Homework 10 Solutions - homework 10 GADHIA TEJAS Due Apr 4...

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homework 10 – GADHIA, TEJAS – Due: Apr 4 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A uniform rod of mass 4 . 4 kg is 17 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 17 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 64 with the horizontal. The rod is released from rest at an angle of 64 with the horizontal, as shown in the figure below The acceleration of gravity is 9 . 8 m / s 2 . Hint: The moment of inertia of the rod about its center-of-mass is I cm = 1 12 m ℓ 2 . 17 m 17 m 4 . 4 kg O 64 What is the magnitude of the horizontal acceleration of the center of mass of the rod at the instant the rod is in a horizontal position? Correct answer: 16 . 2613 rad / s 2 (tolerance ± 1 %). Explanation: Let : = 17 m , d = = 17 m , θ = 64 , and m = 4 . 4 kg . Basic Concepts: K R = 1 2 I ω 2 τ = r × F = I α K i + U i = K f + U f . Solution: I = I cm + m d 2 = 1 12 m ℓ 2 + m ℓ 2 = 13 12 m ℓ 2 (1) = 13 12 (4 . 4 kg) (17 m) 2 = 1377 . 57 kg m 2 . β F y F x m g O Since the rod is uniform, its center of mass is located a distance from the pivot. The vertical height of the center of mass above hor- izontal is sin θ . Using conservation of energy and substituting I from Eq. 1, we have K f = U i 1 2 I ω 2 = m g ℓ sin θ 13 24 m ℓ 2 ω 2 = m g ℓ sin θ ω 2 = 24 13 g sin θ (4) ω = radicalbigg 24 g sin θ 13 = radicalBigg 24 (9 . 8 m / s 2 ) sin(64 ) 13 (17 m) = 0 . 978031 rad / s . Since a x = a r = r ω 2 and Eq. 4, we have a x = r ω 2 = 24 13 g sin θ = 24 13 g sin θ = 24 13 (9 . 8 m / s 2 ) sin(64 ) = 16 . 2613 m / s 2 .
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homework 10 – GADHIA, TEJAS – Due: Apr 4 2007, 4:00 am 2 Question 2 part 1 of 3 10 points A 5 . 5 kg mass is connected by a light cord to a 3 . 3 kg mass on a smooth surface as shown. The pulley rotates about a frictionless axle and has a moment of inertia of 0 . 19 kg · m 2 and a radius of 0 . 85 m. The acceleration of gravity is 9 . 81m / s 2 . 5 . 5kg 3 . 3kg F 1 F 2 R Note: Figure is not drawn to scale Assume that the cord does not slip on the pulley. a) What is the acceleration of the two masses? Correct answer: 5 . 95334 m / s 2 (tolerance ± 1 %). Explanation: F 2 a F 1 (3 . 3kg) g a Note: Figure is not drawn to scale Basic Concepts: Newton’s second law for m 1 and m 2 : m 1 a = m 1 g - F 1 m 2 a = F 2 Newton’s second law for the rotation of the pulley: τ net = = F net R since F net and R are perpendicular. a = αR Given: I = 0 . 19 kg · m 2 R = 0 . 85 m m 1 = 5 . 5 kg m 2 = 3 . 3 kg g = 9 . 81 m / s 2 Solution: F net = F 1 - F 2 = ( m 1 g - m 1 a ) - m 2 a = m 1 g - ( m 1 + m 2 ) a F net R = = I parenleftBig a R parenrightBig so Ia = F net R 2 Ia = [ m 1 g - ( m 1 + m 2 ) a ] R 2 a = m 1 gR 2 I + ( m 1 + m 2 ) R 2 = (5 . 5kg)(9 . 81m / s 2 )(0 . 85m) 2 0 . 19kg · m 2 +(5 . 5kg+3 . 3kg)(0 . 85m) 2 = 5 . 95334 m / s 2 Question 3 part 2 of 3 10 points b) What is the magnitude of the force F 1 ? Correct answer: 21 . 2116 N (tolerance ± 1 %). Explanation: Solution: Using the basic concept: F 1 = m 1 g - m 1 a = (5 . 5kg)(9 . 81m / s 2 - 5 . 95334 m / s 2 ) = 21 . 2116 N Question 4 part 3 of 3 10 points c) What is the magnitude of the force F 2 ?
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