# hw32 - Overton Mays – Homework 32 – Due 4:00 am –...

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Unformatted text preview: Overton, Mays – Homework 32 – Due: Apr 23 2005, 4:00 am – Inst: Turner 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The graph below represents the potential en- ergy U as a function of displacement x for an object on the end of a spring ( F =- k x ) oscillating in simple harmonic motion with amplitude x max . + x max- x max x U U max Which of the following graphs represents the kinetic energy K of the object as a func- tion of displacement x ? 1. + x max- x max x K K max 2. + x max- x max x K K max 3. + x max- x max x K K max 4. + x max- x max x K K max correct 5. + x max- x max x K K max 6. + x max- x max x K K max 7. + x max- x max x K K max Explanation: At the equilibrium point ( x = 0), the ve- locity is maximum and the kinetic energy is U max due to conservation of energy. At the maximum displacement points, (+ x max and- x max ), the velocity is zero and the kinetic energy is zero. From a different perspective, in simple har- monic motion of an object on the end of a spring, the total energy is conserved. At the maximum displacement x max , the kinetic en- ergy is 0, so, E = U ( x max ) = U max . U + K = E = U max K ( x ) = U max- U ( x ) . Thus, K ( x ) looks like an upsidedown U ( x ). 002 (part 1 of 2) 10 points A hoop has a radius R and mass m . It is pivoted at A , a point on the circumference of the hoop. Denote the moment of inertia about the center by I and about the pivot point A by I A . Overton, Mays – Homework 32 – Due: Apr 23 2005, 4:00 am – Inst: Turner 2 The acceleration of gravity is 9 . 8 m / s 2 . A O O' θ The equation of motion about A in the small angle approximation (sin θ ≈ θ ) is given by 1. I A d 2 θ dt 2 = mg Rθ . 2. I d 2 θ dt 2 = mg Rθ . 3. I d 2 θ dt 2 =- 2 mg Rθ . 4. I A d 2 θ dt 2 = 2 mg Rθ . 5. I A d 2 θ dt 2 =- 1 2 mg Rθ . 6. I d 2 θ dt 2 = 2 mg Rθ . 7. I d 2 θ dt 2 =- 1 2 mg Rθ . 8. I A d 2 θ dt 2 =- mg Rθ . correct 9. I d 2 θ dt 2 =- mg Rθ . 10. I A d 2 θ dt 2 =- 2 mg Rθ . Explanation: The torque on the hoop is provided by grav- ity, and its magnitude is given by τ = mg R sin θ ≈ mg Rθ , where we used the small angle approximation in the last step. Note: The distance from the pivot A to the center of mass of the hoop is R ....
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## This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw32 - Overton Mays – Homework 32 – Due 4:00 am –...

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