This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: dang (ttd73) HW01 Tsoi (58020) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points This problem shows how dimensional analysis helps us check and sometimes even find a formula. A rope has a cross section A = 11 . 4 m 2 and density = 2150 kg / m 3 . The linear density of the rope , defined to be the mass per unit length, can be written in the form = x A y . Based on dimensional analysis, determine the powers x and y by choosing an expression below. 1. = A 2. = A 2 2 3. = A 2 4. = A 5. = A correct 6. = 1 A 7. = A 2 8. = A 2 9. = 1 A 2 10. = A 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 [ ] = M L = ML- 1 , and the right hand side has [ 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 = 1 A 1 = A . 002 (part 2 of 2) 10.0 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 radicalbigg mg L 2. T = k mL g 3. T = k mg L 4. T = k radicalBigg L g correct 5. T = k g L 6. T = k L g 7. T = k radicalbigg g L 8. T = k radicalBigg L mg Explanation: dang (ttd73) HW01 Tsoi (58020) 2 Here we proceed in the same way. A pe- riod is a measure of time, thus the correct 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 = radicalBigg L ML/T 2 = M- 1 2 T bracketleftbigg k L g bracketrightbigg = L L/T 2 = T 2 bracketleftBig k g L bracketrightBig = L/T 2 L = T- 2 bracketleftbigg k radicalbigg g L bracketrightbigg = radicalbigg L/T 2 L = T- 1 So they are all incorrect, as they should be....
View Full Document
This note was uploaded on 03/31/2009 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.
- Spring '08