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Unformatted text preview: homework 02 – YOO, HEE – Due: Jan 22 2008, 4:00 am 1 Question 1, chap 1, sect 6. part 1 of 2 10 points This problem shows how dimensional anal ysis helps us check and sometimes even find a formula. A rope has a cross section A = 9 . 57 m 2 and density ρ = 1580 kg / m 3 . The “linear” density of the rope μ , defined to be the mass per unit length, can be written in the form μ = ρ x A y . Based on dimensional analysis, determine the powers x and y by choosing an expression below. 1. μ = A 2 ρ 2. μ = 1 ρA 2 3. μ = 1 ρA 4. μ = A 2 ρ 2 5. μ = A ρ 6. μ = ρA 2 7. μ = ρ A 8. μ = ρ A 2 9. μ = ρA correct 10. μ = A ρ 2 Explanation: Kilogram (kg): a unit of mass ( M ). Meter (m): a unit of length ( L ). [ x ] means ”the units of x ”. The units of both sides of any equation must be the same for the equation to make sense. The units of the left hand side (LHS) are given as [ μ ] = M L = ML 1 , and the right hand side has [ ρ x A y ] = parenleftbigg M L 3 parenrightbigg x × ( L 2 ) y = M x L 3 x L 2 y = M x L 2 y 3 x , thus M +1 L 1 = M x L 2 y 3 x . The powers of the units of mass and length need to be the same as for the LHS above, so x = 1 2 y 3 x = 1 . We can substitute the first equation into the second one to obtain y = 1. Since x = 1 and y = 1, the answer is μ = ρ 1 A 1 = ρA. Question 2, chap 1, sect 6. part 2 of 2 10 points A simple pendulum is made out of a string with length L and a mass m attached to one end of the string. Its period T of oscillation may depend on the gravitational acceleration g , and also depend on L and m . Based on dimensional analysis, check which one of the following expressions is dimension ally acceptable, where k is a dimensionless constant. 1. T = k mL g 2. T = k radicalBigg L g correct 3. T = k radicalBigg L mg 4. T = k g L 5. T = k L g 6. T = k radicalbigg mg L 7. T = k mg L 8. T = k radicalbigg g L Explanation: Here we proceed in the same way. A pe riod is a measure of time, thus the correct homework 02 – YOO, HEE – Due: Jan 22 2008, 4:00 am 2 expression must have units of time. bracketleftBigg k radicalBigg L g bracketrightBigg = radicalBigg L L/T 2 = T is the correct one. As for the others, bracketleftBig k mg L bracketrightBig = ML/T 2 L = MT 2 bracketleftbigg k mL g bracketrightbigg = ML L/T 2 = MT 2 bracketleftbigg k radicalbigg mg L bracketrightbigg = radicalbigg ML/T 2 L = M 1 2 T 1 bracketleftBigg k radicalBigg L mg bracketrightBigg...
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This note was uploaded on 05/04/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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