hw 24 sol - Patel (pnp223) homework 24 Turner (92160) This...

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Patel (pnp223) – homework 24 – Turner – (92160) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±eng and Isaac are riding on a merry-go- round. ±eng rides on a horse at the outer rim oF the circular platForm, twice as Far From the center oF the circular platForm as Isaac, who rides on an inner horse. When the merry-go-round is rotating at a constant angular speed, what is ±eng’s angu- lar speed? 1. halF oF Isaac’s 2. impossible to determine 3. twice Isaac’s 4. the same as Isaac’s correct Explanation: Angular speed is the same For every point on the merry-go-round. 002 (part 1 oF 2) 10.0 points Consider the problem oF the solid sphere rolling down an incline without slipping. The incline has an angle θ , the sphere’s length up the incline is , and its height is h . At the beginning, the sphere oF mass M and radius R rests on the very top oF the incline. M μ θ h What is the acceleration oF the center oF mass? The moment oF inertia oF a sphere with respect to an axis through its center is 2 5 M R 2 . 1. a = 5 7 g cos θ 2. a = 2 7 g sin θ 3. a = 3 7 g cos θ 4. a = 3 7 g sin θ 5. a = 5 7 g sin θ correct 6. a = 3 5 g cos θ 7. a = 3 5 g sin θ 8. a = 2 7 g cos θ Explanation: Consider the Forces acting on the sphere: mg cos θ N f mg sin θ Using the parallel-axis theorem I = 2 5 M R 2 + M R 2 = 7 5 M R 2 , and because the sphere rolls without slipping, α = a R . With the origin at the point oF contact between the sphere and the incline surFace, s τ : M g R sin θ = I α M g R sin θ = 7 5 M R 2 a R a = 5 7 g sin θ .
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This note was uploaded on 07/11/2009 for the course PHY 92160 taught by Professor Turner during the Spring '09 term at University of Texas.

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hw 24 sol - Patel (pnp223) homework 24 Turner (92160) This...

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