Question

In Problems 17 to 30, for the curve $\mathit{y}\mathbf{=}\sqrt{\mathbf{x}}$, between$\mathit{x}\mathbf{=}\mathbf{0}$and$\mathit{x}\mathbf{=}\mathbf{2}$, find:

The moments of inertia about the x axis of a lamina in the shape of the plane area under the curve;

Short answer

The moment of inertia in terms of mass is 0.869M.

Step-by-step solution

Definition of Moments of inertia

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.

Examination of the section of wire

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

Determine the small element.

$$\begin{gathered} \mathrm{dl}=\mathrm{dM}\mathrm{f}{\left(\mathrm{x}\right)}^2 \\ =\mathrm{xdM} \\ =\mathrm{x\lambda ds}\left(\mathrm{x}\right) \\ =\mathrm{\lambda x}\sqrt{1+\left({\mathrm{f}}^{\cancel{\mathrm{c}}}\left(\mathrm{x}\right)\right)\mathrm{dx}} \\ =\mathrm{\lambda }\sqrt{{\mathrm{x}}^2+\frac{\mathrm{x}}{4}\mathrm{dx}} \\ =\mathrm{\lambda }\sqrt{{\left(\mathrm{x}+\frac{1}{8}\right)}^2-\frac{1}{64}\mathrm{dx}} \end{gathered}$$

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

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

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

$l=\lambda {\int }_{1/8}^{17/8}\sqrt{u^2-\frac{1}{64}du}$

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

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

Determination of the Moment of Inertia

Write the expression for the mass of the wire.

$M=\lambda L$

So, the value of L is 2.562.

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