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There is a rod of length L and mass M; one line is passing through the center of the rod and another line is passing through the end of the rod; The radius of gyrations ratio of this rod for the given cases (concerning the line positions) center/end is 

Option: 1

1:2


Option: 2

2:1


Option: 3

12:3

 


Option: 4

3:12


Answers (1)

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Moment of inertia of the rod about center =  \frac{1}{12}ML^{2}

Moment of inertia of the rod about the end =  \frac{1}{3}ML^{2}

 The radius of gyration is    K=\sqrt{\frac{I}{M}}

The ratio of Radius of Gyration                                      

\frac{K_{center}}{K_{end}}= \sqrt{\frac{I_{center}}{I_{end}}}

Now the ratio will become                                                    

\frac{K_{center}}{K_{end}}= \sqrt{\frac{\frac{1}{12}ML^{2}}{\frac{1}{3}ML^{2}}}=2:1

Posted by

Divya Prakash Singh

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