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Q 5CQ

Expert-verifiedFound in: Page 353

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Question****: The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is ${\mathit{M}}{{\mathit{L}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{3}}$. Why is this moment of inertia greater than it would be if you spun a point mass M at the location of the center of mass of the rod (at L/2)? (That would be ${\mathit{M}}{{\mathit{L}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{4}}$.)**

The moment of inertia of a rod which is spun around an axis through one end perpendicular to its length is larger because it is the sum of moment of inertia of point mass spun at the center of mass and point mass spun at an axis passing through the center perpendicular to the length.

The moment of inertia of a point mass is the product of mass and the square of distance from the axis of rotation. The moment of a body is the sum of moment of inertia of all point masses in the body. It is expressed as follows:

$I=\sum m{r}^{2}$

Here, m is the point mass and r is its distance from the axis of rotation.

The moment of inertia of a point mass M which spun at the center of mass of rod with length L will be equal to $\frac{M{L}^{2}}{4}$. Here the distance is $\frac{L}{2}$. The moment of inertia of this rod, spun around an axis through one end perpendicular to its length, is equal to the sum of the moments of inertia of point masses spun at the centre of mass and point masses spun at an axis perpendicular to the rod. It is calculated as follows:

$\begin{array}{rcl}I& =& {I}_{center\text{}of\text{}mass}+\frac{M{L}^{2}}{12}\\ & =& \frac{M{L}^{2}}{4}+\frac{M{L}^{2}}{12}\\ & =& \frac{M{L}^{2}}{3}\end{array}$

Therefore, the moment of inertia at an axis through one end perpendicular to length is greater than at an axis through the center of mass because it consists of a moment of inertia at an axis perpendicular to its length along with at the center of mass.

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