Practice Questions for Final-solutions

Practice Questions for Final-solutions - markowitz(am45362...

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markowitz (am45362) – Practice Questions for Final – distler – (56295) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – ±nd all choices before answering. The solution to Problem 11 makes it sound harder than it is. The key point, as empha- sized in class, is that the net force on the block, when it’s displaced a distance, y, from its equilibrium (²oating) position is F = - k y, for some value of k. So the block oscillates, just as if it were attached to a spring . .. 001 10.0 points A large wheel is coupled to a wheel with half the diameter as shown. r 2 How does the rotational speed of the smaller wheel compare with that of the larger wheel? How do the tangential speeds at the rims compare (assuming the belt doesn’t slip)? 1. The smaller wheel has twice the rotational speed and the same tangential speed as the larger wheel. correct 2. The smaller wheel has half the rotational speed and half the tangential speed as the larger wheel. 3. The smaller wheel has four times the ro- tational speed and the same tangential speed as the larger wheel. 4. The smaller wheel has twice the rotational speed and twice the tangential speed as the larger wheel. Explanation: v = r ω The tangential speeds are equal, since the rims are in contact with the belt and have the same linear speed as the belt. The smaller wheel (with half the radius) rotates twice as fast: p 1 2 r P (2 ω ) = r ω = v 002 10.0 points A cannon ±res a 0 . 371 kg shell with initial velocity v i = 9 . 6 m / s in the direction θ = 58 above the horizontal. Δ x Δ h 9 . 6 m / s 58 Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 381 s the shell is below the straight line by some vertical distance Δ h . Find this distance Δ h in the absence of air resistance. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 711289 m. Explanation: In the absence of gravity, the shell would ²y along the straight line at constant velocity: ˆ x = t v i cos θ , ˆ y = t v i sin θ . The gravity does not a³ect the x coordinate of the shell, but it does pull its y coordinate downware at a constant acceleration a y = g , so x = t v i cos θ, y = t v i sin θ g t 2 2 .
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markowitz (am45362) – Practice Questions for Final – distler – (56295) 2 Thus, x = ˆ x but y = ˆ y 1 2 gt 2 ; in other words, the shell deviates from the straight-line path by the vertical distance Δ h = ˆ y y = g t 2 2 . Note: This result is completely indepen- dent of the initial velocity v i or angle θ of the shell. It is a simple function of the ±ight time t . Δ h = g t 2 2 = (9 . 8 m / s 2 ) (0 . 381 s) 2 2 = 0 . 711289 m .
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This note was uploaded on 02/27/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas.

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Practice Questions for Final-solutions - markowitz(am45362...

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