Analytical Mech Homework Solutions 10

Analytical Mech Homework Solutions 10 - 1.28 ˆ r P = ib...

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Unformatted text preview: 1.28 ˆ r P = ib sin θ + ˆb cos θ j ˆ vrel = ibθ cos θ − ˆbθ sin θ j ˆ arel = ib θ cosθ − θ 2 sin θ − ˆb θ sin θ + θ 2 cosθ j ( ) at the point θ = So, vrel π 2 = bθ = v ) , vrel = −v va = bb v2 v2 ˆ ˆ ˆ ˆ = vrelτ + rel n = a τ + n ρ b θ= Now, arel ( v b θ= 1 2 v4 2 arel = a + 2 b vP = v + vrel and aP = a + arel a a v2 v2 ˆ aP = i a + b cos θ − 2 sin θ − ˆb sin θ + 2 cosθ j b b b b 1 2 v4 2v 2 aP = a 2 + 2 cos θ + 2 2 − si n θ ab ab aP is a maximum at θ = 0 , i.e., at the top of the wheel. 2v 2 −2sin θ − cosθ = 0 ab v2 ab θ = tan −1 − 2 0 0 x -x 0 x x 0 2 x 1 1.29 RR = x x 0 − x x 0 = 0 2 x 2 0 Therefore, x = 2 0 0 1 0 0 1 0 0 1 The transformation represents a rotation of 45o about the z-axis (see Example 1.8.2) 10 ...
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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