Find the moment of inertia of a ring about an axis through the center and normal to the plane of the ring and estimate its error.
The flat circular ring has mass M=0,191+-0,003Kg, outer radius a = 110 +-1mm and inner radius b=15+-1mm
1 Answer | Add Yours
We'll apply the formula of the moment of inertia:
I = (pi*sigma/2)(a^4 - b^4)
Knowing, from enunciation, the mass of the circular ring is:
I = (M/2)(a^2 + b^2)
We'll insert the given values:
I = 0.191(0.110^2 + 0.015^2)/2
I = 1.177*10^-3 Kg*m^2
The absolute error in a is delta a.
The fractional error in a is delta a/a.
The fractional error in a^2 is 2delta a/a.
The absolute error in a^2 is 2adelta a and the absolute error in ab2 is 2bdelta b.
The absolute error in a^2+b^2:
a^2+b^2 = sqrt[4a^2(delta a)^2 + 4b^2(delta b)^2]
a^2+b^2 = 1.23*10^-2 m^2
From this value results a fractional error of 1.8%.
The fractional error in M is :
M = (0.003/0.191)
M = 1.6%
I = sqrt(1.8^2 + 1.6^2) %
I = 2.4%
I = 1.177*10^-3 Kg*m^2 +/- 2.4%
We’ve answered 318,989 questions. We can answer yours, too.Ask a question