Question

The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M,about an axis passing through centre and perpendicular to length of rod is

• A. $\frac {(M+3M)L^2}{12}$
• B. $\frac {(M+6M)L^2}{12}$
• C. $\frac {ML^2}{4}$
• D. $\frac {ML^2}{12}$

Answer 1

Answer: (B)

Moment of inertia of rod about XY
$ I_1=\frac {ML^2}{12}$
Moment of inertia of two masses
$ I_2=m \bigg (\frac {L}{2}\bigg )^2+m\bigg (\frac {L}{2}\bigg )^2= \frac {mL^2}{2}$
$ I=I_1+I_2= \frac {ML^2}{12}+ \frac {mL^2}{2}$
$ = \frac {(M+6m)L^2}{12}$