Discussion Forum



1)

Seven identical circular planar discs, each of mass M and radius R are welded symmetrically as show in figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is

382019798_ammmm.JPG


A) $\frac{19}{2}MR^{2}$

B) $\frac{55}{2}MR^{2}$

C) $\frac{73}{2}MR^{2}$

D) $\frac{181}{2}MR^{2}$

Answer:

Option D

Explanation:

First we found moment of inertia (MI) of system using parallel axis theorem about centre of mass, then we use it to find moment of inertia about given axis.

Moment of inertia of an outer disc about the axis through centre is

= $\frac{MR^{2}}{2}+M(2K)^{2}=MR^{2}(4+\frac{1}{2})=\frac{9}{2}MR^{2}$

   38201919_idea.JPG

For 6 such discs,

  Moment of inertia = $6\times \frac{9}{2}MR^{2}=27MR^{2}$

 So, moment of inertia of system

           = $\frac{MR^{2}}{2}-27MR^{2}=\frac{55}{2}MR^{2}$

Hence,

   $I_{p}=\frac{55}{2}MR^{2}+(7M\times 9R^{2})$

    $I_{p}=\frac{181}{2}MR^{2}$

    $I_{system}=\frac{181}{2}MR^{2}$