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Rotational Motion

Section: Physics

Question : 5 of 24

Marks: +1, -0

L

Solution:

A thin uniform rod of length

L

, and we need the radius of gyration

k

about an axis passing through its centre and perpendicular to its length.

Formulae:

Moment of inertia (M.I.) about the given axis for a uniform rod is:

I=\; \frac{1}{12} M L^2

The radius of gyration

k

is related to the moment of inertia by

I=M k^2

Substituting:

\;M k^2=\; \frac{1}{12} M L^2

\;k^2=\; \frac{L^2}{12}

\;k=\; \frac{L}{\sqrt{12}}

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