91Ó°ÊÓ

The sum of the products of each element of mass in the body and the square of the element's distance from the axis is a measure of a body's resistance to rotational acceleration about a certain axis.

Examine a tiny section of wire at $f\left(x\right)$from the -axis,

Determine the small element.

$$\begin{gathered} \mathrm{dl}=\mathrm{dM}\mathrm{f}{\left(\mathrm{x}\right)}^2 \\ =\mathrm{xdM} \\ =\mathrm{³\ae λ»å²õ}\left(\mathrm{x}\right) \\ =\mathrm{λ³\ae }\sqrt{1+\left({\mathrm{f}}^{\cancel{\mathrm{c}}}\left(\mathrm{x}\right)\right)\mathrm{dx}} \\ =\mathrm{λ}\sqrt{{\mathrm{x}}^2+\frac{\mathrm{x}}{4}\mathrm{dx}} \\ =\mathrm{λ}\sqrt{{\left(\mathrm{x}+\frac{1}{8}\right)}^2-\frac{1}{64}\mathrm{dx}} \end{gathered}$$

Combining all of the bits and integrate in the limits 0 to 2.

$$\begin{gathered} l={∫}_0^{2}dl\left(x\right) \\ =\mathrm{λ}{∫}_0^2\sqrt{{\left(x+\frac{1}{8}\right)}^2-\frac{1}{64}dx} \end{gathered}$$

It is known that$u=x+\frac{1}{8}$ and role="math" localid="1658834364864" $du=dx$. Substitute it in the above expression.

$l=\mathrm{λ}{∫}_{1/8}^{17/8}\sqrt{u^2-\frac{1}{64}du}$

It is known that $u=\frac{1}{8}\cosh \mathrm{θ}$and $du=\frac{1}{8}\sin h\mathrm{θ}d\mathrm{θ}$. Substitute it in the above expression.

$$\begin{gathered} l=\frac{\mathrm{λ}}{64}{∫}_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\sqrt{\cos h^2\mathrm{θ}-1\sin h\mathrm{θ}d\mathrm{θ}} \\ =\frac{\mathrm{λ}}{64}{∫}_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\sin h^2\mathrm{θ}d\mathrm{θ} \\ =\frac{\mathrm{λ}}{128}{∫}_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)}\left(\cos h\left(2\mathrm{θ}\right)-1\right)d\mathrm{θ} \\ =\frac{\mathrm{λ}}{128}{\left[\frac{1}{2}\sin h\left(2\mathrm{θ}\right)-\mathrm{θ}\right]}_{\cos h-1\left(1\right)}^{\cos h-1\left(17\right)} \\ =2.226\mathrm{λ} \end{gathered}$$

Write the expression for the mass of the wire.

$M=\mathrm{λ}L$

So, the value of L is 2.562.

Thus, the moment of inertia in terms of mass is $0.869M$.