Hw2_s - homework 02 – YOO HEE – Due 4:00 am 1 Question 1 chap 1 sect 6 part 1 of 2 10 points This problem shows how dimensional anal ysis helps

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 8

Hw2_s - homework 02 – YOO HEE – Due 4:00 am 1 Question 1 chap 1 sect 6 part 1 of 2 10 points This problem shows how dimensional anal ysis helps

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online