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Unformatted text preview: hÝ × ic×ÙÑÑ eÖ8ÓÑ eÛ Ó Ök# 9 Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 9 Problem 15 ( a ): The flywheel turns 200 revolutions in 30.0 s, which is an average rate of 400 rpm. This is the average , which is one half of the initial plus the final. As the initial rate was 500 rpm, the final rate must be 300 rpm. ( b ): After 30 seconds the flywheel lost 200 rpm. At this rate, it would take 75 s for it to lose the whole 500 rpm and come to a stop. During this time the average rate is 250 rpm, and as it takes 1.25 minutes, this comes to 313 complete revolutions. § 9 Problem 34 ( a ): We ignore the mass of the rods. We find mr 2 for each mass and add them up. For an axis through the center of the square and perpendicular to it, each mass is a distance 0 . 200 √ 2 m from the axis. Thus the moment of inertia is I = 4(0 . 200 kg) bracketleftbig (0 . 200 m) 2 + (0 . 200 m) 2 bracketrightbig = 0 . 064 kg m 2 . ( b ): Now each mass is a distance 0.200 m from the axis, so I = 4(0 . 200 kg)(0 . 200 m) 2 = 0 . 032 kg m 2 . ( c ): Now two of the masses have a distance of 0 from the axis, and two have a distance of 0 . 200 √ 2 m from the axis. Thus I = 2(0 . 200 kg) bracketleftbig (0 . 200 m) 2 + (0 . 200 m) 2 bracketrightbig = 0 . 032 kg m 2 . § 9 Problem 41: The mass of the inner disc is ρπr 2 = 23600 g = 23 . 6 kg ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
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