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Q43P

Expert-verifiedFound in: Page 290

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, $4.7\times {10}^{-4}kg.{m}^{2}$.

Masses and coordinates of four particles are

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

**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.**** **

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

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.

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

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

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