Culculate moment of inertia of a uniform solid cone of mass m, base radius R and height H about

Culculate moment of inertia of a uniform solid cone of mass...

Key Concepts

Moment of Inertia

Moment of inertia is a physical quantity that measures an object's resistance to changes in its rotational motion about a specific axis. It is calculated by integrating the square of the distance from the axis of rotation multiplied by the mass distribution, thereby providing an effective measure of how mass is distributed with respect to the axis.

Uniform Density Distribution

A uniform density distribution implies that the mass is evenly distributed throughout the entire volume of the object. This simplification allows for the mass density to be treated as a constant during integration, making it easier to relate the geometry of the object to its overall mass and subsequently to its moment of inertia.

Volume Integration

Volume integration is the process of summing up the contribution of infinitesimal mass elements across the entire volume of a three-dimensional object. In the context of moment of inertia calculations, this technique involves integrating the product of the mass element and the square of its distance from the axis of rotation over the region occupied by the object.

Integration in Cylindrical Coordinates

Cylindrical coordinates are particularly useful for objects with rotational symmetry, such as cones, because they align with the geometry of the object. By expressing the coordinates in terms of radius, angle, and height, the integration limits become more natural and the moment of inertia calculation simplifies considerably.

Cone Geometry

The geometry of a cone, defined by its base radius and height, dictates how the radius of the cone changes linearly from the tip to the base. Understanding this relationship is crucial when setting up the integral for the moment of inertia, as the variable radius must be accounted for correctly in the integration process.

Transcript

00:01 Hello everyone we are going to understand this question here given in the question given mass of cone that is represented by m mass of cone that is represented by m radius of cone that is r and height of cone that is h now we have to find the movement of inertia of cone about the given axis so volume of cone is equal to volume of cone that is equal to total mass upon volume of cone means 1 upon 3 into pi r square into h now substituting the value in this m upon 1 upon 3 into pi r square h so here 3 will be numerator 3 m square h.

01:34 This is the mass density of cone.

01:37 Now taking the partial cone of radius r, let the radius of partial cone, the radius of partial cone, that is equal to radius of partial cone is r.

02:07 So volume of partial cone, volume of partial cone, dv is equal to pry r square into d z here d is the height of cone height of partial cone so mass of partial cone is equal to mass of partial cone mass of partial cone that is represented by dm which is equal to mass density of cone that is 3m upon pi r square h into into volume of partial cone that is pi r square into d z here pi and pi will get cancelled so dm is equal to 3m upon r square into h into small r square into d z now moment of inertia of partial cone that is part of disk so moment of inertia of dm of dm that is represented by di so di is equal to 1 upon 2 dm into r square so we can write 1 upon 2 into the value of dm is 3m upon r square into h into small r square into d z and here value of r to r square into r square now again we can write 1 by 2, sorry, we will get 3 by 2 m upon capital r square into h into small r to the power 4 into dz.

04:50 That is equal to d .i.

04:53 Now, now integrating both side, integrating both side.

05:13 So we will get i is equal to 3 upon.

05:18 2 into m upon r square into h and integration of r to the power 4 into d z and here integration limit is from 0 to h now doing further integration we will get i is equal to 3 upon 2 into m upon r square plus h into r to 5 upon 5 and integration limit is from 0 to h.

06:04 Now substituting the value in this, we will get i is equal to 3 upon 2 into m upon r square into h and we will get here substituting the value of r is equal to h.

06:24 So h to the power 5 upon 5.

06:26 Now we can write i is equal to 3 upon 2 3 upon 2 m upon r square into h into h to the over 5 upon 5 5 so here h and h will get cancelled sorry here we did a calculation mistake let's correct it here integration is from r square into d z but here we know that 10 theta is equal to for this triangular cone.

07:30 10 theta is equal to r upon z.

07:38 This height is z.

07:43 10 theta is equal to r upon, this is small r and this is capital r...

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