# 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 =...

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

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### 1 Answer

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%**