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Moment of inertia of a thin rod of mass M and length l about an axis perpendicular to the rod and passing through one end is equal to

Option: 1

\frac{Ml^{2}}{12}

 

 

 


Option: 2

\frac{Ml^{2}}{4}


Option: 3

\frac{Ml^{2}}{3}


Option: 4

\frac{Ml^{2}}{2}


Answers (1)

best_answer

As we learned

Paraller Axis Theorem -

I_{b\: b'}=I_{a\: a'}+Mh^{2}

- wherein

b\: b' is axis parallel to a\: a' & a\: a' an axis passing through centre of mass.

Moment of inertia of rod about axis passly through CM and perpendicular to length is \frac{Ml^{2}}{12}

passing through one end by parallel axis theorm

I=I_{c}+md^{2}

=\frac{ml^{2}}{12}+\frac{ml^{2}}{4}

=\frac{ml^{2}}{3}

 

Posted by

Ritika Harsh

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