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# solution_pdf21 - reddy(ar38357 – Rotational Kinetic...

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Unformatted text preview: reddy (ar38357) – Rotational Kinetic Energy – clancy – (SCI411-2) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0 . 25 m power sawblade has a mass of 0 . 5 kg distributed uniformly as in a disc. What is the rotational kinetic energy at 3600 rpm (revolutions per minute)? 1. 3 . 6 × 10 1 J 2. 2 . 6 × 10 3 J 3. 8 . 9 × 10 1 J 4. 1 . 1 × 10 3 J correct 5. 9 . 1 × 10 2 J Explanation: Let : r = 0 . 25 m , m = 0 . 5 kg , and ω = 3600 rpm . The moment of inertia of a disk of uniform density is I = 1 2 mr 2 . The rotational kinetic energy of the saw- blade is then K rot = 1 2 I ω 2 = 1 2 parenleftbigg 1 2 mr 2 parenrightbigg ω 2 = 1 4 (0 . 5 kg)(0 . 25 m) 2 (3600 rpm) 2 × parenleftbigg 2 π rev parenrightbigg 2 parenleftbigg 1 min 60 s parenrightbigg 2 = 1110 . 33 J . 002 10.0 points The net work done in accelerating a propeller from rest to an angular speed of 214 rad/s is 2892 . 2 J....
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solution_pdf21 - reddy(ar38357 – Rotational Kinetic...

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