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43 P

Expert-verifiedFound in: Page 335

Book edition
9th Edition

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

Pages
1624 pages

ISBN
9781133947271

**Question: Three identical thin rods, each of length L and mass m, are welded perpendicular to one another as shown in Figure P10.43. The assembly is rotated about an axis that passes through the end of one rod and is parallel to another. Determine the moment of inertia of this structure about this axis.**

The moment of inertia of this structure about this axis $I=\frac{11}{12}m{L}^{2}$

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.

This problem is solved by calculating the moment of inertia of the three rods about the y-axis. Then, make use of the fact that the rotation axis is parallel to the y-axis a distance of $\frac{L}{2}$ away and apply parallel axis theorem to determine the moment of inertia about the given axis. The diagram is shown below.

The moment of inertia of a long, thin rod with rotation axis is

$I=\frac{1}{12}m{L}^{2}$

The moment of inertia of the rod lying in the x-axis is

${I}_{x}=\frac{1}{12}m{L}^{2}$

Similarly, the moment of inertia of the rod lying in the z-axis is

${I}_{z}=\frac{1}{12}m{L}^{2}$

And the rod lying in the y-axis contributes zero moment about its own axis.

Hence, the total moment of inertia about y-axis which lies at the center of mass of the rod that lies in the x and z axes

$\begin{array}{rcl}{I}_{\text{CM}}& =& {I}_{x}+{I}_{y}+{I}_{z}\\ & =& \frac{1}{12}m{L}^{2}+0+\frac{1}{12}m{L}^{2}\\ & =& \frac{1}{6}m{L}^{2}\end{array}$

Then, according to parallel-axis theorem, the moment of inertia of the system of the three rods about the y-axis is

$I={I}_{\text{CM}}+M{\left(\frac{L}{2}\right)}^{2}$

where (M) is the total mass of the three rods which are identical.

$\begin{array}{rcl}I& =& \frac{1}{6}m{L}^{2}+3m{\left(\frac{L}{2}\right)}^{2}\\ & =& \frac{1}{6}m{L}^{2}+\frac{3}{4}m{L}^{2}\\ & =& \frac{11}{12}m{L}^{2}\end{array}$

Thus, the solution is $I=\frac{11}{12}\u200am{L}^{2}$.

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