hw21 - alexander (jra2623) homework 21 Turner (92510) This...

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alexander (jra2623) – homework 21 – Turner – (92510) 1 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 . 7 kg is 7 m long. The rod is pivoted about a horizontal, Frictionless pin at the end oF a thin extension (oF negligi- ble mass) a distance 7 m From the center oF mass oF the rod. Initially the rod makes an angle oF 54 with the horizontal. The rod is released From rest at an angle oF 54 with the horizontal, as shown in the fgure. 7 m 7 m 2 . 7 kg O 54 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: 1 . 44603 rad / s. Explanation: Let : = 7 m , θ = 54 , and m = 2 . 7 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 54 13 (7 m) = 1 . 44603 rad / s . keywords: 002 10.0 points A circular-shaped object oF mass 11 kg has an inner radius oF 15 cm and an outer radius oF 28 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 20 below the horizontal. 12 N 13 N 24 N 20 ω
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alexander (jra2623) – homework 21 – Turner – (92510) 2 Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 3 . 4 N · m. Explanation: Let : a = 15 cm = 0 . 15 m , b = 28 cm = 0 . 28 m , F 1 = 12 N , F 2 = 24 N , F 3 = 13 N , and θ = 20 . F 1 F 3 F 2 θ ω The total torque is τ = a F 2 - bF 1 - bF 3 = (0 . 15 m) (24 N) - (0 . 28 m) (12 N + 13 N) = - 3 . 4 N · m , with a magnitude of 3 . 4 N · m . 003
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.

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hw21 - alexander (jra2623) homework 21 Turner (92510) This...

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