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Question

The radius of gyration of rod of length L and mass M about an axis perpendicular to its length and passing through a point at a distance L/3 from one of its ends is

A
76L
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B
L29
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C
L3
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D
52L
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Solution

The correct option is C L3
Moment of inertia of a rod about its centre of mass is

Icm=ML212

To calculate the moment of inertia about a point at a distance of L3 from the end we need to find the value of x which is the distance between the COM and this point.

As the COM for a rod lies at the centre of the rod at L2 we have,


We have from parallel axis theorem,

I=Icm+Mx2

I=ML212+M(L6)2

I=ML2[112+136]

I=ML2[436]

If K is the radius of gyration,

MK2=I=ML2[436]

K2=L2[436]

K=L×26=L3

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