Rotating Half-Parabola

Assuming that the wire rotates about the Z-axis, the moment of inertia of the wire can be calculated as such:

$dI_z = \sigma (ds)\left(\sqrt{x^2 + y^2}\right)^2$

Where $ds$ is the arc length element.

$ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx = \sqrt{1+4x^2}dx$

This gives:

$dI_z = (x^2 + x^4)\sqrt{1+4x^2}dx$

Therefore:

$I_z = \int_{0}^{1} (x^2 + x^4)\sqrt{1+4x^2}dx$

The location of the point mass $M$ is:

$\vec{r}_1 = \hat{i} + 0 \hat{j} + 0 \hat{k}$

The location of an element arc length of mass $\sigma (ds)$ of the wire is:

$\vec{r}_w = x\hat{i} + x^2\hat{j} + 0 \hat{k}$

$\vec{r} = \vec{r}_1 -\vec{r}_w$

Applying the law of gravitation gives:

$d\vec{F} = \left(\frac{GM\sigma(ds)}{\mid \vec{r} \mid ^3}\right)\vec{r}$

The torque due to this force about the Z - axis is:

$d\vec{T} = \vec{r}_w \times d\vec{F}$

Substituting all expressions results in the following:

$d\vec{T} = -\left(\frac{x^2\,\sqrt{4\,x^2+1}}{{\left({\left(x-1\right)}^2+x^4\right)}^{3/2}}\right) dx\hat{k}$

The magnitude of the net torque about the Z axis is then:

$T_z = \int_{0}^{1} \left(\frac{x^2\,\sqrt{4\,x^2+1}}{{\left({\left(x-1\right)}^2+x^4\right)}^{3/2}}\right)dx$

The angular acceleration at this instant is, therefore:

$\boxed{\alpha_z = \frac{T_z}{I_z} \approx 1.9651}$