hw25 - homework 25 Turner (58185) This print-out should...

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This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A uniForm rod oF mass 2 . 1 kg is 17 m long. The rod is pivoted about a horizontal, Frictionless pin at the end oF a thin extension (oF negligible mass) a distance 17 m From the center oF mass oF the rod. Initially the rod makes an angle oF 60 with the horizontal. The rod is released From rest at an angle oF 60 with the horizontal, as shown in the fgure. 17 m 17 m 2 . 1 kg O 60 What is the angular speed oF the rod at the instant the rod is in a horizontal position? The acceleration oF 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 mℓ 2 . Correct answer: 0 . 960037 rad / s. Explanation: Let : = 17 m , θ = 60 , and m = 2 . 1 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 mℓ 2 + mℓ 2 = 13 12 mℓ 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 ℓ sin θ 13 24 mℓ 2 ω 2 = mg ℓ sin θ ω 2 = 24 13 g sin θ ω = r 24 g sin θ 13 = R 24 (9 . 8 m / s 2 ) sin 60 13 (17 m) = 0 . 960037 rad / s . keywords: 002 10.0 points A circular-shaped object has an inner radius oF 7 cm and an outer radius oF 30 cm. Three Forces (acting perpendicular to the axis oF rotation) oF magnitudes 12 N, 24 N, and 13 N act on the object, as shown. The Force oF magnitude 24 N acts 40 below the horizontal. 12 N 13 N 24 N 15 kg 40 ω 0 . 07 m 0 . 3 m ±ind the magnitude oF the net torque on the wheel about the axle through the center oF the object. Correct answer: 0 . 0582 N · m. homework 25 – Turner – (58185) 1
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hinojosa (jlh3938) – homework 25 – Turner – (58185) 2 Explanation: Let : a = 7 cm = 0 . 07 m , b = 30 cm = 0 . 3 m , F 1 = 12 N , F 2 = 24 N , F 3 = 13 N , and θ = 40 . F 1 F 3 F 2 M θ ω a b The total torque is τ = a F 2 b F 1 b F 3 = (0 . 07 m) (24 N) (0 . 3 m) (12 N + 13 N) = 0 . 0582 N · m b v τ b = 0 . 0582 N · m . 003
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw25 - homework 25 Turner (58185) This print-out should...

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