12 - obennoskey (blo242) hw12 Demkov (59910) This print-out...

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obennoskey (blo242) – hw12 – Demkov – (59910) 1 This print-out should have 44 questions. Multiple-choice questions may continue on the next column or page – fnd 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 > 0 . 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 = 0 . 85 m , ω = 17 rad / s , t 0 = 0 s , and R = 6 . 4 m , Since the amplitude oF the particle’s mo- tion equals the radius oF the circle and ω = 17 rad / s , we have x = A cos( ω t + φ ) = (6 . 4 m) cos b (17 rad / s) t + φ B . We can fnd φ using the initial condition that x 0 = 0 . 85 m at t = 0 (0 . 85 m) = (6 . 4 m) cos(0 + φ ) , which implies φ = arccos x 0 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 b ω t + φ B = (6 . 4 m) cos b (17 rad / s) (2 . 87 s) + (1 . 43759 rad) B = 6 . 39541 m . Note: The angles in the cosine are in radians. 002 (part 2 oF 3) 10.0 points ±ind the x component oF the particle’s veloc- ity at t = 2 . 87 s. Correct answer: 4 . 12169 m / s. Explanation: Di²erentiating the Function x ( t ) with re- spect to t , we fnd the x component oF the particle’s velocity at any time t v x = dx dt = - ω A sin( ω t + φ ) , so at t = 2 . 87 s , the argument oF the sine is φ 2 ω t + φ = (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 = - ω R sin( φ 2 ) = - (17 rad / s) (6 . 4 m) sin(50 . 2276 rad) = 4 . 12169 m / s . 003 (part 3 oF 3) 10.0 points ±ind the x component oF the particle’s accel- eration at t = 2 . 87 s. Correct answer: - 1848 . 27 m / s 2 . Explanation: Di²erentiating the x component oF the par- ticle’s velocity with respect to t , we fnd the
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obennoskey (blo242) – hw12 – Demkov – (59910) 2 x component of the particle’s acceleration at any time t a x = dv x dt = - ω 2 A cos( ω t + φ ) , so at t = 2 . 87 s , the x component of the particle’s acceleration is a x = - ω 2 R cos φ 2 = - (17 rad / s) 2 (6 . 4 m) cos(50 . 2276 rad) = - 1848 . 27 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. the reciprocal of its speed. 4. inversely proportional to its accelera- tion. 5. proportional to its amplitude. Explanation: The period of simple harmonic motion is the same regardless of the amplitude.
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12 - obennoskey (blo242) hw12 Demkov (59910) This print-out...

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