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hw1solutions

# hw1solutions - homework 01 VENNES ROSS Due Sep 7 2007 1:00...

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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. 31 cm 4 . 5 cm 2 . 9 cm 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 ρ = 8 g / cm 3 . Basic Concepts: The volume of the pipe will be the cross- sectional area times the length. Solution: V = ( π r 2 1 π r 2 2 ) = π [ r 2 1 r 2 2 ] = π [(4 . 5 cm) 2 (2 . 9 cm) 2 ] (31 cm) = 1153 . 09 cm 3 . Thus the density is ρ = m V so m = ρ V = ρ π [ r 2 1 r 2 2 ] = (8 g / cm 3 ) π [(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 0 , 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.

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homework 01 – VENNES, ROSS – Due: Sep 7 2007, 1:00 am 2 Which one is dimensionally incorrect ? 1. s = s 0 + v t + v 2 a 2. t = k radicalbigg s g + a v correct 3. a = g + k v t + v 2 s 0 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 0 [ a ] = L T 2 bracketleftbigg g + k v t + v 2 s 0 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. (4): v 2 = 2 a s + k s v t [ v 2 ] = L 2 T 2 bracketleftbigg 2 a s + k s v t bracketrightbigg = L T 2 L + L L T 1 T 1 = L 2 T 2 is also consistent. (5): s = s 0 + v t + v 2 a [ s ] = L bracketleftbigg s 0 + v t + v 2 a bracketrightbigg = L + L T 1 T + L 2 T 2 L T 2 = L is also consistent. So only t = k radicalbigg s g + a v is dimensionally incorrect . Question 4, chap 1, sect 6. part 1 of 2 10 points s is a distance with unit [L], t is a time with unit [T] and θ is an angle in radians.
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