Question

the radius of gyration of a thin ring of radius R about a tangent perpendicular to its plane is..........​

Answer 1

Answer:

About the tangent perpendicular to the axis,

k= √2 r

Answer 2

The radius of gyration of a thin ring of radius R about a tangent perpendicular to its plane is $\sqrt{2}R\ .$

We know ,

Moment of inertia of ring about its center and perpendicular to its plane is :

$I=MR^2\ .$

So, moment of inertia of ring about a tangent perpendicular to its plane is :

$I'=MR^2+MR^2=2MR^2\ .$

Now, we need to find radius of gyration .

Let , k be radius of gyration.

So,

$Mk^2=2MR^2\\\\k=\sqrt{2}R\ .$    { Product of mass and radius of gyration is equal to moment                        

                          of inertia }

Hence , this is the required solution .

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Rotational Motion

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