solution_05pdf - yoon(jy4326 – HW05 – markert –(58840...

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Unformatted text preview: yoon (jy4326) – HW05 – markert – (58840) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the following situations. A) An object moves with uniform circular motion. B) An object travels as a projectile in a gravitational field with negligible air re- sistance. C) An object moves in a straight line at con- stant speed. In which of the situations would the object be accelerated? 1. All exhibit acceleration. 2. A and B only correct 3. B and C only 4. A and C only 5. A only 6. None exhibits acceleration. 7. B only 8. C only Explanation: A) The direction of the velocity constantly changes; the centripetal acceleration is di- rected toward the center of the motion. B) The projectile undergoes gratitational acceleration. C) The velocity of the object (its direction and magnitude) is unchanged, so it is not accelerated. 002 10.0 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the figure is 3.13 m/s. The acceleration of gravity is 9.8 m/s 2 . v At the position shown in the figure, which of the labeled arrows best represents the di- rection of the acceleration of the mass? 1. 2. 3. correct 4. 5. 6. 7. The mass is traveling at a constant veloc- ity, therefore it has no acceleration. 8. 9. Explanation: The magnitude of the centripetal is a r = v 2 r = (3 . 13 m / s) 2 1 m ≈ 9 . 8 m / s 2 . The centripetal acceleration is inward- ˆ r and gravity is down- ˆ k . yoon (jy4326) – HW05 – markert – (58840) 2 a a r g 003 10.0 points Calculate the speed at the edge of a disc of radius 2 . 5 cm that rotates at the rate of 10 rev / s. Correct answer: 1 . 5708 m / s. Explanation: Let : R = 2 . 5 cm and f = 10 rev / s . T = 1 f = 1 10 rev / s = 0 . 1 s . v = D T = 2 π R T = 2 π (2 . 5 cm) . 1 s × 1 m 100 cm = 1 . 5708 m / s . 004 (part 1 of 2) 10.0 points A dog sits 1.7 m from the center of a merry- go-round. a) If the dog undergoes a 1.9 m/s 2 cen- tripetal acceleration, what is the dog’s linear speed? Correct answer: 1 . 79722 m / s. Explanation: Basic Concept: a c = v t 2 r Given: a c = 1 . 9 m / s 2 r = 1 . 7 m Solution: v t = √ a c r = radicalBig (1 . 9 m / s 2 )(1 . 7 m) = 1 . 79722 m / s 005 (part 2 of 2) 10.0 points b) What is the angular speed of the merry-go- round? Correct answer: 1 . 05719 rad / s. Explanation: Basic Concept: a c = rω 2 Given: a c = 1 . 9 m / s 2 r = 1 . 7 m Solution: ω = radicalbigg a c r = radicalbigg 1 . 9 m / s 2 1 . 7 m = 1 . 05719 rad / s 006 10.0 points A ball rolls around a circular wall, as shown in the figure below. The wall ends at point X ....
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solution_05pdf - yoon(jy4326 – HW05 – markert –(58840...

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