Moment of Inertia of Hollow Cone

In science, a moment of inertia, quantitative measure of a body's rotational inertia— that is, the reaction that the body demonstrates to getting its rotational speed around an axis altered by applying a torque (turning force). The axis can be internal or external and can be fixed or not.

However, the moment of inertia (I) is always specified about that axis and is defined as the sum of the products obtained by multiplying the mass of each particle of matter in a given body by the square of its distance from the axis. When determining angular momentum for a rigid body the moment of inertia is equal to the mass at linear momentum.

The momentum p is equal to the mass m times the velocity v for linear momentum; whereas for angular momentum, the angular momentum L is equal to the moment of inertia I times the angular velocity.

The moment of hollow cone inertia can be determined using the expression below;

\[I = \frac {MR^2}{2}\]

What is Inertia?

Simply explained, an object continues to be in motion or rest until an external source acts on it and this tends to be in motion or the rest of an object is called inertia. For example, as soon as the bus or a vehicle starts moving, all the passengers feel a backward push as they were in a state of rest until the external force, in this case, movement of the bus acted on them. The law of inertia also known as the first law of motion was coined by Sir Issac Newton. Inertia is further divided into the inertia of rest, inertia of motion and finally inertia of direction. 

Moment of Inertia of Hollow Cone Formula Derivation

We should basically follow certain general guidelines which are to extract the moment of inertia formula of a hollow cone

Determining the mass density per unit area.

Defines parameters and discovers the mass of small parts.

Consider the moment of inertia corresponding to a ring for the small parts.

1. A hollow cone with radius R, height H, and mass M is located.

Now, at a slant height l and radius r having thickness dl and mass dm, we'll take the product disk. 

Using the similarity of the triangle we get;  

\[\frac {r}{x} = \frac {R}{\sqrt {R^2 + H^2}}\]

\[r=\frac {R}{R} = \frac {R}{\sqrt {R^2 + H^2}}\]

\[r=\frac{\frac{R}{R}}{\sqrt {R^2 +H^2 \times x}}\]

The weight of the elementary disk mass is given by;

dm = \[\frac {M}{\pi R \sqrt {R^2 +H^2 \times \pi rdx} }\]

dm= \[\frac{\frac{M}{\pi R \sqrt {R^2 + H^2 \times 2 \times R}}}{R \sqrt {R^2 + H^2 \times xdx}}\]

dm =  \[\frac {2Mxdx}{R^2 + H^2}\]

2. Calculating the time of elementary disk inertia. It is given as;

dI = r2 dm

dI = 2Mxdx / R2 + H2 X R2 / R2 + H2X x2

3. Finding the moment of inertia through integration.

I = \[\frac{\frac{\int 2 mx dx}{R^2 + H^2 \times R^2}}{R^2 + H^2 \times x^2}\]

I = \[\frac {2MR^2}{(R^2 + H^2)^2} \int x^3 dx\] 

Using the limits of x where x usually varies from o to \[\sqrt {R^2 + H^2}\]

We now get;  

I = \[\frac{\frac{2MR^2}{(R^2 + H^2)^2 \times (R^2 + H^2)^2}}{4}\]

I = \[\frac {1}{2} MR^2\]

Thus, \[I = \frac {MR^2}{2}\]

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