Question

two rods each of mass M and length L joined at the centre to form a cross. the moment of inertia of disc about an axis passing through the common centre of roads and perpendicular to the plane formed by them,is​

Answer 1

The moment of inertia of disc about an axis passing through the common center of rods and perpendicular to the plane formed by them is ml²/6.

Given-

• Mass of each rod = M
• Length of each rod = L

We know that moment of inertia about center of one rod = ml²/12

So, Moment of inertial about center of other rod is also = ml²/12

As both moment of inertia are on the common axis. So,

Net moment of inertia = ml²/12 + ml²/12 = ml²/6

Answer 2

Net Moment of Inertia is $\frac {Ml^2}{6}$

Explanation:

Consider the line perpendicular to the plane of the figure through the center of the cross.

• The moment of inertia of each rod about this line (say about z-axis)  

$I' = \frac {Ml^2}{12}$

• Hence the moments of inertia of the cross $I_z = 2I'_z = \frac {Ml^2}{6}$

• The moments of inertia of the cross about the bisector are equal by symmetry

$I_x = I_y$

according to the theorem of perpendicular axis,

$I_Z = I_X + I_Y$

the moment of inertia of the cross about the bisector is

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

Net Moment of Inertia = $\frac {Ml^2}{12} + \frac {Ml^2}{12} = \frac {Ml^2}{6}$