x

along the ideal thin rod from its centre of mass. Its perpendicular distance from the inclined axis is

∣ x ∣ sin α

M / L

. The moment of inertia is therefore

I = \frac{M}{L} \int_{- L /2}^{L /2} x^{2} sin^{2} α d x = \frac{M L ^{2}}{12} sin^{2} α .

This tends to zero for an axis along an infinitely thin rod and to

M L^{2} /12

for a perpendicular central axis. A finite rod thickness would add a contribution absent from the stated thin-rod model.

b

Solution

The parallel axis through the centre is the rod's own long axis, with zero moment of inertia in the thin-rod approximation. By the parallel axis theorem, the displaced parallel axis has

I_{D} = 0 + M D^{2} = M D^{2} .

Equivalently, every mass element is the same perpendicular distance

D

from that axis, so the defining integral immediately gives this result.

There are no discussions about this article yet.

There are currently no matching articles.