Question

Two masses each of mass M are attached to the end of a rigid massless rod of length L.The moment of inertia of the system about an axis passing centre of mass and perpendicular to its length is :

• A. $\frac{M{L}^2}{4}$
• B. $\frac{M{L}^2}{2}$
• C. $M{L}^2$
• D. $2M{L}^2$

Answer 1

The correct option is B $\frac{M{L}^2}{2}$

Given,

Both objects have mass $=M$

The separation between them $=L$

Both objects has equal mass, canter of mass will be at the axis.

Distance of object from center of mass $d=\frac{L}{2}$

Moment of Inertia = Mass X (Distance from axis)²

Total moment of Inertia = Moment of Inertia of object 1 + Moment of Inertia of object 1

$I=M{d}^2+M{d}^2$

$I=M{\left(\frac{L}{2}\right)}^2+M{\left(\frac{L}{2}\right)}^2$

$I=\frac{M{L}^2}{2}$

Moment of Inertia of system $I=\frac{M{L}^2}{2}$