Question

Prove that moment of inertia of uniform ring of
mass M and radius R about its geometric axis is
MR2.
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Answer 1

We will assume the mass of the ring to be M and radius be R.

Now we need to cut an elemental ring (dx) at the circumference of the ring. Hence, the mass (dm) of the elemental ring will be;

dm = m / 2πR dx

Next, we calculate I = (dm) R2

Substituting the values, we get;

I = m / 2πR dx R2

Using integration;

I = m / 2π o∫2πR dx

I = mR / 2π 2πR

I = mR2

Answer 2

Given : Uniform ring of  mass M and radius R about its geometric axis.

To Prove : Moment of inertia of uniform ring of  mass M and radius R about its geometric axis is  $\[m{R^2}\]$.

Solution :

We will assume the mass of the ring to be M and radius be R.

Now we need to cut an elemental ring (dx) at the circumference of the ring. Hence, the mass (dm) of the elemental ring will be ;  

dm = m/2πR dx

Next, we calculate I = (dm) R2

Substituting the values, we get;

I = m / 2πR dx R2

Using integration;

I = m / 2π o∫2πR dx

I = $\[\frac{{mR}}{{{\text{ }}2\pi {\text{ }}2\pi R}}\]$

I = $\[m{R^2}\]$

Hence Proved

Hence, moment of inertia of uniform ring of  mass M and radius R about its geometric axis is  $\[m{R^2}\]$.

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