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**Unformatted text preview: **homework 07 – KIM, JI – Due: Oct 17 2007, 4:00 am 1 Question 1, 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 62 . 3 cm apart on the same axle. From the angular displacement 13 . 7 ◦ of the two bul- let holes in the disks and the rotational speed 1130 rev / min of the disks, we can determine the speed v of the bullet. 13 . 7 ◦ v 1130 rpm 62 . 3 cm What is the speed of the bullet? Correct answer: 308 . 317 m / s (tolerance ± 1 %). Explanation: Let : ω = 1130 rev / min , d = 62 . 3 cm , and θ = 13 . 7 ◦ . From θ = ω t the time to pass through an angle θ is t = θ ω = (13 . 7 ◦ ) (118 . 333 rad / s) π rad 180 ◦ = 0 . 00202065 s . Then the speed of the bullet is v = d t = (62 . 3 cm) (0 . 01 m / cm) . 00202065 s = 308 . 317 m / s . Question 2, chap 12, sect 2. part 1 of 2 10 points A wheel starts from rest and rotates with constant angular acceleration to an angular speed of 7 . 72 rad / s in 2 . 3 s. Find the magnitude of the angular acceler- ation of the wheel. Correct answer: 3 . 35652 rad / s 2 (tolerance ± 1 %). Explanation: Angular acceleration is defined by, α = dω dt When the angular acceleration is constant, we can replace the differentials with simply differences, α = Δ ω Δ t = ω f − ω i t The angle through which the wheel rotates during this time interval is, θ = 1 2 αt 2 Question 3, chap 12, sect 2. part 2 of 2 10 points Find the angle in radians through which it rotates in this time. Correct answer: 8 . 878 rad (tolerance ± 1 %). Explanation: Question 4, chap 12, sect 2. part 1 of 1 10 points The tub of a washer goes into its spin- dry cycle, starting from rest and reaching an angular speed of 4 . 1 rev / s in 8 . 2 s. At this point the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 10 . 1 s. Through how many revolutions does the tub turn? Assume constant angular accelera- tion while it is starting and stopping. homework 07 – KIM, JI – Due: Oct 17 2007, 4:00 am 2 Correct answer: 37 . 515 rev (tolerance ± 1 %). Explanation: Given : ω = 4 . 1 rev / s , Δ t 1 = 8 . 2 s , and Δ t 2 = 10 . 1 s . θ = ω av Δ t = ω i + ω f 2 Δ t = ω f 2 Δ t since ω i = 0, so while speeding up, θ 1 = 4 . 1 rev / s 2 (8 . 2 s) = 16 . 81 rev , and while decelerating, θ 2 = 4 . 1 rev / s 2 (10 . 1 s) = 20 . 705 rev . Thus θ = θ 1 + θ 2 = 37 . 515 rev . Question 5, chap 12, sect 5. part 1 of 1 10 points A circular disk with a mass m and radius R is mounted at its center, about which it can rotate freely. The disk has moment of inertia I = (1 / 2) mR 2 . A light cord wrapped around it supports a weight mg . Find the total kinetic energy of the system, when the weight is moving at a speed v ....

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