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Unformatted text preview: Alvarado, Patrick – Homework 4 – Due: Oct 5 2006, 10:00 am – Inst: Andrei Sirenko 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A flywheel of radius 0 . 4 m and moment of inertia of 19 . 8 kg · m 2 rotates initially at a rate of 4 . 2 revolutions / sec. If a force of 3 . 2 N is applied tangentially to the flywheel to slow it down, how much work will be done by this force in bringing the flywheel to a stop? 1. 174 . 636 J 2. 1 . 28 J 3. 13788 . 7 J 4. 6894 . 35 J correct 5. 83 . 16 J 6. 522 . 51 J 7. 349 . 272 J Explanation: Use workenergy principle, we have W = Δ K = 1 2 I ω 2 = 1 2 I (2 π ) 2 f 2 . So, W = 1 2 (19 . 8 kg · m 2 ) × 4 π 2 (4 . 2 revolutions / sec) 2 = 6894 . 35 J . keywords: 002 (part 1 of 3) 10 points A flywheel in the form of a heavy circular disk of diameter 0 . 334 m and mass 270 kg is mounted on a frictionless bearing. A motor connected to the flywheel accelerates it from rest to 1360 rev / min. What is the moment of inertia of the fly wheel? Correct answer: 3 . 76501 kg m 2 . Explanation: The moment of inertia is I = 1 2 m r 2 = 1 2 m µ d 2 ¶ 2 = 1 2 (270 kg) µ . 334 m 2 ¶ 2 = 3 . 76501 kg m 2 . 003 (part 2 of 3) 10 points How much work is done on it during this acceleration? Correct answer: 38183 . 2 J. Explanation: The work done by the motor to accelerate the flywheel is equal to the final kinetic energy of the flywheel W = K f K i = 1 2 I ω 2 1 = 1 2 (3 . 76501 kg m 2 ) (142 . 419 rad / s) 2 = 38183 . 2 J . 004 (part 3 of 3) 10 points After 1360 rev / min is achieved, the motor is disengaged. A friction brake is used to slow the rotational rate to 942 rev / min. What is the magnitude of the energy dissi pated as heat from the friction brake? Correct answer: 19864 . 4 J. Explanation: The energy loss to friction is equal to the difference between the final and the initial kinetic energy of the flywheel Δ E = K f K i = 1 2 I ω 2 2 K i = 1 2 (3 . 76501 kg m 2 ) (98 . 646 rad / s) 2 38183 . 2 J = 19864 . 4 J . Alvarado, Patrick – Homework 4 – Due: Oct 5 2006, 10:00 am – Inst: Andrei Sirenko 2 Or  Δ E  = 19864 . 4 J keywords: 005 (part 1 of 1) 10 points An automobile engine develops a torque of 290 Nm and is rotating at a speed of 4000 rev / min. What horsepower does the engine generate? 1 hp = 746 W Correct answer: 162 . 835 hp. Explanation: ω = (4000 rev / min) µ 2 π rad 1 rev ¶ µ 1 min 60 sec ¶ = 418 . 879 rad / s . We have P = τ ω = (290 Nm) (418 . 879 rad / s) = (121475 W) µ 1 hp 746 W ¶ = 162 . 835 hp . keywords: 006 (part 1 of 1) 10 points A string is wound around a uniform disc of radius R = 0 . 39 m and mass M = 2 . 5 kg....
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This note was uploaded on 09/04/2008 for the course PHYS 106 taught by Professor Opyt during the Spring '08 term at NJIT.
 Spring '08
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