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PHYS310_homework_06_solutions

# PHYS310_homework_06_solutions - Homework set 6 due...

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Homework set 6, due Wednesday, October 5, 11:30 am Following problems from Analytical Mechanics by Fowles & Cassiday 1. 4.7 For a point on the rim measured from the center of the wheel: ˆ ˆ cos sin r ib jb θ θ = - v v t t b θ ϖ = = o , so ˆ ˆ sin cos r iv jv θ θ = - - o o v & Relative to the ground, ( 29 ˆ ˆ 1 sin cos v iv jv θ θ = - - o o v For a particle of mud leaving the rim: sin y b θ = - o and cos y v v θ = - o o So cos y y v v gt v gt θ = - = - - o o and 2 1 sin cos 2 y b v t gt θ θ = - - - o At maximum height, 0 y v = : cos v t g θ = - o 2 cos 1 cos sin cos 2 v v h b v g g g θ θ θ θ = - - - - - o o o 2 2 cos sin 2 v h b g θ θ = - + o Maximum h occurs for 2 2 cos sin 0 cos 2 dh v b d g θ θ θ θ = = - - o 2 sin gb v θ = - o 4 2 2 2 2 4 cos 1 sin v g b v θ θ - = - = o o 2 4 2 2 2 2 max 2 2 2 2 2 2 gb v g b gb v h v gv v g - = + = + o o o o o Measured from the ground, 2 2 max 2 2 2 gb v h b v g = + + o o The mud leaves the wheel at 1 2 sin gb v θ - = - o

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2. 4.8 (20 points) cos x R φ = and ( 29 cos x x v t v t α = = o o so cos cos R t v φ α = o sin y R φ = and ( 29 2 2 1 1 sin 2 2 y y v t gt v t gt α = - = - o
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PHYS310_homework_06_solutions - Homework set 6 due...

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