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Unformatted text preview: Muraj, Hamza – Homework 8 – Due: Feb 8 2006, 4:00 am – Inst: Florin 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points A ball tied to the end of a string swings in a vertical arc under the influence of gravity, as in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 9 . 8 m / s 2 . 7 8 5 m 1 . 3 8 m / s 2 3 . 8 ◦ Find the magnitude of the radial accelera tion. Correct answer: 2 . 42599 m / s 2 . Explanation: Let : v = 1 . 38 m / s , r = 0 . 785 m , θ = 23 . 8 ◦ , and g = 9 . 8 m / s 2 . g r a r a θ a φ v 6 = 0 v = 1 . 38 m / s θ Since v = 1 . 38 m / s and r = 0 . 785 m, we find that the magnitude of the radial acceler ation at this instant is a r = v 2 r = (1 . 38 m / s) 2 (0 . 785 m) = 2 . 42599 m / s 2 . 002 (part 2 of 3) 10 points When the ball is at an angle 23 . 8 ◦ to the vertical, it has a tangential acceleration of magnitude g sin θ (produced by the tangential component of the force m~g ). Therefore, at θ = 23 . 8 ◦ , using the formula a θ = g sin θ , we obtain a θ = 3 . 95475 m / s 2 . Find the magnitude of the total accelera tion at θ = 23 . 8 ◦ . Correct answer: 4 . 63955 m / s 2 . Explanation: Since ~a = ~a r + ~a t , and the vectors ~a r and ~a t are perpendicular, the magnitude of k ~a k at θ = 23 . 8 ◦ is a = k ~a k = q a 2 r + a 2 θ = q (2 . 42599 m / s 2 ) 2 + (3 . 95475 m / s 2 ) 2 = 4 . 63955 m / s 2 . 003 (part 3 of 3) 10 points Calculate the angle between the total acceler ation ~a and the string at θ = 23 . 8 ◦ . Correct answer: 58 . 4735 ◦ . Explanation: From the figure we see that a θ a r = tan φ, so φ = arctan µ a θ a r ¶ = arctan µ 3 . 95475 m / s 2 2 . 42599 m / s 2 ¶ = (1 . 02056 rad) (57 . 2958 deg / rad) = 58 . 4735 ◦ . Muraj, Hamza – Homework 8 – Due: Feb 8 2006, 4:00 am – Inst: Florin 2 004 (part 1 of 1) 10 points A boat moves through the water of a river at 9 . 63 m / s relative to the water, regardless of the boat’s direction....
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 Spring '09
 Turner
 Acceleration, Force, Friction, Correct Answer

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