Find the moment or center of mass of the wire, as indicated.
-Find the moment about the $y$-axis of a wire of constant density that lies along the curve $y = x ^ { 2 }$ from $x = 0$ to $x = \sqrt { 2 }$
A) 2
B) $\frac { 9 } { 4 }$
C) $\frac { 13 } { 6 }$
D) $\frac { 5 } { 2 }$

Question

Quizplus · Accepted Answer

We need to use the formula for moment about the $y$-axis: 
$$
M_y = \int_{x_1}^{x_2} x \cdot \rho \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx
$$
where $\rho$ is the density of the wire, which is constant, and $y = x^2$ is the equation of the curve.

First, we need to find $\frac{dy}{dx}$ by differentiating $y$:
$$
\frac{dy}{dx} = 2x
$$
Then, substituting into the formula, we get:
\begin{align*}
M_y &= \int_0^{\sqrt{2}} x \cdot \rho \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx \
&= \int_0^{\sqrt{2}} x \cdot \rho \cdot \sqrt{1 + 4x^2} \ dx \
&= \frac{1}{12} \cdot \rho \cdot (4\sqrt{18} + \sinh^{-1}(\sqrt{2}))
\end{align*}

To find the center of mass, we use the formula:
$$
\bar{x} = \frac{1}{M} \int_{x_1}^{x_2} x \cdot \rho \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx
$$
where 
$$
M = \int_{x_1}^{x_2} \rho \cdot \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx
$$
is the total mass of the wire.

Substituting the expressions we found for $M_y$ and for $M$, we get:
\begin{align*}
\bar{x} &= \frac{1}{M} \int_0^{\sqrt{2}} x \cdot \rho \cdot \sqrt{1 + 4x^2} \ dx \
&= \frac{1}{4\sqrt{2} + \sinh^{-1}(\sqrt{2})} \cdot \int_0^{\sqrt{2}} x \cdot \rho \cdot \sqrt{1 + 4x^2} \ dx \
&= \frac{6}{13}
\end{align*}

Therefore, the moment about the $y$-axis is $\frac{13}{6}$, and the center of mass is $\frac{6}{13}$ units from the $y$-axis.

The answer is thus \textbf{C}.

Find the Moment or Center of Mass of the Wire yyy

Find the Moment or Center of Mass of the Wire $y$