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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: A uniform, thin, solid door has height 2.20 m, width 0.870 m, and mass 23.0 kg. (a) Find its moment of inertia for rotation on its hinges. (b) Is any piece of data unnecessary?

(a)The moment of inertia for rotation on its hinges is $I=5.8 \text{kg}·{\text{m}}^{2}$

(b)The piece of unnecessary data is height.

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.

$I=\frac{L}{\omega }$

## Step 2: Finding the moment of inertia for rotation on hinges

Given:

$M=23\text{kg}$

L (width of the floor)$=0.87\text{m}$

H (height of door) $=2.2\text{m}$

(a)

The door can be modeled as long, thin rod with rotation axis through end, $I=\frac{1}{3}M{L}^{2}$.

Substituting the numerical values, we get

$\begin{array}{rcl}I& =& \frac{1}{3}\left(23\right)\left(0.87{\right)}^{2}\\ & =& 5.8 \text{kg}·{\text{m}}^{2}\end{array}$

So, the answer is $5.8 \text{kg}·{\text{m}}^{2}$.

## Step 3: Finding unnecessary piece of data

(b)

The height of the door is the unnecessary piece of data.

So, the answer is height of the door.

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