Question

The radius of gyration of a uniform rod of length L about an axis passing
through its centre of mass and perpendicular to its length is :​

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

The radius of gyration of a uniform rod of length L about an axis passing through its centre of mass and perpendicular to its length is :

$\huge\underline{\underline{\bf \green{Solution-}}}$

Moment of inertia ( I ) of a uniform rod of length L about axis is

$\large{\boxed{\bf I = \dfrac{ML^2}{12}}}$

Now ,

$\implies{\sf MK^2=\dfrac{ML^2}{12}}$

Here , K = Radius of Gyration

$\implies{\sf K^2=\dfrac{L^2}{12}}$

$\implies{\bf \red{K=\dfrac{L}{\sqrt{12}}}}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Radius of Gyration is ${\bf \red{\dfrac{L^2}{\sqrt{12}}}}$

Answer 2

The radius of gyration is  $k=\dfrac{L}{\sqrt{12}}$  .

Explanation:

Given :

Length of rod = L .

Mass of rod =  M.

Let , radius of gyration is k .

We know product of square radius of gyration and mass is equal to moment of inertia.

So,  $Mk^2=I\\\\k=\sqrt{\dfrac{I}{M}}$    ...... equation 1.

Now, We know moment of inertia of a uniform rod about an axis passing

through its center of mass and perpendicular to its length is :

$I=\dfrac{ML^2}{12}$

Putting above value of I in equation 1 .

We get ,

$k=\sqrt{\dfrac{\dfrac{ML^2}{12}}{M}}\\\\k=\dfrac{L}{\sqrt{12}}$

Hence , this is the required solution.

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