Homework 10 Solutions

# Homework 10 Solutions - homework 10 – ALIBHAI ZAHID –...

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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 work-energy principle, we have Work = Δ K = 1 2 I ω 2 = 1 2 I (2 π ) 2 f 2 . So, W = 1 2 (17 . 5 kg · m 2 ) × 4 π 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 disk-shaped flywheel is attached to an electric motor, which brings the flywheel’s 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 Iω 2 = 1 2 parenleftbigg 1 2 MR 2 parenrightbigg ω 2 Given: R = 1 . 75 m M = 571 . 5 kg ω = 1721 . 0 rev / min Solution: KE = 1 4 MR 2 ω 2 = 1 4 (571 . 5 kg)(1 . 75 m) 2 · (1721 rev / min) 2 parenleftbigg 1 min 60 s parenrightbigg 2 · parenleftbigg 2 π 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 Δ t = KE Δ t Given: P = 7311 W Solution: Δ 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 ◦ ω 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 θ = 33 ◦ . F 1 F 3 F 2 M θ ω a b The total torque is τ = 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 τ bardbl = 3 . 512 N m ....
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Homework 10 Solutions - homework 10 – ALIBHAI ZAHID –...

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