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Expert-verified Found in: Page 326 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A uniform block of granite in the shape of a book has face dimensions of 20 cm and 15 cm and a thickness of 1.2 cm. The density (mass per unit volume) of granite is 2.64 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?

Angular kinetic energy is 0.62 J.

See the step by step solution

## Step 1: Given Data

$\mathrm{l}=20\mathrm{cm}=0.2\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{w}=15\mathrm{cm}=0.15\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{t}=1.2\mathrm{cm}=0.012\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{\rho }=2.64\mathrm{g}/{\mathrm{cm}}^{3}=2.64×{10}^{-3}×{10}^{6}\mathrm{kg}/{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{L}=0.104\mathrm{kg}.{\mathrm{m}}^{2}/\mathrm{s}\phantom{\rule{0ex}{0ex}}$

## Step 2: 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, L is angular momentum and I is moment of inertia.

## Step 3: Determining the angular kinetic energy

Mass,

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

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

$\mathrm{r}=\sqrt{{\left(\frac{\mathrm{L}}{4}\right)}^{2}+{\left(\frac{\mathrm{W}}{4}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\sqrt{{\left(\frac{0.2}{4}\right)}^{2}+{\left(\frac{0.15}{4}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{r}=0.0625\mathrm{m}$

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,

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

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. ### Want to see more solutions like these? 