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Unformatted text preview: Bodet (ngb299) – hw 22 optional – turner – (57340) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uniform rod of mass 1 . 8 kg is 7 m long. The rod is pivoted about a horizontal, frictionless pin at the end of a thin extension (of negligible mass) a distance 7 m from the center of mass of the rod. The rod is released from rest at an initial angle of 69 ◦ with respect to the horizontal, as shown. 7 m 7 m 1 . 8 kg O 69 ◦ What is the angular speed of the rod at the instant the rod is in a horizontal position? The acceleration due to gravity is 9 . 8 m / s 2 and the moment of inertia of the rod about its center of mass is I cm = 1 12 m 2 . Correct answer: 1 . 55337 rad / s. Explanation: Let : = 7 m , θ = 69 ◦ , and m = 1 . 8 kg . Rotational kinetic energy is K R = 1 2 I ω 2 and gravitational kinetic energy is K trans = mg d. The inertia of the system is I = I CM + md 2 = 1 12 m 2 + m 2 = 13 12 m 2 . Since the rod is uniform, its center of mass is located a distance from the pivot. The vertical height of the center of mass above the horizontal is sin θ . Using conservation of energy, K i + U i = K f + U f K f = U i 1 2 I ω 2 = mg sin θ 13 24 m 2 ω 2 = mg sin θ ω 2 = 24 13 g sin θ ω = 24 g sin θ 13 = 24(9 . 8 m / s 2 ) sin 69 ◦ 13(7 m) = 1 . 55337 rad / s . 002 10.0 points The hour and minute hands of a tower clock like Big Ben in London are 2 . 78 m and 4 . 57 m long and have masses of 61 . 1 kg and 94 kg, respectively. Calculate the total rotational kinetic en ergy of the two hands about the axis of rota tion. Model the hands as long thin rods. Correct answer: 0 . 000998363 J. Explanation: Let : L h = 2 . 78 m , L m = 4 . 57 m , M h = 61 . 1 kg , and M m = 94 kg . The hour hand makes 1 rev in 24 hours and the minute hand in 60 minutes. The moment of inertia of a thin rod about an axis through one end is I = M L 2 3 , Bodet (ngb299) – hw 22 optional – turner – (57340) 2 so the total rotational kinetic energy is K tot = I h ω 2 h 2 + I m ω 2 m 2 = m h L 2 h 6 ω 2 h + m m L 2 m 6 ω 2 m = (61 . 1 kg) (2 . 78 m) 2 6 × 2 π rad 12 h 2 1 h 3600 s 2 + (94 kg) (4 . 57 m) 2 6 × 2 π rad 60 min 2 1 min 60 s 2 = . 000998363 J ....
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Mass

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