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Found in: Page 290

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The uniform solid block in Fig 10-38 has mass 0.172kg and edge lengths a = 3.5cm, b = 8.4cm, and c = 1.4cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

The rotational inertia about an axis through one corner and perpendicular to the large faces is, $4.7×{10}^{-4}kg.{m}^{2}$.

See the step by step solution

## Step 1: Listing the given quantities

Masses and coordinates of four particles are

1. The mass of block is, M = 0.172kg.
2. The edge length are, a = 3.5cm, b = 3.5cm, and c = 1.4cm.

## Step 2: Concept and Formula used for the given question

By using the concept of moment of inertia we can calculate the moment of inertia about its axis. For the slab, if the axis is not passing through the center, then we can use the parallel axis theorem to calculate the moment of inertia.

1. For the slab when axis is passing through its center then, ${{\mathbit{l}}}_{\mathbf{c}\mathbf{o}\mathbf{m}}{\mathbf{=}}\frac{\mathbf{M}}{\mathbf{12}}\left({a}^{2}+{b}^{2}\right)$
2. When the axis is not passing through its center, then according to the parallel axis theorem,${\mathbit{l}}{\mathbf{=}}{{\mathbit{l}}}_{\mathbf{c}\mathbf{o}\mathbf{m}}{\mathbf{+}}{\mathbit{m}}{{\mathbit{h}}}^{{\mathbf{2}}}$

## Step 3: Calculation of rotational inertia about an axis through one corner and perpendicular to the large faces

For solving parallel axis theorem, wecalculate the perpendicular distance from the axis passing through its center and corner.

$h=\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}}$

According to the parallel axis theorem,

localid="1661141288241" $\begin{array}{rcl}l& =& {l}_{com}+M{h}^{2}\\ & =& \frac{M}{12}\left({a}^{2}+{b}^{2}\right)+M\left[{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}\right]\\ & =& \frac{M}{12}\left({a}^{2}+{b}^{2}\right)+M\left[{\left(a\right)}^{2}+{\left(b\right)}^{2}\right]\\ l& =& \frac{M}{3}\left({a}^{2}+{b}^{2}\right)\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}l& =& \frac{0.172kg}{3}\left[{\left(3.5×{10}^{-2}m\right)}^{2}+{\left(8.4×{10}^{-2}m\right)}^{2}\right]\\ & =& 4.7×{10}^{-4}kg.{m}^{2}\end{array}$

Therefore, rotational inertia about an axis through one corner and perpendicular to the large faces is, $4.7×{10}^{-4}kg.{m}^{2}$ .