fl - bello (rtb473) hw12 Demkov (59910) 1 This print-out...

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Unformatted text preview: bello (rtb473) hw12 Demkov (59910) 1 This print-out should have 44 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points A particle rotates counterclockwise in a circle of radius 5 . 9 m with a constant angular speed of 7 rad / s . At t = 0, the particle has an x coordinate of 4 . 5 m and y > . x y (4 . 5 m ,y ) Figure: Not drawn to scale. radius 5 . 9 m 7 rad / s Determine the x coordinate of the particle at t = 1 . 14 s . Correct answer:- 4 . 35108 m. Explanation: Let : x = 4 . 5 m , = 7 rad / s , t = 0 s , and R = 5 . 9 m , Since the amplitude of the particles mo- tion equals the radius of the circle and = 7 rad / s , we have x = A cos( t + ) = (5 . 9 m) cos bracketleftBig (7 rad / s) t + bracketrightBig . We can find using the initial condition that x = 4 . 5 m at t = 0 (4 . 5 m) = (5 . 9 m) cos(0 + ) , which implies = arccos x R = arccos (4 . 5 m) (5 . 9 m) = 40 . 2962 = 0 . 7033 rad . Therefore, at time t = 1 . 14 s , the x coordinate of the particle is x = R cos bracketleftBig t + bracketrightBig = (5 . 9 m) cos bracketleftBig (7 rad / s) (1 . 14 s) + (0 . 7033 rad) bracketrightBig =- 4 . 35108 m . Note: The angles in the cosine are in radians. 002 (part 2 of 3) 10.0 points Find the x component of the particles veloc- ity at t = 1 . 14 s. Correct answer:- 27 . 8931 m / s. Explanation: Differentiating the function x ( t ) with re- spect to t , we find the x component of the particles velocity at any time t v x = dx dt =- A sin( t + ) , so at t = 1 . 14 s , the argument of the sine is 2 t + = (7 rad / s) (1 . 14 s) + (0 . 7033 rad) = 8 . 6833 rad , and the x component of the velocity of the particle is v x =- R sin( 2 ) =- (7 rad / s) (5 . 9 m) sin(8 . 6833 rad) =- 27 . 8931 m / s . 003 (part 3 of 3) 10.0 points Find the x component of the particles accel- eration at t = 1 . 14 s. Correct answer: 213 . 203 m / s 2 . Explanation: Differentiating the x component of the par- ticles velocity with respect to t , we find the bello (rtb473) hw12 Demkov (59910) 2 x component of the particles acceleration at any time t a x = dv x dt =- 2 A cos( t + ) , so at t = 1 . 14 s , the x component of the particles acceleration is a x =- 2 R cos 2 =- (7 rad / s) 2 (5 . 9 m) cos(8 . 6833 rad) = 213 . 203 m / s 2 . 004 10.0 points When a body executes simple harmonic mo- tion, its period is 1. proportional to its acceleration. 2. independent of its amplitude. correct 3. inversely proportional to its accelera- tion. 4. the reciprocal of its speed. 5. proportional to its amplitude....
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This note was uploaded on 12/28/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas at Austin.

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