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Unformatted text preview: This print-out should have 14 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 8 . 6 m with a constant angular speed of 14 rad / s . At t = 0, the particle has an x coordinate of 4 m and y > . x y (4 m ,y ) Figure: Not drawn to scale. radius 8 . 6 m 14 rad / s Determine the x coordinate of the particle at t = 2 . 66 s . Correct answer: 6 . 95957 m. Explanation: Let : x = 4 m , = 14 rad / s , t = 0 s , and R = 8 . 6 m , Since the amplitude of the particles mo- tion equals the radius of the circle and = 14 rad / s , we have x = A cos( t + ) = (8 . 6 m) cos bracketleftBig (14 rad / s) t + bracketrightBig . We can find using the initial condition that x = 4 m at t = 0 (4 m) = (8 . 6 m) cos(0 + ) , which implies = arccos x R = arccos (4 m) (8 . 6 m) = 62 . 2823 = 1 . 08703 rad . Therefore, at time t = 2 . 66 s , the x coordinate of the particle is x = R cos bracketleftBig t + bracketrightBig = (8 . 6 m) cos bracketleftBig (14 rad / s) (2 . 66 s) + (1 . 08703 rad) bracketrightBig = 6 . 95957 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 . 66 s. Correct answer:- 70 . 7304 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 . 66 s , the argument of the sine is 2 t + = (14 rad / s) (2 . 66 s) + (1 . 08703 rad) = 38 . 327 rad , and the x component of the velocity of the particle is v x =- R sin( 2 ) =- (14 rad / s) (8 . 6 m) sin(38 . 327 rad) =- 70 . 7304 m / s . 003 (part 3 of 3) 10.0 points Find the x component of the particles accel- eration at t = 2 . 66 s. Correct answer:- 1364 . 08 m / s 2 . Explanation: Differentiating the x component of the par- ticles velocity with respect to t , we find the oldhomework 32 Turner (58185) 1 hinojosa (jlh3938) oldhomework 32 Turner (58185) 2 x component of the particles acceleration at any time t a x = dv x dt =- 2 A cos( t + ) , so at t = 2 . 66 s , the x component of the particles acceleration is a x =- 2 R cos 2 =- (14 rad / s) 2 (8 . 6 m) cos(38 . 327 rad) =- 1364 . 08 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 4 L from the lower end parenrightbigg , as shown....
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oldhw32 - This print-out should have 14 questions....

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