A uniform-density disk of mass 13 kg, thickness 0.5 m. and radius 0.2 m make one complete rotation every 0.6 s. What is the rotational kinetic energy of the disk?

The given data can be listed below as,

• The mass of disk is, m=13 kg.
• The thickness of disk is, t=0.5 m .
• The radius of disk is,r=0.2 m .
• The rotation period is, T=0.6 s.

The rotational kinetic energy is the rotational kinetic energy of a rotating rigid body or particle system.

The relation of rotational kinetic energy is expressed as,

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

Here $K_{rot}$is the rotational kinetic energy, $\omega$is the angular speed and $I$is the moment of inertia.

The value of the moment of inertia and angular velocity for the disk is expressed as,

$I=\frac{1}{2}m{r}^2$

And

$\omega =\frac{2\pi }{T}$

Here m is the mass of disk, r is the radius of disk and T is the rotation period.

Substitute the value of $T$and $\omega$in the equation (i).

$$\begin{gathered} K_{rot}=\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{2\pi }{T}\right)}^2 \\ {K}_{rot}=\frac{1}{4}m{r}^2{\left(\frac{2\pi }{T}\right)}^2 \end{gathered}$$

Substitute $13\mathrm{kg}\mathrm{for}m,0.2\mathrm{m}\mathrm{for}r,\mathrm{and}0.6\mathrm{s}\mathrm{for}T$in the above equation.

$$\begin{gathered} K_{rot}=\frac{1}{4}\times 13\mathrm{kg}\times {\left(0.2\mathrm{m}\right)}^2{\left(\frac{2\pi }{T}\right)}^2 \\ =14.26\mathrm{J} \end{gathered}$$

Hence the rotational kinetic energy is,$14.26\mathrm{J}$.