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# solution_pdf2 - garcia(jg46766 Problem Set 10...

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garcia (jg46766) – Problem Set 10 – lacosse – (S210-7) 2 10. v = radicalbig 3 g Δ h Explanation: The application of work-energy equation gives m g Δ h = 1 2 m v 2 + 1 2 I ω 2 . But 1 2 I ω 2 = 1 2 parenleftbigg 1 2 m r 2 parenrightbigg ω 2 = 1 4 m v 2 . So m g Δ h = parenleftbigg 1 2 + 1 4 parenrightbigg m v 2 = v = radicalbigg 4 3 g Δ h . 003(part1of2)10.0points A particle is located at the vector position vectorr = (1 . 6 m)ˆ ı + (4 . 2 m)ˆ and the force acting on it is vector F = (2 . 5 N)ˆ ı + (3 . 2 N)ˆ  . What is the magnitude of the torque about the origin? Correct answer: 5 . 38 N m. Explanation: BasicConcept: vector τ = vectorr × vector F Solution: Since neither position of the par- ticle, nor the force acting on the particle have the z -components, the torque acting on the particle has only z -component: vector τ = [ x F y - y F x ] ˆ k = [(1 . 6 m) (3 . 2 N) - (4 . 2 m) (2 . 5 N)] ˆ k = [ - 5 . 38 N m] ˆ k . 004(part2of2)10.0points What is the magnitude of the torque about the point having coordinates [ a, b ] = [(3 m) , (8 m)]? Correct answer: 5 . 02 N m. Explanation: Reasoning similarly as we did in the pre- vious section, but with the difference that relative to the point [(3 m) , (8 m)] the y - component of the particle is now [ y - (8 m)], we have vector τ = { [ x - a ] F y - [ y - b ] F x } ˆ k = { [(1 . 6 m) - (3 m)] [3 . 2 N] - [(4 . 2 m) - (8 m)] [2 . 5 N] } ˆ k = { 5 . 02 N m } ˆ k .
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solution_pdf2 - garcia(jg46766 Problem Set 10...

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