Question

A uniform block of granite in the shape of a book has face dimensions of 20 cm and 15 cmand a thickness of 1.2 cm. The density (mass per unit volume) of granite is2.64 g/cm³. The block rotates around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that axis is 0.104 kg.m²/s. What is its rotational kinetic energy about that axis?

Short answer

Angular kinetic energy is 0.62 J.

Step-by-step solution

Step 1: Given Data

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

Determining the concept

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.

Determining the angular kinetic energy

Mass,

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

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,

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

So,

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

Hence,angular kinetic energy is 0.62 J.

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