Question

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the other half. this bent object is rotating at an angular velocity of 9.3 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?

Answer 1

The angular velocity of the straight rod is 4.65 rad/s

Explanation:

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the other half. this bent object is rotating at an angular velocity of 9.3 rad/s about an axis that is perpendicular to the left end of the rod and parallel to the rod's upward half (see the drawing). without the aid of external torques, the rod suddenly assumes its straight shape. what is the angular velocity of the straight rod?

The initial moment of inertia is

$I = \frac{ML^2}{3} + ML^2 = \frac{4ML^2}{3}$ because the vertical portion could be treated as a point mass when we rotate a parallel axis

where  is the length of each half of the rod.

When it straightened,

$I' = \frac{ML^2}{3} + \frac{ML^2}{12} + M \frac{3L}{2}^2 = \frac{ML^2}{12} (4 + 1 + 27) = \frac{8ML^2}{3}$

by the parallel axis theorem on the distant segment.

Then after the conservation of angular momentum:

$\frac{4ML^2}{3} * 9.3 rad/s = \frac{8ML^2}{3} * \omega$

$\omega = 4.65 rad/s$

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Answer 2

Answer: initial I = (m/2)L²/3 + (m/2)L² where L = ½ the length of the rod, and the vertical half can be treated as a point mass. initial I = mL²(1/6 + 1/2) = 2mL²/3 final I = m(2L)²/3 = 4mL²/3 Since I has doubled and momentum is conserved, ω has halved. ω = 3.9 rad/s. By formulae: 2mL²/3 * 7.8rad/s = 4mL²/3 * ω