10 - homework 11 JACOBS, AMBER Due: Apr 22 2008, 1:00 am 1...

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Unformatted text preview: homework 11 JACOBS, AMBER Due: Apr 22 2008, 1:00 am 1 Question 1, chap 15, sect 6. part 1 of 3 10 points A pendulum that moves through its equilib- rium position once every 1.000 s is sometimes called a seconds pendulum. a) What is the period of any seconds pen- dulum? 1. 4.000 s 2. None of these 3. 2.000 s correct 4. 0.250 s 5. 1.000 s 6. 0.500 s Explanation: Basic Concept: A pendulum passes through its equilibrium position twice each cycle. Solution: T = 2(1 . 000 s) = 2 . 000 s Question 2, chap 15, sect 6. part 2 of 3 10 points In Cambridge, England, a seconds pendu- lum is 0.9942 m long. b) What is the free-fall acceleration in Cam- bridge? 1. None of these 2. 9 . 81236 m / s 2 correct 3. 9 . 81341 m / s 2 4. 9 . 79552 m / s 2 5. 9 . 81 m / s 2 6. 9 . 79657 m / s 2 Explanation: Basic Concept: T = 2 radicalBigg L g Solution: parenleftbigg T 2 parenrightbigg 2 = L g g 1 = L 1 parenleftbigg 2 T parenrightbigg 2 = (0 . 9942 m) parenleftbigg 2 2 s parenrightbigg 2 = 9 . 81236 m / s 2 Question 3, chap 15, sect 6. part 3 of 3 10 points In Tokyo, Japan, a seconds pendulum is 0.9926 m long. c) What is the free-fall acceleration in Tokyo? 1. None of these 2. 9 . 81 m / s 2 3. 9 . 79552 m / s 2 4. 9 . 79657 m / s 2 correct 5. 9 . 81341 m / s 2 6. 9 . 81236 m / s 2 Explanation: Solution: g 2 = L 2 parenleftbigg 2 T parenrightbigg 2 = (0 . 9926 m) parenleftbigg 2 2 s parenrightbigg 2 = 9 . 79657 m / s 2 Question 4, chap 15, sect 7. part 1 of 3 10 points homework 11 JACOBS, AMBER Due: Apr 22 2008, 1:00 am 2 A particle rotates counterclockwise in a cir- cle of radius 6 . 6 m with a constant angular speed of 18 rad / s . At t = 0, the particle has an x coordinate of 5 . 3 m and y > . x y (5 . 3 m , y ) Figure: Not drawn to scale. radius 6 . 6 m 18 rad / s Determine the x coordinate of the particle at t = 1 . 71 s . Correct answer: 6 . 59998 m (tolerance 1 %). Explanation: LET : x = 5 . 3 m , = 18 rad / s , t = 0 s , and R = 6 . 6 m , Since the amplitude of the particles mo- tion equals the radius of the circle and = 18 rad / s , we have x = A cos( t + ) = (6 . 6 m) cos bracketleftBig (18 rad / s) t + bracketrightBig . We can find using the initial condition that x = 5 . 3 m at t = 0 (5 . 3 m) = (6 . 6 m) cos(0 + ) , which implies = arccos x R = arccos (5 . 3 m) (6 . 6 m) = 36 . 5796 = 0 . 638433 rad . Therefore, at time t = 1 . 71 s , the x coordinate of the particle is x = R cos bracketleftBig t + bracketrightBig = (6 . 6 m) cos bracketleftBig (18 rad / s) (1 . 71 s) + (0 . 638433 rad) bracketrightBig = 6 . 59998 m . Note: The angles in the cosine are in radians. Question 5, chap 15, sect 7....
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This note was uploaded on 04/28/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.

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10 - homework 11 JACOBS, AMBER Due: Apr 22 2008, 1:00 am 1...

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