Chapter5 - Mastering Physics Assignment 2 Is available on Mastering Physics website Seven practice problems six for credit on material from chapter

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Unformatted text preview: Mastering Physics Assignment 2 Is available on Mastering Physics website Seven practice problems + six for credit on material from chapter 3 Due Wednesday, October 10 at 11 pm Penalty of 25% per day late Assignment available for practice until end of year On Campus Machines Use Firefox if problems with Internet Explorer! 1 Friday, October 12, 2007 Experiment 2: Measurement of g by free fall Week of October 15: Tutorial and test 2: ch. 4, 5 Tuesday, October 23, 7-9 pm, midterm: ch. 1-5 (20 multiple-choice questions) 2 Friday, October 12, 2007 Chapter 5: Uniform Circular Motion • Motion at constant speed in a circle • Centripetal acceleration • Banked curves • Orbital motion • Weightlessness, artificial gravity • Vertical circular motion 3 Friday, October 12, 2007 Uniform Circular Motion • An object is travelling at constant speed in a circular path. • The velocity is changing because the direction of the speed is changing and so the object is accelerated . • The period, T, of the motion is the time to go once around the circle. • For an object travelling at speed v around a circle of radius r – T = circumference speed = 2 ! r v r v 4 Friday, October 12, 2007 Centripetal Acceleration The object is accelerated toward the centre of the circle – this is the centripetal acceleration. Work out the change in velocity in a short time interval... a c Centripetal acceleration, a c = Δ v Δ t 5 Friday, October 12, 2007 Centripetal Acceleration ! v v = " radians, if time interval ! t is short Object travels v ! t in time ! t So, ! = " v v = v " t r v ! t r = " radians = centripetal acceleration a c = ! v ! t = v 2 r 6 Friday, October 12, 2007 A car is driven at a constant speed of 34 m/s (122 km/h). What is the centripetal acceleration in the two turns? Centripetal acceleration, a c = v 2 r = 34 2 33 a c = 35 . 0 m/s 2 = 3 . 6 × g = 3 . 6 g a c = 34 2 24 = 48 . 2 m/s 2 = 4 . 9 g First turn: r = 33 m Second turn, r = 24 m 7 Friday, October 12, 2007 5.7/6 : Lettuce drier: spin a container containing the lettuce, water is forced out through holes in the sides of the container. Radius = 12 cm, rotated at 2 revolutions/second. What is the centripetal acceleration of the wall of the container? Centripetal acceleration, a c = v 2 r v = 2 × 2 ! r m/s = 1 . 51 m/s a c = 1 . 51 2 . 12 = 18 . 9 m/s 2 = 1 . 9 g What is v? 8 Friday, October 12, 2007 a c Centripetal Force • the force that causes the centripetal acceleration • acts toward the centre of the circular path – in the direction of the acceleration • generated by tension in a string, gravity (planetary motion), friction (driving around a curve)... As F = ma, centripetal force is: F c = ma c = mv 2 r 9 Friday, October 12, 2007 5.C11 : A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain in place? Centripetal force is supplied by friction between the penny and turntable....
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This note was uploaded on 11/11/2011 for the course PHYSICS 32455 taught by Professor Penly during the Summer '10 term at University of Leeds.

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Chapter5 - Mastering Physics Assignment 2 Is available on Mastering Physics website Seven practice problems six for credit on material from chapter

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