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76P

Expert-verifiedFound in: Page 326

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/cm ^{3}**

Angular kinetic energy is 0.62 J.

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

**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.

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,

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

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,

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

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.

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