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Unformatted text preview: sousse (res2468) – HW 8 – Kleinman – (58225) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This HW covers Ch. 13. 001 10.0 points A simple pendulum consists of a 4.8 kg point mass hanging at the end of a 2.1 m long light string that is connected to a pivot point. Calculate the magnitude of the torque (due to the force of gravity) around this pivot point when the string makes a 5 . 4 ◦ angle with the vertical. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 9 . 30588 N · m. Explanation: Let : m = 4 . 8 kg , d = 2 . 1 m , θ = 5 . 4 ◦ , and g = 9 . 81 m / s 2 . Solution: τ = F d sin θ = mg d sin θ = (4 . 8 kg) ( 9 . 81 m / s 2 ) (2 . 1 m) sin5 . 4 ◦ = 9 . 30588 N · m . 002 10.0 points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and per pendicular to the page. Four forces are ex erted tangentially to the rims of the wheels, as shown below. F 2 F F F 2 R 3 R What is the magnitude of the net torque on the system about the axis? 1. τ = 14 F R 2. τ = F R 3. τ = 5 F R 4. τ = 2 F R correct 5. τ = 0 Explanation: The three forces F apply counterclockwise torques while the other force 2 F applies a clockwise torque, so τ = summationdisplay F i R i = ( − 2 F ) (3 R ) + F (3 R ) + F (3 R ) + F (2 R ) = 2 F R . 003 10.0 points A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum speed of 3820 rpm by an electric motor. The flywheel is a solid cylinder of mass 453 kg and diameter 1 . 57 m. If the bus requires an average power of 9 . 26 kW, how long does the flywheel rotate if all of the energy of the flywheel is used by the bus? Correct answer: 1206 . 01 s. Explanation: The maximum energy of the flywheel is E = I ω 2 2 = 1 2 parenleftbigg M R 2 2 parenrightbigg ω 2 = 1 2 parenleftbigg 453 kg (1 . 57 m) 2 2 parenrightbigg (400 . 03 rad) 2 = 1 . 11677 × 10 7 J . t = E P = 1 . 11677 × 10 7 J 9260 W = 1206 . 01 s . sousse (res2468) – HW 8 – Kleinman – (58225) 2 keywords: 004 10.0 points Hint: The kinetic energy of a rolling object equals the kinetic energy from the linear mo tion of its center of mass plus the kinetic energy from from the rotation of the object. A bowling ball has a mass of 6 . 8 kg, a moment of inertia of 0 . 4352 kg · m 2 , and a radius of 0 . 4 m. It rolls along the lane without slipping at a linear speed of 3 . 4 m / s. What is the kinetic energy of the rolling ball? Correct answer: 55 . 0256 J. Explanation: Let : v = 3 . 4 m / s , r = 0 . 4 m , m = 6 . 8 kg , and I = 0 . 4352 kg · m 2 ....
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This note was uploaded on 04/08/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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