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FINALLLLLLLLSOUL

FINALLLLLLLLSOUL - final 01 SWEDLUND EDWARD Due 2:00 pm...

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final 01 – SWEDLUND, EDWARD – Due: May 14 2007, 2:00 pm 1 Question 1, chap 13, sect 1. part 1 of 1 10 points A 25 kg mass and a 12 kg mass are sus- pended by a pulley that has a radius of 13 cm and a mass of 5 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 3 . 1 m apart. Treat the pulley as a uniform disk. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 1 m 13 cm 5 kg ω 25 kg 12 kg Determine the speeds of the two masses as they pass each other. Correct answer: 3 . 16204. Explanation: Let : M = 5 kg , R = 13 cm , m 1 = 25 kg , m 2 = 12 kg , h = 3 . 1 m , v = ω R , I = 1 2 M R 2 , and K disk = 1 2 I ω 2 = 1 4 M v 2 . From conservation of energy K 1 + K 2 + K disk = Δ U or m 1 v 2 2 + m 2 v 2 2 + M v 2 4 = ( m 1 m 2 ) g h 2 parenleftbigg m 1 + m 2 + M 2 parenrightbigg v 2 = ( m 1 m 2 ) g h , where h 2 is the height. Taking no slipping into account, we can solve for v v = radicalBigg 2 ( m 1 m 2 ) g h 2 ( m 1 + m 2 ) + M (1) = radicalBigg 394 . 94 J 39 . 5 kg = 3 . 16204 m / s . Alternative Solution: The forces in the vertical directions for m 1 , m 2 , and the torque on the pulley give us m 1 g T 1 = + m 1 a m 2 g T 2 = m 2 a , so T 1 = m 1 ( g a ) T 2 = m 2 ( g + a ) , and I α = [ T 1 T 2 ] R 1 2 M R 2 parenleftBig a R parenrightBig = bracketleftBig m 1 ( g a )] R [ m 2 ( g + a ) bracketrightBig R 1 2 M a = m 1 g m 1 a m 2 a m 2 g a = 2 ( m 1 m 2 ) g M + 2 ( m 1 + m 2 ) . The velocity is v 2 f = v 2 i + 2 a Δ y v f = radicalbigg 2 a h 2 = radicalBigg 2 ( m 1 m 2 ) g h 2 ( m 1 + m 2 ) + M , which is the same as Eq. 1. Question 2, chap 17, sect 4. part 1 of 1 10 points The speed of the observer in the truck, and the speed of the source , the police car are shown in the figure below.

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final 01 – SWEDLUND, EDWARD – Due: May 14 2007, 2:00 pm 2 A police car is traveling at a speed v c to the left. A truck is traveling at a speed v t to the left. When the police car is stationary its siren’s has a wavelength of λ c = 0 . 343 m and a frequency of f c = 1000 Hz . A wind is blowing in the same direction as that of the truck with a speed, 5 m / s, to the left. The frequency of the siren on the police car is 1000 Hz. Police 23 m / s 36 m / s Truck 5 m / s wind What is the frequency f t heard by an ob- server on the moving truck? Correct answer: 1036 . 01. Explanation: Let : v w = 5 m / s .v c = 23 m / s , v t = 36 m / s , λ c = 0 . 343 m , f c = 1000 Hz , v a λ f = (0 . 343 m) (1000 Hz) = 343 m / s . The problem must be worked in the frame of reference relative to the air. “ v t v w ” is the relative velocity v o of the truck (observer), and “ v c v w ” is the relative velocity v s of the car (source), so f t = v a ± ( v t v w ) v a ( v c v w ) f c . (2) The relative velocity of the observer is to- wards the source so the upper sign is used in the numerator ( ± → +), and the relative ve- locity of the source is away from the observer so the lower sign is used in the denominator ( ∓ → +). Therefore Eq. 2 becomes f t = v a + ( v t v w ) v a + ( v c v w ) f c , so = v a + v t v w v a + v c v w f c = (343 m / s) + (36 m / s) (5 m / s) (343 m / s) + (23 m / s) (5 m / s) (1000 Hz) = 1036 . 01 Hz .
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