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Unformatted text preview: quiz 03 – BAUTISTA, ALDO – Due: Apr 5 2006, 9:00 pm 1 Version number encoded for clicker entry: V1:1, V2:1, V3:5, V4:5, V5:1. Question 1 Part 1 of 1. 10 points. A uniform bar of length L and weight W is attached to a wall with a hinge that exerts on the bar a horizontal force H x and a vertical force H y . The bar is held by a cord that makes a 90 ◦ angle with respect to bar and angle θ with respect to wall. The acceleration of gravity g = 9 . 8 m / s 2 . L 90 ◦ W θ What is the magnitude of the horizontal force H x on the pivot? 1. H x = 1 2 W tan θ 2. H x = 1 2 W cos θ 3. H x = 1 2 W sin 2 θ 4. H x = 1 2 W sin θ cos θ correct 5. H x = 1 2 W sin θ 6. H x = 1 2 W cos 2 θ Explanation: Analyzing the torques on the bar, with the hinge at the axis of rotation, we have X τ = L T- L 2 sin θ W = 0 , so, T = 1 2 W sin θ . Analyzing the force on the bar, we have X F x = H x- T cos θ = 0 . Put T into this equation and get H x = 1 2 W sin θ cos θ . Question 2 Part 1 of 1. 10 points. A massless rod of length L has a mass 2 m fastened at its center and another mass m fastened at one end. On the opposite end from the mass m , the rod is hinged with a frictionless hinge. The rod is released from rest from an initial horizontal position; then it swings down. What is the angular velocity ω max as the rod swings through its lowest (vertical) posi- tion? 1. ω max = r 2 g L 2. ω max = r 3 g 2 L 3. ω max = r g L 4. ω max = r 12 g 5 L 5. ω max = r 8 g 3 L correct 6. ω max = r 4 g 5 L Explanation: The mechanical energy of the system is con- served. Measuring heights from the point at quiz 03 – BAUTISTA, ALDO – Due: Apr 5 2006, 9:00 pm 2 the bottom of the rod when it is vertical, the initial potential energy of the system is U i = (3 m ) g L . The potential energy at the bottom of the swing, U f = 2 m g ( L/ 2) = m g L . To calculate the final kinetic energy, we need the moment of inertia of the system, I = 2 m ( L/ 2) 2 + m L 2 = 3 m L 2 / 2 . There- fore, 3 m g L = m g + 1 2 3 2 m L 2 ω 2 , ω max = r 8 g 3 L . Question 3 Part 1 of 2. 10 points. A student sits on a rotating stool holding two 3 . 8 kg masses. When his arms are ex- tended horizontally, the masses are 0 . 78 m from the axis of rotation, and he rotates with an angular velocity of 2 . 4 rad / sec. The stu- dent then pulls the weights horizontally to a shorter distance 0 . 24 m from the rotation axis and his angular velocity increases to ω 2 . ω i ω f For simplicity, assume the student himself plus the stool he sits on have constant com- bined moment of inertia I s = 4 . 8 kg m 2 . Find the new angular velocity ω 2 of the student after he has pulled in the weights. 1. 3 . 93273 rad / s 2. 4 . 06179 rad / s 3. 4 . 18764 rad / s 4. 4 . 31811 rad / s correct 5. 4 . 45277 rad / s 6. 4 . 59392 rad / s Explanation: Let : M = 3 . 8 kg , R 1 = 0 . 78 m , ω 1 = 2 . 4 rad / sec , R 2 = 0 . 24 m , and ω 2 = 4 . 31811 rad / sec , As the student moves his arms, his moment of inertia changes from...
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08