Answer to Question #107682 in Mechanics | Relativity for Caylin

Question

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.

Answer 1

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)

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

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

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

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