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Unformatted text preview: bello (rtb473) hw12 Demkov (59910) 1 This printout should have 44 questions. Multiplechoice 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.
 Spring '07
 Swinney
 mechanics

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