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hw9 solutions - homework 09 VENNES ROSS Due Nov 2 2007 1:00...

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homework 09 – VENNES, ROSS – Due: Nov 2 2007, 1:00 am 1 Question 1, chap 12, sect 2. part 1 of 2 10 points A record has an angular speed of 44 . 2 rev / min. What is its angular speed? Correct answer: 4 . 62862 rad / s (tolerance ± 1 %). Explanation: 1 rev = 6 . 28319 rad, and 1 min = 60 s , Therefore, ω 1 = (44 . 2 rev / min) parenleftbigg 2 π rad rev parenrightbigg parenleftbigg 1 min 60 s parenrightbigg = 4 . 62862 rad / s . Question 2, chap 12, sect 2. part 2 of 2 10 points Through what angle, in radians, does it rotate in 1 . 41 s? Correct answer: 6 . 52635 rad (tolerance ± 1 %). Explanation: θ = ω t = (4 . 62862 rad / s) (1 . 41 s) = 6 . 52635 rad . Question 3, chap 12, sect 2. part 1 of 2 10 points A racing car travels on a circular track of radius 335 m. The car moves with a constant linear speed of 42 . 4 m / s. Find its angular speed. Correct answer: 0 . 126567 rad / s (tolerance ± 1 %). Explanation: The linear speed v and the angular speed ω are related by, v = R ω ω = v R . Question 4, chap 12, sect 2. part 2 of 2 10 points Find the magnitude of its acceleration. Correct answer: 5 . 36645 m / s 2 (tolerance ± 1 %). Explanation: If the car is moving at a constant speed, there is no tangential acceleration, thus the acceleration is purely radial, a r = v 2 R . Question 5, chap 12, sect 2. part 1 of 1 10 points A woman passes through a revolving door with a tangential speed of 1.8 m/s. If she is 0.78 m from the center of the door, what is the door’s angular speed? Correct answer: 2 . 30769 rad / s (tolerance ± 1 %). Explanation: Basic Concept: v t = Given: v t = 1 . 8 m / s r = 0 . 78 m Solution: ω = v t r = 1 . 8 m / s 0 . 78 m = 2 . 30769 rad / s Question 6, chap 12, sect 2. part 1 of 1 10 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 64 . 5 cm apart on the same axle. From the
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homework 09 – VENNES, ROSS – Due: Nov 2 2007, 1:00 am 2 angular displacement 48 of the two bullet holes in the disks and the rotational speed 866 rev / min of the disks, we can determine the speed v of the bullet. 48 v 866 rpm 64 . 5 cm What is the speed of the bullet? Correct answer: 69 . 8212 m / s (tolerance ± 1 %). Explanation: Let : ω = 866 rev / min , d = 64 . 5 cm , and θ = 48 . From θ = ω t the time to pass through an angle θ is t = θ ω = (48 ) (90 . 6873 rad / s) π rad 180 = 0 . 00923788 s . Then the speed of the bullet is v = d t = (64 . 5 cm) (0 . 01 m / cm) 0 . 00923788 s = 69 . 8212 m / s . Question 7, chap 12, sect 2. part 1 of 3 10 points A car accelerates uniformly from rest and covers a distance of 53 m in 6 . 9 s. If the diameter of a tire is 49 cm, find the angular acceleration of the wheel. Correct answer: 9 . 08744 rad / s 2 (tolerance ± 1 %). Explanation: Find the wheel’s linear acceleration x = 1 2 a t 2 a = 2 x t 2 . Then, angular acceleration is α = a R = 2 x R t 2 = 2 (53 m) 1 2 (49 cm) (6 . 9 s) 2 = 9 . 08744 rad / s 2 , where x is the distance traveled, t is the time taken and R = 1 2 (49 cm) is the radius of the wheel. Question 8, chap 12, sect 2.
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