Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

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

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:

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

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

Step 3: Determining the angular kinetic energy

Mass,

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

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

$r = \sqrt{{(\frac{L}{4})}^{2} + {(\frac{W}{4})}^{2}} r = \sqrt{{(\frac{0.2}{4})}^{2} + {(\frac{0.15}{4})}^{2}} r = 0.0625 m$

Rotational inertia of block about axis is,

$I_{axis} = \frac{1}{12} m (L^{2} + w^{2})$

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

$I = I_{axis} + I_{m} I = \frac{1}{12} m (L^{2} + w^{2}) + {mr}^{2} I = \frac{1}{12} \times 0.9504 \times (0 . 2^{2} + 0 . 15^{2}) + 0.9504 \times 0 . 0625^{2} I = 8.6625 \times 10^{- 3} kg . m^{2}$

So,

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

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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