91Ó°ÊÓ

$$\begin{gathered} \mathrm{l}=20\mathrm{cm}=0.2\mathrm{m} \\ \mathrm{w}=15\mathrm{cm}=0.15\mathrm{m} \\ \mathrm{t}=1.2\mathrm{cm}=0.012\mathrm{m} \\ \mathrm{ÒÏ}=2.64\mathrm{g}/{\mathrm{cm}}^3=2.64×{10}^{-3}×{10}^6\mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{L}=0.104\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$

Using the volume and density, find mass. Find the distance from the axis of rotation using the given information. Using the parallel axis theorem, find the rotational inertia of the system. And finally, find the rotational kinetic energy.

Formula is as follow:

$\mathrm{K}.\mathrm{E}.=0.5\frac{{\mathrm{L}}^2}{\mathrm{I}}$

Where, Lis angular momentum and I is moment of inertia.

Mass,

$\begin{array}{rcl}\mathrm{m}& =& \mathrm{ÒÏV}\\ & =& \mathrm{ÒÏ}×\mathrm{I}×\mathrm{w}×\mathrm{t}\\ & =& 2.64×{10}^3×0.2×0.15×0.012\\ & =& 0.9504\mathrm{kg}\end{array}$

Now, distance from center to point about which block spins is given by,

$$\begin{gathered} \mathrm{r}=\sqrt{{\left(\frac{\mathrm{L}}{4}\right)}^2+{\left(\frac{\mathrm{W}}{4}\right)}^2} \\ \mathrm{r}=\sqrt{{\left(\frac{0.2}{4}\right)}^2+{\left(\frac{0.15}{4}\right)}^2} \\ \mathrm{r}=0.0625\mathrm{m} \end{gathered}$$

Rotational inertia of block about axis is,

${\mathrm{I}}_{\mathrm{axis}}=\frac{1}{12}\mathrm{m}\left({\mathrm{L}}^2+{\mathrm{w}}^2\right)$

Now, M.I. about axis of spin is given by parallel axis theorem,

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_{\mathrm{axis}}+{\mathrm{I}}_{\mathrm{m}} \\ \mathrm{I}=\frac{1}{12}\mathrm{m}\left({\mathrm{L}}^2+{\mathrm{w}}^2\right)+{\mathrm{mr}}^2 \\ \mathrm{I}=\frac{1}{12}×0.9504×\left(0.2^2+0.{15}^2\right)+0.9504×0.{0625}^2 \\ \mathrm{I}=8.6625×{10}^{-3}\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

So,

$\begin{array}{rcl}\mathrm{KE}& =& 0.5\frac{{\mathrm{L}}^2}{\mathrm{I}}\\ & =& 0.5×\frac{0.{104}^2}{8.6625×{10}^{-3}}\\ & =& 0.62\mathrm{J}\end{array}$

Hence,angular kinetic energy is 0.62 J.

Therefore, the kinetic energy of rotation can be found using the formula of the energy.