Tilting Parabola

Consider an arc length element of the parabola at a horizontal distance $x$ from the origin. The elementary arc length is:

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

Since the parabola is $y = x^2$ , the arc length element is:

$ds = \sqrt{1 + 4x^2} \ dx$

Therefore, the mass of this arc length element is:

$dm = \sigma \ ds = \sigma \sqrt{1 + 4x^2} \ dx$

At the initial orientation, the gravitational potential energy of the element is:

$dV = dm \ g y= \left(\sigma \sqrt{1 + 4x^2} \ dx\right)gx^2$

The total initial potential energy of the parabolic wire is therefore:

$\boxed{V_{initial} = \int_{-2}^{2} \sigma gx^2\sqrt{1 + 4x^2} \ dx}$

Now, the moment of inertia of the element arc length about the Z-axis is:

$dI = dm \ \left(x^2 + y^2\right) = \left(\sigma \sqrt{1 + 4x^2} \ dx\right) \left(x^2 + x^4\right)$

$\boxed{I = \int_{-2}^{2} \sigma \left(x^2 + x^4\right)\sqrt{1 + 4x^2} \ dx}$

Now, as the wire rotates starting from rest, about the Z-axis, its axis of symmetry moves from being vertically upwards to horizontal along positive X. At this instant, curve describing the parabola is $x = y^2$ . The potential energy of the wire at this configuration can be calculated in the same manner above and the value will come out to be zero. Thus, $V_{final} = 0$ . Now, applying the energy conservation principle:

$V_{initial} + KE_{initial} = V_{final} + KE_{final}$

$V_{initial} + 0 = 0 + \frac{1}{2}I \omega^2$

$\implies \boxed{\omega = \sqrt{\frac{2V_{initial}}{I}}}$

The integrals are computed using Wolfram -Alpha and the steps of computation are left out here.

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Final potential energy calculation:

Given the curve: $x = y^2$

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

$ds = \sqrt{1 + 4y^2} \ dy$

Therefore, the mass of this arc length element is:

$dm = \sigma \ ds = \sigma \sqrt{1 + 4y^2} \ dy$

At the final orientation, the gravitational potential energy of the element is:

$dV = dm \ g y= \left(\sigma \sqrt{1 + 4y^2} \ dy\right)gy$

The total final potential energy of the parabolic wire is therefore:

$V_{final} = \int_{-2}^{2} \sigma gy\sqrt{1 + 4y^2} \ dy = -\int_{-2}^{2} \sigma gy\sqrt{1 + 4y^2} \ dy \implies 2V_{final} = 0 \implies \boxed{V_{final}=0}$