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 mass of a wire bent in the shape of the arc if its density (mass per unit length) is.

Short answer

The mass of a wire bent in the shape of the arc is $\frac{13}{6}$.

Step-by-step solution

Definition of Mass

A physical body's mass is the amount of matter it contains. It is also a measure of the body's inertia, or resistance to acceleration (velocity change) when a net force is applied.

Determination of the Mass of the wire

Calculate the mass of the wire.

$$\begin{gathered} M=\int \lambda ds \\ ={\int }_0^2\sqrt{x}\sqrt{1+y^{c2}}dx \\ ={\int }_0^2\sqrt{x}\sqrt{1+\frac{1}{4x}}dx \\ ={\int }_0^2\sqrt{x+\frac{1}{4}d\left(x+\frac{1}{4}\right)} \\ =\frac{2}{3}{\left(x+\frac{1}{4}\right)}^{3/2} \\ =\frac{2}{3}\left[\frac{27}{8}-\frac{1}{8}\right] \\ =\frac{13}{6} \end{gathered}$$

Thus, the mass of the bent wire is $\frac{13}{6}$.