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Unformatted text preview: hernandez (ejh742) oldhomework 30 Turner (58120) 1 This printout should have 14 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 4 . 8 m with a constant angular speed of 13 rad / s . At t = 0, the particle has an x coordinate of 1 . 4 m and y > . x y (1 . 4 m ,y ) Figure: Not drawn to scale. radius 4 . 8 m 13 rad / s Determine the x coordinate of the particle at t = 1 . 44 s . Correct answer: 1 . 98143 m. Explanation: Let : x = 1 . 4 m , = 13 rad / s , t = 0 s , and R = 4 . 8 m , Since the amplitude of the particles mo tion equals the radius of the circle and = 13 rad / s , we have x = A cos( t + ) = (4 . 8 m) cos bracketleftBig (13 rad / s) t + bracketrightBig . We can find using the initial condition that x = 1 . 4 m at t = 0 (1 . 4 m) = (4 . 8 m) cos(0 + ) , which implies = arccos x R = arccos (1 . 4 m) (4 . 8 m) = 73 . 0423 = 1 . 27483 rad . Therefore, at time t = 1 . 44 s , the x coordinate of the particle is x = R cos bracketleftBig t + bracketrightBig = (4 . 8 m) cos bracketleftBig (13 rad / s) (1 . 44 s) + (1 . 27483 rad) bracketrightBig = 1 . 98143 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 . 44 s. Correct answer: 56 . 8353 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 . 44 s , the argument of the sine is 2 t + = (13 rad / s) (1 . 44 s) + (1 . 27483 rad) = 19 . 9948 rad , and the x component of the velocity of the particle is v x = R sin( 2 ) = (13 rad / s) (4 . 8 m) sin(19 . 9948 rad) = 56 . 8353 m / s . 003 (part 3 of 3) 10.0 points Find the x component of the particles accel eration at t = 1 . 44 s. Correct answer: 334 . 862 m / s 2 . Explanation: hernandez (ejh742) oldhomework 30 Turner (58120) 2 Differentiating the x component of the par ticles velocity with respect to t , we find the x component of the particles acceleration at any time t a x = dv x dt = 2 A cos( t + ) , so at t = 1 . 44 s , the x component of the particles acceleration is a x = 2 R cos 2 = (13 rad / s) 2 (4 . 8 m) cos(19 . 9948 rad) = 334 . 862 m / s 2 . 004 (part 1 of 3) 10.0 points Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O parenleftbigg 5 8 L from the lower end parenrightbigg , as shown....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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