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# Hw1 - homework 01 RAMSEY TAYLOR Due Sep 7 2006 4:00 am...

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homework 01 – RAMSEY, TAYLOR – Due: Sep 7 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:3, V4:5, V5:2. Question 1 part 1 of 3 10 points Stokes law says F = 6 πrηv. F is a force r the radius and v the velocity. The parameter η has the dimension of 1. [ η ] = T 2 L M 2. [ η ] = M T 3. [ η ] = M T 2 L 2 4. [ η ] = T L 2 M 5. [ η ] = T 2 L 2 M 6. [ η ] = M T L 2 7. [ η ] = T M 8. [ η ] = M T 2 L 9. [ η ] = T L M 10. [ η ] = M T L correct Explanation: Recall that the dimension of force is [ F ] = [ m a ] = M L T 2 . The number 6 π is dimensionless. The dimen- sions of the radius and velocity are [ r ] = L ; [ v ] = L T so [ F ] = [6 π r η v ] M L T 2 = L L T [ η ] Solving for [ η ] yields [ η ] = M LT . Question 2 part 2 of 3 10 points Consider a simple pendulum which consists of a string with a bob attached to its end. Its period, ( i.e. , the time interval taken for the bob to complete one cycle of motion), may be written in the form T = k m x g y b z , where k is a dimensionless constant, g the magnitude of the gravitational acceleration, m the mass of the bob, and b the length of the string. The appropriate x , y , and z values are given respectively by 1. ( x, y, z ) = ± 1 , 1 2 , 1 2 ² 2. ( x, y, z ) = ± 0 , - 1 2 , - 1 2 ² 3. ( x, y, z ) = ± 1 , - 1 2 , 1 2 ² 4. ( x, y, z ) = (1 , - 1 , 1) 5. ( x, y, z ) = ± 0 , 1 2 , 1 2 ² 6. ( x, y, z ) = ± 0 , 1 2 , - 1 2 ² 7. ( x, y, z ) = ± 1 , 1 2 , - 1 2 ² 8. ( x, y, z ) = ± 1 , - 1 2 , - 1 2 ² 9. ( x, y, z ) = ± 0 , - 1 2 , 1 2 ² correct 10. ( x, y, z ) = (0 , - 1 , 1) Explanation: [ T ] = [ k m x g y b z ] T = M x L y T 2 y L z T = M x L y + z T - 2 y

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homework 01 – RAMSEY, TAYLOR – Due: Sep 7 2006, 4:00 am 2 Hence x = 0, y + z = 0, - 2 y = 1 y = - 1 2 , z = 1 2 ( x, y, z ) = ± 0 , - 1 2 , 1 2 ² . Question 3 part 3 of 3 10 points Consider a piece of string which is placed along the x -axis. Let Δ m be the mass of a segment of the string and Δ x the length of this segment. The linear mass density, μ , of a piece of string is deFned as μ = Δ m Δ x . Denote ρ to be its mass density deFned as ρ = mass volume and A its cross sectional area. Let us write μ = ρ x A y . Use dimensional analysis to determine the equations for x and y . 1. x = - 1 , - 2 y - 3 x = - 1 2. x = 1 , 2 y - 3 x = - 1 correct 3. x = 1 , 2 y + 3 x = - 1 4. x = - 1 , 2 y + 3 x = - 1 5. x = 1 , 3 x + 2 y = 1 6. x = - 1 , 2 y - 3 x = - 1 7. x = - 1 , 3 x + 2 y = 1 8. x = - 1 , 2 y - 3 x = 1 9. x = 1 , 2 y - 3 x = 1 10. x = 1 , - 2 y - 3 x = - 1 Explanation: [ μ ] = [ ρ x A y ] M L - 1 = M x L - 3 x L 2 y = M x L 2 y - 3 x Equating both sides yields the equations x = 1 , 2 y - 3 x = - 1 . Question 4 part 1 of 1 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. Which one is dimensionally incorrect ? 1. v 2 = 2 a s + k s v t 2. s = s 0 + v t + v 2 a 3. t = k r s g + a v correct 4. t = v a + x v 5. a = g + k v t + v 2 s 0 Explanation: ±or an equation to be dimensionally cor- rect, all its terms must have the same units. (1) t = v a + x v [ t ] = T h v a i + h x v i = LT - 1 LT - 2 + L LT - 1 = T + T = T It is consistent. (2) a = g + k v t + v 2 s 0 [ a ] = LT - 2 ³ g + kv t + v 2 s 0 ´ = LT - 2 + LT - 1 T + L 2 T - 2 L = LT - 2
homework 01 – RAMSEY, TAYLOR – Due: Sep 7 2006, 4:00 am 3 It is also consistent.

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Hw1 - homework 01 RAMSEY TAYLOR Due Sep 7 2006 4:00 am...

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