(a) Consider an elementry disc of thickness d x. Moment of...

(a) Consider an elementry disc of thickness $d x$. Moment of inertia of this element about the $z$ -axis, passing through its $\mathrm{C} . \mathrm{M}$. $d I_{z}=\frac{(d m) R^{2}}{2}=\rho S d x \frac{R^{2}}{2}$ where $\rho=$ density of the material of the plate and $S=$ area of cross section of the plate. Thus the sought moment of inertia $=\frac{\pi}{2} \rho b R^{4}\left(\right.$ as $\left.S=\pi R^{2}\right)$putting all the vallues we get, $I_{z}=2 \cdot \mathrm{gm} \cdot \mathrm{m}^{2}$ (b) Consider an element disc of radius $r$ and thickness $d x$ at a distance $x$ from the point O. Then $r=x \tan \alpha$ and volume of the disc $=\pi x^{2} \tan ^{2} \alpha d x$ Hence, its mass $d m=\pi x^{2} \tan \alpha d x \cdot \rho$ (where $\rho=$ density of the cone $\left.=m / \frac{1}{3} \pi R^{2} h\right)$ Moment of inertia of this element, about the axis $O A$, $d I=d m \frac{r^{2}}{2}$ $=\left(\pi x^{2} \tan ^{2} \alpha d x\right) \frac{x^{2} \tan ^{2} \alpha}{2}$ $=\frac{\pi \rho}{2} x^{4} \tan ^{4} \alpha d x$ Thus the sought moment of inertia $=\frac{\pi \rho R^{4} \cdot h^{5}}{10 h^{4}}\left(\right.$ as $\left.\tan \alpha=\frac{R}{h}\right)$ Hence $\quad I=\frac{3 m R^{2}}{10} \quad\left(\right.$ putting $\left.\rho=\frac{3 m}{\pi R^{2} h}\right)$

(a) Consider an elementry disc of thickness $d x$. Moment of inertia of this element about the $z$ -axis, passing through its $\mathrm{C} . \mathrm{M}$. $d I_{z}=\frac{(d m) R^{2}}{2}=\rho S d x \frac{R^{2}}{2}$
where $\rho=$ density of the material of the plate and $S=$ area of cross section of the plate. Thus the sought moment of inertia
$=\frac{\pi}{2} \rho b R^{4}\left(\right.$ as $\left.S=\pi R^{2}\right)$putting all the vallues we get, $I_{z}=2 \cdot \mathrm{gm} \cdot \mathrm{m}^{2}$
(b) Consider an element disc of radius $r$ and thickness $d x$ at a distance $x$ from the point
O. Then $r=x \tan \alpha$ and volume of the disc $=\pi x^{2} \tan ^{2} \alpha d x$
Hence, its mass $d m=\pi x^{2} \tan \alpha d x \cdot \rho$ (where $\rho=$ density of the cone $\left.=m / \frac{1}{3} \pi R^{2} h\right)$ Moment of inertia of this element, about the axis $O A$, $d I=d m \frac{r^{2}}{2}$
$=\left(\pi x^{2} \tan ^{2} \alpha d x\right) \frac{x^{2} \tan ^{2} \alpha}{2}$
$=\frac{\pi \rho}{2} x^{4} \tan ^{4} \alpha d x$
Thus the sought moment of inertia
$=\frac{\pi \rho R^{4} \cdot h^{5}}{10 h^{4}}\left(\right.$ as $\left.\tan \alpha=\frac{R}{h}\right)$
Hence $\quad I=\frac{3 m R^{2}}{10} \quad\left(\right.$ putting $\left.\rho=\frac{3 m}{\pi R^{2} h}\right)$

Key Concepts

Geometric Representation and Slicing of Solids

The slicing method involves dividing complex solids into simpler shapes whose properties are easier to evaluate. By understanding the geometry of each slice—for example, a thin disc or a circular element of a cone—one can express dimensions such as radius and thickness in terms of a variable coordinate and then integrate these expressions to find the overall moment of inertia or other related quantities.

Density and Mass Distribution

Density, which is mass per unit volume (or area in two-dimensional objects), links the geometric properties of a shape with its mass. In problems involving continuous mass distributions, density facilitates the conversion from differential geometric elements (such as area or volume) to differential mass elements, ensuring that the integration accounts for the material’s mass distribution.

Integration Method in Rotational Dynamics

Integration is employed to sum the contributions of all infinitesimal mass elements to a physical property, such as the moment of inertia. This method involves establishing an appropriate differential element, relating its physical dimensions with the geometry of the object, and then integrating over the relevant limits to achieve the total value.

Moment of Inertia

This concept represents a body's resistance to angular acceleration about an axis. It is calculated by summing or integrating the products of mass elements and the square of their distances from the rotation axis. Understanding moment of inertia is crucial when analyzing rotational dynamics, as it directly affects the torque and angular acceleration relationships.

Infinitesimal Mass Elements

The use of differential mass elements (dm) allows for the calculation of physical quantities, like moment of inertia, for bodies with continuous mass distributions. By slicing a body into infinitesimally small pieces, one can integrate over the entire volume or area to obtain the aggregate effect, such as the total moment of inertia.

Transcript

00:01 In this problem drawing the diagram first so the diagram looks something like this just look at it carefully.

00:08 This is b here it is x -axis in this side it is y -axis this side length is a this point it is c now consider an element elementary disk of thickness d x moment of energy of the element about the z -axis passing through its center of mass is given by d i .z is equal to d m r square by 2 which is equal to row s d x r square by 2.

00:54 Also i can write the value of i z is equal to row s r square by 2 integration of 0 to b d x which is equal to r square by 2 row s b which is equal to r square by 2 row as b which is is equal to pi by 2 row b r to the power 4 as the value of s is equal to pi r square so this is the answer for part a putting all the value we will get the value of i z as 2 gram meter square now i will solve part b so here i can draw the diagram something like this just look at it carefully after that i will simplify the problem so i'm just drawing a rough diagram so that you can learn the concept so this is of thickness d x this is r this is z axis and this length is b so here i can write r is equal to tan alpha and the volume of ddd is equal to pi x square 10 square alpha d x.

02:25 Also i can write d i is equal to d m r square by 2.

02:31 Which on further simplification i can write it as pi x square 10 square alpha d x x squared 10 alpha by 2 which is equal to pi row by 2 x to the power 4 10 to the power 4 alpha d x this out of the moment of inertia, i is equal to pi row by 2, 10 to the power 4 alpha integration of 0 to h, h to the power 4 d x.

03:08 It is because here i just draw a rough diagram.

03:13 Just look at it carefully...

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