In this article we would like to explain the derivation of moment of inertia of a hollow sphere.

**Moment of inertia**, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. For every geometrical object we can find moment of inertia by an axis using integration..

First of all consider hollow sphere with radius R. We need to calculate the moment of inertia by its center axis. Consider a differential element ring at an angle theta from its center horizontal axis,

Figure left to this will graphically represent the required data for derivation.

Basic mathematic process of finding moment of inetia of a hollow sphere is adding moment of inertial of all small rings, over entire sphere surface.

Mass of sphere

=\quad M

Mass per unit surface area

=\frac { M }{ 4\pi { R }^{ 2 } }

Surface area of the differential ring element considered

=2\pi R\cos { \theta } .Rd\theta

From above data we can derive the mass of the ring as dM

=\frac { M }{ 4\pi { R }^{ 2 } } 2\pi R\cos { \theta } .Rd\theta

=\frac { M }{ 2 } \cos { \theta } d\theta

Hence, moment of inertia on the ring dI

=dM\times { \left( R\cos { \theta } \right) }^{ 2 }

=\frac { M }{ 2 } \cos { \theta } d\theta \times { \left( R\cos { \theta } \right) }^{ 2 }

=\frac { M{ R }^{ 2 } }{ 2 } \cos ^{ 3 }{ \theta } d\theta

Moment of inertia of a hollow sphere I

I skip the integration part and directly replace with answer,

I

=\frac { M{ R }^{ 2 } }{ 8 } \left\lfloor \frac { 16 }{ 3 } \right\rfloor

So, moment of inertia of a hollow sphere, is I

=\frac { 2M{ R }^{ 2 } }{ 3 }

From above derivation you can clearly understand the step by step procedure. If you have any inquiries or suggestion please feel free to write us on contact us page.

*Also Read: Sins to confess*

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