Practice Homework 10 Solutions

# Practice Homework 10 Solutions - practice 10 – ALIBHAI...

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Unformatted text preview: practice 10 – ALIBHAI, ZAHID – Due: Apr 1 2007, 4:00 am 1 Question 1 part 1 of 1 10 points A spool (similar to a yo-yo) is pulled in three ways, as shown below. There is suffi- cient friction for rotation. a b c In what direction will the spool move in each case (in the order spool a, spool b, spool c)? 1. right; right; left 2. left; right; right 3. right; left; left 4. None of these is correct. 5. right; left; right 6. right; right; right correct Explanation: In all three cases the spool moves to the right. In (a) there is a torque about the point of contact with the table that rotates the spool clockwise, so the spool rolls to the right. In (b) the line of action extends through the point of table contact, yielding no lever arm and therefore no torque; with a force component to the right, the spool slides to the right without rolling. In (c) the torque produces clockwise rota- tion so the spool rolls to the right. Question 2 part 1 of 1 10 points A uniform rod of mass 2 . 2 kg is 4 m long. The rod is pivoted about a horizontal, fric- tionless pin at the end of a thin extension (of negligible mass) a distance 4 m from the cen- ter of mass of the rod. Initially the rod makes an angle of 43 ◦ with the horizontal. The rod is released from rest at an angle of 43 ◦ 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 mℓ 2 . 4 m 4 m 2 . 2 kg O 43 ◦ 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: 12 . 3389 rad / s 2 (tolerance ± 1 %). Explanation: Let : ℓ = 4 m , d = ℓ = 4 m , θ = 43 ◦ , and m = 2 . 2 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 (2 . 2 kg) (4 m) 2 = 38 . 1333 kg m 2 . practice 10 – ALIBHAI, ZAHID – Due: Apr 1 2007, 4:00 am 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(43 ◦ ) 13 (4 m) = 1 . 75634 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(43 ◦ ) = 12 . 3389 m / s 2 . Question 3 part 1 of 3 10 points A 4 . 9 kg mass is connected by a light cord to a 1 . 6 kg mass on a smooth surface as shown....
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Practice Homework 10 Solutions - practice 10 – ALIBHAI...

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