Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

Question: Figure P10.41 shows a side view of a car tire before itis mounted on a wheel. Model it as having two side-walls of uniform thickness 0.635 cm and a tread wall of uniform thickness 2.50 cm and width 20.0 cm. Assume the rubber has uniform density 1.10 3 103 kg/m3. Find its moment of inertia about an axis perpendicular to the page through its center.

$I_{t o t a l} = 1.28 kg \cdot m^{2}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of moment of inertia

Moment of inertia about a given axis of rotation resists a change in its rotational motion; it can be regarded as a measure of rotational inertia of the body.

Step 2: Finding the wire consists of three parts

1-Two side walls which are modeled as hollow cylinder with inner radius $(R_{s 1} = 16.5 cm)$ and outer radius $(R_{s 2} = 30.5 cm)$ and thickness $(t_{r} = 2.5 cm)$

2-Tread wall which is modeled as hollow cylinder of inner radius $(R_{t 1} = 30.5 cm)$ outer radius localid="1663670015744" $(R_{t 2} = 33 cm)$ width $(w = 20 cm)$ and thickness $(t_{r} = 2.5 cm)$ .

Step: 3 Calculations for the side wall

First: The moment of inertia of the side wall is

$I_{s} = \frac{1}{2} m_{s} (R_{s 2} + R_{s 1})$ → (1)

The mass of the side wall $(m_{s})$ is found by

$\begin{array}{rcl} m_{s} & = & ρ V_{s} \\ = & ρ π (R_{s 2}^{2} - R_{s 1}^{2}) t \end{array}$

Substituting the numerical values, we get

$\begin{array}{rcl} m_{s} & = & π (1.1 \times 10^{3}) ({0.305}^{2} - {0.165}^{2}) (0.635 \times 10^{- 2}) \\ = & 1.44 kg \end{array}$ → (2)

Substituting for from Equation (2) into Equation (1) and substituting numerical values, we get

$\begin{array}{rcl} I_{s} & = & \frac{1}{2} (1.44) ({0.165}^{2} + {0.305}^{2}) \\ = & 8.68 \times 10^{- 2} kg \cdot m^{2} \end{array}$

Step: 4 Calculation for the tread wall.

Second: The moment of inertia of the tread wall is

$I_{t} = \frac{1}{2} m_{t} (R_{t 2}^{2} + R_{t 1}^{2})$ → (3)

The mass of the tread wall $(m_{t})$ is found by

$\begin{array}{rcl} m_{t} & = & ρ V_{s} \\ = & ρ π (R_{t 2}^{2} - R_{t 1}^{2}) t_{r} \end{array}$

Substituting the numerical values, we get

$\begin{array}{rcl} m_{t} & = & π (1.1 \times 10^{3}) ({0.33}^{2} - {0.305}^{2}) (20 \times 10^{- 2}) \\ = & 11 kg \end{array}$ → (4)

Substituting for from Equation (4) into Equation (3) and substituting numerical values, we get

$\begin{array}{rcl} I_{t} & = & \frac{1}{2} (11) ({0.33}^{2} + {0.305}^{2}) \\ = & 1.11 kg \cdot m^{2} \end{array}$

Step: 5 Calculating inertia for the entire tire.

Hence, the total inertia for the entire tire is

$I_{total} = 2 I_{s} + I_{t}$

Substituting the numerical values, we get

$I_{t o t a l} = 2 (8.68 \times 10^{- 2}) + 1.11$

$= 1.28 kg \cdot m^{2}$

The solution is $I_{t o t a l} = 1.28 kg \cdot m^{2}$ .

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Definition of moment of inertia

Step 2: Finding the wire consists of three parts

Step: 3 Calculations for the side wall

Step: 4 Calculation for the tread wall.

Step: 5 Calculating inertia for the entire tire.

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Physics For Scientists & Engineers

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Short Answer

Step by Step Solution

Step 1: Definition of moment of inertia

Step 2: Finding the wire consists of three parts

Step: 3 Calculations for the side wall

Step: 4 Calculation for the tread wall.

Step: 5 Calculating inertia for the entire tire.

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