A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

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A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

nd its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

Physics

1 answer:

inna [3.1K] · Answer 1 · 2026-03-11T19:20:41+03:00

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

The moment of inertia for a rod with mass M and length L about its center is represented as 1/12ML².

(a) When the rod is bent at its center, all points maintain equal distance from the axis. Therefore, as the moment of inertia depends on the distance of every mass to this axis, it stays unchanged, yielding I_A=1/12ML².

(b) The two ends along with the intersection point create a right triangle. The distance between the ends, d, can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, forming another right triangle using the meeting point, the midpoint to the two ends of the rod, and one end itself allows us to compute the distance between the two axes, x, also using the Pythagorean Theorem:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, applying the Parallel Axis Theorem helps us in calculating I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$