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Unformatted text preview: 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 centerofmass is I cm = 1 12 mℓ 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 : ℓ = 17 m , d = ℓ = 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 + md 2 = 1 12 mℓ 2 + mℓ 2 = 13 12 mℓ 2 (1) = 13 12 (4 . 4 kg) (17 m) 2 = 1377 . 57 kg m 2 . β F y F x mg O Since the rod is uniform, its center of mass is located a distance ℓ from the pivot. The vertical height of the center of mass above hor izontal is ℓ sin θ . Using conservation of energy and substituting I from Eq. 1, we have K f = U i 1 2 I ω 2 = mg ℓ sin θ 13 24 mℓ 2 ω 2 = mg ℓ sin θ ω 2 = 24 13 g sin θ ℓ (4) ω = radicalbigg 24 g sin θ 13 ℓ = 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 = ℓ 24 13 g sin θ ℓ = 24 13 g sin θ = 24 13 (9 . 8 m / s 2 ) sin(64 ◦ ) = 16 . 2613 m / s 2 . 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 . b 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 = Iα = 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 = Iα = 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 . 19kg · m 2 +(5 . 5kg+3 . 3kg)(0 . 85m) 2 = 5 . 95334 m / s 2 Question 3 part 2 of 3...
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Friction, Mass, Work

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