Question

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

The moments of inertia of a wire bent along the arc of the curve.

Short answer

The moments of inertia of a wire bent along the arc of the curve is $\frac{2}{3}M$.

Step-by-step solution

Definition of Moment of inertia of circular ring

The moment of inertia of a circular ring along an axis perpendicular to its plane and passing through its center is given as follows,

$$\begin{gathered} \mathbf{l}\mathbf{=}{\mathbf{Mr}}^{\mathbf{2}} \\ \mathbf{}\mathbf{=}\left(2\mathrm{\pi r\lambda }\right){\mathbf{r}}^{\mathbf{2}} \end{gathered}$$

Here, $\mathbf{\lambda }$ is the mass per unit length.

Determination of inertia along an axis perpendicular to its plane

Determine the value of moment of inertia of circular lamina by taking the sum of the rings.

$$\begin{gathered} l={\int }_0^R{r}^2\left(2\mathrm{\pi r\delta dr}\right) \\ =2\mathrm{\pi \delta }{\int }_0^{\mathrm{R}}{\mathrm{r}}^3\mathrm{dr} \\ =\frac{\mathrm{\pi }}{2}{\mathrm{\delta R}}^4 \\ =\frac{{\mathrm{MR}}^4}{2} \end{gathered}$$

Here, $\delta$is mass per unit area and the value of mass is $M=\delta R^2\mathrm{\pi }$.

Determination of moment of inertia

Find the value of $dl$and $dM$.

$$\begin{gathered} dl=\frac{dM{\left(f\left(x\right)\right)}^2}{2} \\ dM=pdx{\left(f\left(x\right)\right)}^2\mathrm{\pi } \end{gathered}$$

Take sum of the moment of inertia.

$$\begin{gathered} l={\int }_0^2\frac{1}{2}p\mathrm{\pi }{\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2\mathrm{dx} \\ =\frac{p\pi }{6}{\mathrm{x}}^3 \\ ={\overline{)\frac{p\pi }{6}{\mathrm{x}}^3}}_0^2 \\ =\frac{4}{3}p\pi \end{gathered}$$

Write the expression for the mass of the volume.

$$\begin{gathered} M=Vp \\ =2\mathrm{\pi p} \end{gathered}$$

Thus, the moment of inertia of a wire bent along the arc of the curve is $\frac{2}{3}M$.