Question

The moment of inertia of a uniform thin rod of length L and mass M about an axis passing through a point at a distance of L//3 from one of its ends and perpendicular to the rod is

Answer 1

Answer:

ML^2/12+4ML^2/9

Explanation:

As the rod is at a distance L/3 therefore at 2L/3 from center

from par alle axis theorem

I at centre +I at 2L/3 from center

ML^2/12+M(2L/3)^2

=ML^2/12+4ML^2/9

MAY IT HELPS U

Answer 2

Answer:

Given:

Mass of rod = M

Length of rod = L

To find:

Moment of Inertia through an axis L/3 distance from the end.

Calculation:

We know that , Moment of Inertia of rod along an axis passing through the centre of rod is :

$I_{1} \: = \dfrac{M {L}^{2} }{12}$

The new axis is L/3 from end. So the position of the axis from the centre is :

$= \dfrac{L}{2} - \dfrac{L}{3} = \dfrac{L}{6}$

As per Parallel Law , we can say that the new Moment of Inertia along the new axis will be :

$I_{2} \: = I_{1} + M { \bigg(\dfrac{L}{6} \bigg) }^{2}$

$I_{2} \: =\dfrac{M {L}^{2} }{12} + M { \bigg(\dfrac{L}{6} \bigg) }^{2}$

$I_{2} \: =\dfrac{M {L}^{2} }{12} +\dfrac{M {L}^{2} }{36}$

$I_{2} \: =\dfrac{4M {L}^{2} }{36}$

$I_{2} \: =\dfrac{M {L}^{2} }{9}$

So final answer :

$\boxed{ \sf{ \green{ \huge{ I_{2} \: =\dfrac{M {L}^{2} }{9}}}}}$