1 - homework 01 ARMSTRONG DOMINIC Due 3:00 am Question 1...

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homework 01 – ARMSTRONG, DOMINIC – Due: Sep 11 2007, 3:00 am 1 Question 1, chap 1, sect 5. part 1 of 1 10 points A newly discovered giant planet has an av- erage radius 13 times that of the Earth and a mass 971 times that of the Earth. Calculate the ratio of the new planet’s den- sity to the Earth’s density. Correct answer: 0 . 441966 (tolerance ± 1 %). Explanation: Let : R n = 13 R E and m n = 971 m E . Density is the ratio of mass to volume, ρ = m V . A spherical planet of average radius R has volume 4 3 π R 3 and hence density ρ = m 4 3 π R 3 . For two planets of respective radii R 1 and R 2 and masses m 1 and m 2 we have ρ 1 ρ 2 = m 1 4 3 π R 3 1 m 2 4 3 π R 3 2 = parenleftbigg m 1 m 2 parenrightbigg parenleftbigg R 1 R 2 parenrightbigg 3 = 971 (13) 3 = 0 . 441966 . Question 2, 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 . 37 m 2 and density ρ = 2470 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 = 1 ρ A 2. μ = A 2 ρ 2 3. μ = ρ A 4. μ = A ρ 5. μ = A 2 ρ 6. μ = ρ A 2 7. μ = ρ A 2 8. μ = 1 ρ A 2 9. μ = ρ A 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 .
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homework 01 – ARMSTRONG, DOMINIC – Due: Sep 11 2007, 3:00 am 2 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 = ρ A . Question 3, 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 radicalBigg L g correct 2. T = k g L 3. T = k radicalbigg m g L 4. T = k m L g 5. T = k radicalbigg g L 6. T = k L g 7. T = k radicalBigg L m g 8. T = k m g L Explanation: 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 m g L bracketrightBig = ML/T 2 L = MT - 2 bracketleftbigg k m L g bracketrightbigg = ML L/T 2 = MT 2 bracketleftbigg k radicalbigg m g L bracketrightbigg = radicalbigg ML/T 2 L = M 1 2 T - 1 bracketleftBigg k radicalBigg L m g bracketrightBigg = radicalBigg L ML/T 2 = M - 1 2 T bracketleftbigg k L g
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