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Define 'Radius of Gyration'. Derive an expression for it.

Vidyadip · Accepted Answer

The radius of gyration is the root mean square of the perpendicular distances of the constituent masses. It is the perpendicular distance from the axis, square of which when multiplied by the total mass of the body gives the same value of moment of inertia as is given by the actual distribution of mass. It is written as

$\text{K}=\sqrt{\frac{\text{r}^2_1+\text{r}^2_2+\dots+\text{r}^2_{\text{n}}}{\text{n}}}$

$=\sqrt{\frac{\sum\text{r}^2_{\text{i}}}{\text{n}}}$

Suppose a rigid body consists of n particles, each of mass m. Let r₁, r₂, r₃,..., r_n, be the perpendicular distances of these particles from the axis of rotation in the figure.

By the definition, the moment of inertia of the body about the given axis is calculated as

$\text{I}=\text{mr}_1^2+\text{mr}^2_2+\text{mr}^2_3+\dots\text{mr}^2_{\text{n}}$

$=\text{m}(\text{r}^2_1+\text{r}^2_2+\text{r}^2_3+\dots+\text{r}^2_{\text{n}}$

$\text{I}=\text{m}\times\text{n}\Big(\frac{\text{r}^2_1+\text{r}^2_2+\text{r}^2_3+\dots+\text{r}^2_{\text{n}}}{\text{n}}\Big)$

$\text{I}=\text{M}\Big(\frac{\text{r}^2_1+\text{r}^2_2+\text{r}^2_3+\dots+\text{r}^2_{\text{n}}}{\text{n}}\Big)$

where M= m × n = total mass of the body.

If the total mass of the body M which is concentrated at P at a perpendicular distance K from the axis, its moment of inertia would be

$\therefore\text{Mk}^2=\frac{\text{M}(\text{r}^2_1+\text{r}^2_2+\dots+\text{r}^2_{\text{n}}}{\text{n}}$

$\Rightarrow\text{K}=\sqrt{\frac{\text{r}^2_1+\text{r}^2_2+\dots+\text{r}^2_{\text{n}}}{\text{n}}}$

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