The moment of inertia of a uniform -density disk rotating about an axle through its center can be shown to be $\frac{1}{2}M{R}^2$. This result is obtained by using integral calculus to add up the contributions of all the atoms in the disk. The factor of $\frac{1}{2}$reflects the fact that some of the atoms are near the center and some are far from the center; the factor of $\frac{1}{2}$is an average of the square distance. A uniform-density disk whose mass is $16kg$and radius is localid="1668665053754" $0.15m$makes one complete rotation every$0.5s$(a) what is the moment of inertia of this disk? (b) what is its rotational kinetic energy? (c)what is the magnitude of its rotational angular momentum?

The moment of inertia is a quantitative measure of a body's rotational inertia—that is, the body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force).

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

Mass of the uniform disk,$M=16kg$

Radius of the disk,$R=0.15m$

Period of rotation,$T=0.5s$

Angular speed of the disk, $\omega =1rev/0.5s$

$=\left(1rev/0.5s\right)\left(\frac{2\pi rad}{1rev}\right)$

$=12.57rad/s$

The moment of inertia of a uniform disc rotating around an axle via its centre is calculated as follows:

$I=\frac{1}{2}M{R}^2.....\left(1\right)$

On substituting the known values in the equation (1), the moment of inertia of the disk can be calculated as

$I=\frac{1}{2}\left(16kg\right){\left(0.15m\right)}^2$

$=0.18kg·m^2.$

The rotational kinetic energy of an object rotating with angular speed $\left(\omega \right)$is

$K_{rot}=\frac{1}{2}I{\omega }^2.......\left(2\right)$

Here,$I\to$moment of inertia of the object.

On substituting the numerical values in the equation (1), the rotational kinetic energy of the disk can be calculated as

$$\begin{gathered} K_{rot}=\frac{1}{2}\left(0.18kg·m^2\right){\left(12.57rad/s\right)}^2 \\ =14.22J \end{gathered}$$

The rotational angular momentum of an object rotating with angular speed $\left(\omega \right)$in terms of its moment of inertia is

$L_{rot}=I\omega .....\left(3\right)$

Here,$I\to$moment of inertia of the object

On substituting the given values in the equation (3) , the magnitude of the rotational angular momentum can be calculated as

$L_{rot}=\left(0.18kg·m^2\right)\left(12.57rad/s\right)$

$=2.27kg·m^2/s$