hw22 - Overton, Mays Homework 22 Due: Mar 25 2005, 4:00 am...

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1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points The speed oF a moving bullet can be deter- mined by allowing the bullet to pass through two rotating paper disks mounted a distance 96 . 9 cm apart on the same axle. ±rom the angular displacement 39 . 9 oF the two bul- let holes in the disks and the rotational speed 1360 rev / min oF the disks, we can determine the speed v oF the bullet. 39 . 9 v 1360 rpm 96 . 9 cm What is the speed oF the bullet? Correct answer: 198 . 171 m / s. Explanation: Let : ω = 1360 rev / min , d = 96 . 9 cm , and θ = 39 . 9 . ±rom θ = ω t the time to pass through an angle θ is t = θ ω = (39 . 9 ) (142 . 419 rad / s) π rad 180 = 0 . 00488971 s . Then the speed oF the bullet is v = d t = (96 . 9 cm) (0 . 01 m / cm) 0 . 00488971 s = 198 . 171 m / s . 002 (part 1 oF 3) 10 points A 5 . 3 kg mass is connected by a light cord to a 2 . 4 kg mass on a smooth surFace as shown. The pulley rotates about a Frictionless axle and has a moment oF inertia oF 0 . 37 kg · m 2 and a radius oF 0 . 86 m. The acceleration oF gravity is 9 . 81m / s 2 . 5 . 3kg 2 . 4kg F 1 F 2 R Note: ±igure is not drawn to scale Assume that the cord does not slip on the pulley. a) What is the acceleration oF the two masses? Correct answer: 6 . 3404 m / s 2 . Explanation: F 2 a F 1 (2 . 4kg) g a Note: ±igure is not drawn to scale Basic Concepts: Newton’s second law For m 1 and m 2 : m 1 a = m 1 g - F 1 m 2 a = F 2 Newton’s second law For the rotation oF the pulley: τ net = = F net R since F net and R are perpendicular. a
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw22 - Overton, Mays Homework 22 Due: Mar 25 2005, 4:00 am...

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