Question

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?

Short answer

The moment of inertia of the disk is-$0.18kg·m^2.$

The rotational kinetic energy is-$14.22J$

The magnitude of rotational angular momentum is-$2.27kg·m^2/s$

Step-by-step solution

Definition of Inertia and 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.

The given data is-

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, $\omega =1rev/0.5s$

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

$=12.57rad/s$

Find the moment of inertia and the rotational kinetic energy of the disc.

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

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

Find the rotational angular momentum of the disc.

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

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

$=2.27kg·m^2/s$