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 -axis of a lamina in the shape of the plane area under the curve; of a wire bent along the arc of the curve.

Short answer

The moment of inertia in terms of the mass of the lamina is: $l=\frac{2}{5}M$.

Step-by-step solution

Expression for the moment of inertia

The expression for the moment of inertia of a stick about its end is given as follows,

$\mathbf{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{ML}}^{\mathbf{2}}$

Representation of the area bounded by the curve

Draw the bounded region for the curve $y=\sqrt{x}$, between $x=0$and $x=2$.

Calculation of the stick element

Determine the value of the element.

$$\begin{gathered} dl=\frac{1}{2}dM{\left(f\left(x\right)\right)}^2 \\ =\frac{1}{3}xdM \\ =\frac{1}{3}x\delta f\left(x\right)dx \\ =\frac{1}{3}\delta x^{3/2}dx \end{gathered}$$

Calculation of moment of inertia

Separate the lamina into such sticks and integrate over the sticks.

Perform integration in the limits 0 to 2.

$$\begin{gathered} l={\int }_0^{2}dl\left(x\right) \\ ={\int }_0^2\frac{1}{3}\delta x^{3/2}dM \\ {\overline{)=\frac{1}{3}\delta \frac{2}{5}{x}^{5/2}}}_0^2 \\ =\frac{8\sqrt{2}}{15}\delta \end{gathered}$$

Calculation of moment of inertia in terms of the mass of the lamina

Write the expression for the area of the lamina.

$A=\frac{4\sqrt{2}}{3}$

Write the expression for the mass of the lamina.

$M=\delta A$

Substitute $\frac{4\sqrt{2}}{3}$for A in the above expression.

$M=\delta \frac{4\sqrt{2}}{3}$

So, the expression for the moment of inertia in terms of the mass of the lamina is as follows,

role="math" localid="1658832561544" $l=\frac{2}{5}M$

Thus, the moment of inertia in terms of the mass of the lamina is$l=\frac{2}{5}M$ .