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Found in: Page 335

### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271

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# 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.

${I}_{total}=1.28 \text{kg}·{\text{m}}^{2}$

See the step by step solution

## 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 $\left({R}_{\text{s}1}=16.5\text{cm}\right)$ and outer radius $\left({R}_{\text{s}2}=30.5\text{cm}\right)$ and thickness $\left({t}_{r}=2.5\text{cm}\right)$

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

## Step: 3 Calculations for the side wall

First: The moment of inertia of the side wall is

${I}_{\text{s}}=\frac{1}{2}{m}_{\text{s}}\left({R}_{\text{s}2}+{R}_{\text{s}1}\right)$ → (1)

The mass of the side wall $\left({m}_{\text{s}}\right)$ is found by

$\begin{array}{rcl}{m}_{\text{s}}& =& \rho {V}_{s}\\ & =& \rho \pi \left({R}_{\text{s}2}^{2}-{R}_{\text{s}1}^{2}\right)t\end{array}$

Substituting the numerical values, we get

$\begin{array}{rcl}{m}_{\text{s}}& =& \pi \left(1.1×{10}^{3}\right)\left({0.305}^{2}-{0.165}^{2}\right)\left(0.635×{10}^{-2}\right)\\ & =& 1.44\text{kg}\end{array}$ → (2)

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

$\begin{array}{rcl}{I}_{\text{s}}& =& \frac{1}{2}\left(1.44\right)\left({0.165}^{2}+{0.305}^{2}\right)\\ & =& 8.68×{10}^{-2}\text{kg}·{\text{m}}^{2}\end{array}$

## Step: 4 Calculation for the tread wall.

Second: The moment of inertia of the tread wall is

${I}_{\text{t}}=\frac{1}{2}{m}_{\text{t}}\left({R}_{\text{t}2}^{2}+{R}_{\text{t}1}^{2}\right)$ → (3)

The mass of the tread wall $\left({m}_{\text{t}}\right)$ is found by

$\begin{array}{rcl}{m}_{\text{t}}& =& \rho {V}_{s}\\ & =& \rho \pi \left({R}_{\text{t}2}^{2}-{R}_{\text{t}1}^{2}\right){t}_{r}\end{array}$

Substituting the numerical values, we get

$\begin{array}{rcl}{m}_{\text{t}}& =& \pi \left(1.1×{10}^{3}\right)\left({0.33}^{2}-{0.305}^{2}\right)\left(20×{10}^{-2}\right)\\ & =& 11 \text{kg}\end{array}$ → (4)

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

$\begin{array}{rcl}{I}_{\text{t}}& =& \frac{1}{2}\left(11\right)\left({0.33}^{2}+{0.305}^{2}\right)\\ & =& 1.11 \text{kg}·{\text{m}}^{2}\end{array}$

## Step: 5 Calculating inertia for the entire tire.

Hence, the total inertia for the entire tire is

${I}_{\text{total}}=2{I}_{\text{s}}+{I}_{\text{t}}$

Substituting the numerical values, we get

${I}_{total}=2\left(8.68×{10}^{-2}\right)+1.11$

$=1.28 \text{kg}·{\text{m}}^{2}$

The solution is ${I}_{total}=1.28\text{kg}·{\text{m}}^{2}$.

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