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...

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%