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Unformatted text preview: homework 10 ALIBHAI, ZAHID Due: Apr 4 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A flywheel of radius 0 . 53 m and moment of inertia of 17 . 5 kg m 2 rotates initially at a rate of 7 . 3 revolutions / sec. If a force of 3 . 1 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. 1 . 643 J 2. 18408 . 3 J correct 3. 127 . 75 J 4. 466 . 287 J 5. 932 . 575 J 6. 36816 . 6 J 7. 802 . 677 J Explanation: Use workenergy principle, we have Work = K = 1 2 I 2 = 1 2 I (2 ) 2 f 2 . So, W = 1 2 (17 . 5 kg m 2 ) 4 2 (7 . 3 revolutions / sec) 2 = 18408 . 3 J . Question 2 part 1 of 2 10 points A car is designed to get its energy from a rotating flywheel with a radius of 1 . 75 m and a mass of 571 . 5 kg. Before a trip, the diskshaped flywheel is attached to an electric motor, which brings the flywheels rotational speed up to 1721 . 0 rev / min. a) Find the kinetic energy stored in the flywheel. Correct answer: 1 . 42119 10 7 J (tolerance 1 %). Explanation: Basic Concept: KE = 1 2 I 2 = 1 2 parenleftbigg 1 2 MR 2 parenrightbigg 2 Given: R = 1 . 75 m M = 571 . 5 kg = 1721 . 0 rev / min Solution: KE = 1 4 MR 2 2 = 1 4 (571 . 5 kg)(1 . 75 m) 2 (1721 rev / min) 2 parenleftbigg 1 min 60 s parenrightbigg 2 parenleftbigg 2 rad 1 rev parenrightbigg 2 = 1 . 42119 10 7 J Question 3 part 2 of 2 10 points b) If the flywheel is to supply as much energy to the car as a 7311 W motor would, find the length of time the car can run before the flywheel has to be brought back up to speed again. Correct answer: 1943 . 9 s (tolerance 1 %). Explanation: Basic Concept: P = W t = KE t Given: P = 7311 W Solution: t = KE P = 1 . 42119 10 7 J 7311 W = 1943 . 9 s homework 10 ALIBHAI, ZAHID Due: Apr 4 2007, 4:00 am 2 Question 4 part 1 of 1 10 points Given: A circular shaped object with an inner radius of 8 . 3 cm and an outer radius of 21 cm. There are three forces (acting perpen dicular to the axis of rotation) whose magni tudes are 12 N, 26 N, and 15 N acting on the object, as shown in the figure. The force of magnitude 26 N is 33 below horizontal. 12 N 15 N 26 N 11 kg 33 8 . 3 cm 21 cm Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 3 . 512 N m (tolerance 1 %). Explanation: Let : a = 8 . 3 cm , b = 21 cm , F 1 = 12 N , F 2 = 26 N , F 3 = 15 N , and = 33 . F 1 F 3 F 2 M a b The total torque is = aF 2 bF 1 bF 3 = (8 . 3 cm) (26 N) (21 cm) bracketleftBig (12 N) + (15 N) bracketrightBig = 3 . 512 N m bardbl vector bardbl = 3 . 512 N m ....
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 Spring '08
 Turner
 Force, Inertia, Work

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