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


My answers:

1) 8,4375 * 10 ( to the power -6)

2) Inertia I got 73,125 kg.m (squared) and standard error i struggled with.


Please assist and confirm whether or not my answers are correct.

Expert's answer

1)


Δρρ=Δaa+Δbb+Δcc+Δmm\frac{\Delta \rho}{\rho}=\frac{\Delta a}{a}+\frac{\Delta b}{b}+\frac{\Delta c}{c}+\frac{\Delta m}{m}

Δρρ=180+110+0.120+0.1135=0.11824\frac{\Delta \rho}{\rho}=\frac{1}{80}+\frac{1}{10}+\frac{0.1}{20}+\frac{0.1}{135}=0.11824

ρ=mabc=0.135(0.08)(0.01)(0.02)=8437.5kgm3\rho=\frac{m}{abc}=\frac{0.135}{(0.08)(0.01)(0.02)}=8437.5\frac{kg}{m^3}

Δρ=8437.5(0.11824)=1000kgm3\Delta\rho=8437.5(0.11824)=1000\frac{kg}{m^3}

2)


I=0.1350.082+0.01212=7.3125105 kgm2I=0.135\frac{0.08^2+0.01^2}{12}=7.3125\cdot 10^{-5}\ kgm^2

ΔII=2Δaa+2Δbb+Δmm\frac{\Delta I}{I}=2\frac{\Delta a}{a}+2\frac{\Delta b}{b}+\frac{\Delta m}{m}

ΔII=2180+2110+0.1135=0.22574\frac{\Delta I}{I}=2\frac{1}{80}+2\frac{1}{10}+\frac{0.1}{135}=0.22574

ΔI=(0.22574)7.3125105=2105 kgm2\Delta I=(0.22574)7.3125\cdot 10^{-5}=2\cdot 10^{-5}\ kgm^2


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