is observed in the figure, consider the masses to be point particles. 8 m - 3 m - 4 m 2 . 87 rad / s x 8 kg 3 kg 2 kg Find the moment of inertia about the x axis. Correct answer: 571 kg m 2 . Explanation: Let : m 1 = 8 kg , m 2 = 3 kg , m 3 = 2 kg , y 1 = 8 m , y 2 = - 3 m , y 3 = - 4 m , and ω = 2 . 87 rad / s . The total rotational inertia of the system about the x axis is I = X m i r 2 i = 571 kg m 2 , where, r i = | y i | . 003 (part 2 of 3) 10 points Find the total rotational energy of the sys- tem. Correct answer: 2351 . 63 J. Explanation:
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Muraj, Hamza – Homework 23 – Due: Mar 27 2006, 4:00 am – Inst: Florin 2 Since ω = 2 . 87 rad / s , the total rotational energy is E = 1 2 I ω 2 = 1 2 (571 kg m 2 ) (2 . 87 rad / s) 2 = 2351 . 63 J . 004 (part 3 of 3) 10 points Find the linear speed of the top particle of mass 8 kg in the figure.
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