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ps02_soln

# ps02_soln - Physics 116 Fall 2007 Reading Sec 1.8 1.9 notes...

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Physics 116, Fall 2007 Problem Set 2 due Sep 12, 2007 Reading: Sec 1.8, 1.9, notes 1.1, 2.1-2.3 1. Circular motion K&K 1.14. SOLUTION: From rolling without slipping, we have v cm = ωR . Taking derivative of time on both sides, dv cm dt = d ( ωR ) dt a = αR α = a R To connect even more concretely with what we’ve done in class, consider our starting point for the point on the rolling tire in problem 5 of PS01. There we began with v p = v translation + v rotation = 0 . (1.1) Again taking the time derivative of both sides and re-arranging, we obtain a translation + a rotation = 0 . (1.2) Now note that a rotation is the tangential acceleration, and that at the point of contact, a translation = a ˆ θ . So we have a ˆ θ = - (2 ˙ r ˙ θ + R ¨ θ ) ˆ θ. (1.3) But, ˙ r = 0 and ¨ θ = - α , so we once again have a = Rα. (1.4) 2. Rolling drum K&K 1.15. SOLUTION: a) See graph. For position vectors, the point in coordinate system A has a position vector r A , which becomes r B in system B. They are related via relation r A = r B + R . Again taking time derivatives on both sides, we have v B = v A - d R dt . b) See graph again. We use the notation v ab to represent velocity of a relative to b . v ab = v a - v b = d R ab dt , and v a is 0 in this case (it’s not moving in its own frame moving with itself). So, v ab = d

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ps02_soln - Physics 116 Fall 2007 Reading Sec 1.8 1.9 notes...

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