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Unformatted text preview: sanchez (ds28677) oldhomework 30 Turner (58220) 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 6 . 4 m with a constant angular speed of 17 rad / s . At t = 0, the particle has an x coordinate of 0 . 85 m and y > . x y (0 . 85 m ,y ) Figure: Not drawn to scale. radius 6 . 4 m 17 rad / s Determine the x coordinate of the particle at t = 2 . 87 s . Correct answer: 6 . 39541 m. Explanation: Let : x = 0 . 85 m , = 17 rad / s , t = 0 s , and R = 6 . 4 m , Since the amplitude of the particles mo tion equals the radius of the circle and = 17 rad / s , we have x = A cos( t + ) = (6 . 4 m) cos bracketleftBig (17 rad / s) t + bracketrightBig . We can find using the initial condition that x = 0 . 85 m at t = 0 (0 . 85 m) = (6 . 4 m) cos(0 + ) , which implies = arccos x R = arccos (0 . 85 m) (6 . 4 m) = 82 . 3679 = 1 . 43759 rad . Therefore, at time t = 2 . 87 s , the x coordinate of the particle is x = R cos bracketleftBig t + bracketrightBig = (6 . 4 m) cos bracketleftBig (17 rad / s) (2 . 87 s) + (1 . 43759 rad) bracketrightBig = 6 . 39541 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 = 2 . 87 s. Correct answer: 4 . 12169 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 = 2 . 87 s , the argument of the sine is 2 t + = (17 rad / s) (2 . 87 s) + (1 . 43759 rad) = 50 . 2276 rad , and the x component of the velocity of the particle is v x = R sin( 2 ) = (17 rad / s) (6 . 4 m) sin(50 . 2276 rad) = 4 . 12169 m / s . 003 (part 3 of 3) 10.0 points Find the x component of the particles accel eration at t = 2 . 87 s. Correct answer: 1848 . 27 m / s 2 . Explanation: sanchez (ds28677) oldhomework 30 Turner (58220) 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 = 2 . 87 s , the x component of the particles acceleration is a x = 2 R cos 2 = (17 rad / s) 2 (6 . 4 m) cos(50 . 2276 rad) = 1848 . 27 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 3 5 L from the lower end parenrightbigg , as shown....
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 Spring '08
 Turner
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