A uniform-density disk whose mass is 10 kg and radius is 0.4 m makes one complete rotation every 0.2 s. What is the rotational kinetic energy of the disk?

The given data can be listed below as,

• The mass of disk is, m = 10 kg.
• The radius of disk is, r =0.4 m.
• The rotation period is, T =0.2 s.

The rotational energy, or angular kinetic energy, is the kinetic energy due to the rotation of an object and is a fraction of the total kinetic energy.

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 10 kg for m, 0.4 m for r ,and 0.2 s for Tin the above equation.

$$\begin{gathered} K_{rot}=\frac{1}{4}\times 10\mathrm{kg}\times {\left(0.4\mathrm{m}\right)}^2{\left(\frac{2\pi }{T}\right)}^2 \\ =394.78\mathrm{J} \end{gathered}$$

Hence the rotational kinetic energy is, $394.78\mathrm{J}$.