HW #9-1-solutions - pasha (sep635) HW 9-1 Antoniewicz...

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pasha (sep635) – HW 9-1 – Antoniewicz – (56445) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points Amanwho s ema s si s80kgandaw oman whose mass is 50 kg sit at opposite ends oF a canoe 5 m long, whose mass is 35 kg. Assume the man is seated at x =0and the boat extends along the positive x axis with the woman at the other end. Where is the center oF mass oF the system consisting oF man, woman, and canoe? Correct answer: 2 . 04545 m. Explanation: With the man ( m 1 )attheorigin,wehave x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = 0+(50kg)(5m)+(35kg)(2 . 5m) 80 kg + 50 kg + 35 kg =2 . 04545 m . 002 (part 2 of 2) 10.0 points Suppose that the man moves quickly to the center oF the canoe and sits down there. How Far does the canoe move in the water? Correct answer: 1 . 21212 m. Explanation: Assume that the net Force on the canoe is zero. Then its center oF mass momentum (and velocity) remains constant, which is zero in this case. So the center oF mass oF the system remains stationary. Place the origin at the leFt end oF the canoe (where the man was sitting in part 1) and the new center oF mass is at x CM = m 1 x 1 + m 2 x 2 + m 3 x 3 m 1 + m 2 + m 3 = (80 kg)(2 . 5m)+(50kg)(5m) 80 kg + 50 kg + 35 kg + (35 kg)(2 . 5m) 80 kg + 50 kg + 35 kg =3 . 25758 m . At frst, the center oF mass was 2 . 04545 m From the leFt end. Now the center oF mass is 3 . 25758 m From the leFt end. But the center oF mass never moved! This means the boat moved 3 . 25758 m - 2 . 04545 m = 1 . 21212 m to the leFt as the man walked 1 . 21212 m to the right. 003 (part 1 of 3) 7.0 points A string is wound around a uniForm disc whose mass is 3 . 5kgandrad iu si s0 . 58 m , see fgure below. The disc is released From rest with the string vertical and its top end tied to a fxed support. The acceleration oF gravity is 9 . 8m / s 2 . h 0 . 58 m 3 . 5kg ω As the disc descends, calculate the tension in the string. Correct answer: 11 . 4333 N. Explanation: Let : R =0 . 58 m , M =3 . 5kg , and g =9 . 8m / s 2 . Basic Concepts X ~ F = m ~ a X = I Δ U + Δ K rot + Δ K trans =0 Solution X F =
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pasha (sep635) – HW 9-1 – Antoniewicz – (56445) 2 X τ = TR = I α = 1 2 MR 2 ± a R ² . (2) Solving for a in (2), a = 2 T M . (3) Using a from Eq. (3) and solving for T in (1), T = M ( g - a ) = M ³ g - 2 T M ´ = Mg - 2 T 3 T = Mg T = Mg 3 (4) = (3 . 5kg)(9 . 8m / s 2 ) 3 = 11 . 4333 N . 004 (part 2 of 3) 7.0 points
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HW #9-1-solutions - pasha (sep635) HW 9-1 Antoniewicz...

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