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Unformatted text preview: adamkiewicz (ja28724) – Rotation – Morozova – (11102) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A large wheel is coupled to a wheel with half the diameter as shown. r 2 r 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 half the rotational speed and half the tangential speed as the larger wheel. 2. The smaller wheel has twice the rotational speed and the same tangential speed as the larger wheel. correct 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: parenleftbigg 1 2 r parenrightbigg (2 ω ) = r ω = v 002 10.0 points Harry and Sue cycle at the same speed. The tires on Harry’s bike have a larger diameter than those on Sue’s bike. Which tires have the greater rotational speed? 1. Harry’s tires 2. Sue’s tires correct 3. The rotational speeds are the same. 4. It depends on the speed. Explanation: v = r ω . Tires with a smaller radius needs a larger rotational speed to obtain the same linear speed. 003 10.0 points The speed of a moving bullet can be deter mined by allowing the bullet to pass through two rotating paper disks mounted a distance 90 cm apart on the same axle. From the angular displacement 36 . 6 ◦ of the two bul let holes in the disks and the rotational speed 1056 rev / min of the disks, we can determine the speed of the bullet. 36 . 6 ◦ v 1056 rev / min 90 cm What is the speed of the bullet? Correct answer: 155 . 803 m / s. Explanation: Let : ω = 1056 rev / min , d = 90 cm , and θ = 36 . 6 ◦ . adamkiewicz (ja28724) – Rotation – Morozova – (11102) 2 θ = ω t t = θ ω , so the speed of the bullet is v = d t = dω θ = (90 cm) (1056 rev / min) 36 . 6 ◦ × 360 ◦ 1 rev 1 m 100 cm 1 min 60 s = 155 . 803 m / s ....
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This note was uploaded on 04/14/2009 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.
 Spring '08
 moro
 Physics

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