# Answer to Question #107682 in Mechanics | Relativity for Caylin

Question #107682
A rectangular brass bar of mass M has dimensions a,b,c. The moment of inertia I is given by
I = M(a squared + b squared)/12 and the following measurements are made:
M= 135.0 +/- 0.1 g
a= 80 +/- 1 mm
b= 10 +/- 1 mm
c= 20.00 +/- 0.1 mm

Calculate the standard error in
1) Density p of material
2) Moment of Inertia I

1) 8,4375 * 10 ( to the power -6)
2) Inertia I got 73,125 kg.m (squared) and standard error i struggled with.

1
2020-04-06T08:45:33-0400

1)

"\\frac{\\Delta \\rho}{\\rho}=\\frac{\\Delta a}{a}+\\frac{\\Delta b}{b}+\\frac{\\Delta c}{c}+\\frac{\\Delta m}{m}"

"\\frac{\\Delta \\rho}{\\rho}=\\frac{1}{80}+\\frac{1}{10}+\\frac{0.1}{20}+\\frac{0.1}{135}=0.11824"

"\\rho=\\frac{m}{abc}=\\frac{0.135}{(0.08)(0.01)(0.02)}=8437.5\\frac{kg}{m^3}"

"\\Delta\\rho=8437.5(0.11824)=1000\\frac{kg}{m^3}"

2)

"I=0.135\\frac{0.08^2+0.01^2}{12}=7.3125\\cdot 10^{-5}\\ kgm^2"

"\\frac{\\Delta I}{I}=2\\frac{\\Delta a}{a}+2\\frac{\\Delta b}{b}+\\frac{\\Delta m}{m}"

"\\frac{\\Delta I}{I}=2\\frac{1}{80}+2\\frac{1}{10}+\\frac{0.1}{135}=0.22574"

"\\Delta I=(0.22574)7.3125\\cdot 10^{-5}=2\\cdot 10^{-5}\\ kgm^2"

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