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Unformatted text preview: homework 01 VENNES, ROSS Due: Sep 7 2007, 1:00 am 1 Question 1, chap 1, sect 6. part 1 of 1 10 points A piece of pipe has an outer radius, an inner radius, and length as shown in the figure below. 3 1 c m 4 . 5 cm 2 . 9 cm b The density is 8 g / cm 3 . What is the mass of this pipe? Correct answer: 9 . 22472 kg (tolerance 1 %). Explanation: Let : r 1 = 4 . 5 cm , r 2 = 2 . 9 cm , = 31 cm , and = 8 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross sectional area times the length. Solution: V = ( r 2 1 r 2 2 ) = [ r 2 1 r 2 2 ] = [(4 . 5 cm) 2 (2 . 9 cm) 2 ] (31 cm) = 1153 . 09 cm 3 . Thus the density is = m V so m = V = [ r 2 1 r 2 2 ] = (8 g / cm 3 ) [(4 . 5 cm) 2 (2 . 9 cm) 2 ] (31 cm) = 9224 . 72 g = 9 . 22472 kg . Question 2, chap 1, sect 6. part 1 of 2 10 points Consider the dimension content of the cen tripetal force. The force should depend on the mass m , the tangential speed v and the radius r . We write it in the form F = m x v y r z . Based on dimensional analysis, determine the powers x , y and z . 1. x = 1, y = 2, z = 1. 2. x = 1, y = 2, z = 1. 3. x = 1, y = 2, z = 1. 4. x = 1, y = 2, z = 1. 5. x = 1, y = 2, z = 1. 6. x = 1, y = 2, z = 1. 7. x = 1, y = 2, z = 1. 8. x = 1, y = 2, z = 1. correct Explanation: In order that F = m x v y r z to be dimen sionally correct: [ F ] = M 1 L 1 T 2 [ m x v y r z ] = M x parenleftBig L T parenrightBig y L z = M x L y + z T y . By equating the powers of M , L , and T x = 1, y + z = 1, and y = 2. Therefore, x = 1, y = 2, and z = 1. Question 3, chap 1, sect 6. part 2 of 2 10 points Consider the following set of equations, where s , s , x and r have units of length, t has units of time, v has units of velocity, g and a have units of acceleration, and k is dimensionless. homework 01 VENNES, ROSS Due: Sep 7 2007, 1:00 am 2 Which one is dimensionally incorrect ? 1. s = s + v t + v 2 a 2. t = k radicalbigg s g + a v correct 3. a = g + k v t + v 2 s 4. v 2 = 2 a s + k s v t 5. t = v a + x v Explanation: For an equation to be dimensionally cor rect, all its terms must have the same units. (1): t = v a + x v [ t ] = T bracketleftBig v a bracketrightBig + bracketleftBig x v bracketrightBig = L T 1 L T 2 + L L T 1 = T + T = T is consistent. (2): a = g + k v t + v 2 s [ a ] = L T 2 bracketleftbigg g + k v t + v 2 s bracketrightbigg = L T 2 + L T 1 T + L 2 T 2 L = L T 2 is also consistent. (3): t = k radicalbigg s g + a v [ t ] = T bracketleftbigg k radicalbigg s g + a v bracketrightbigg = radicalbigg L L T 2 + L T 2 L T 1 = T + T 1 is not dimensionally consistent....
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