Question

Seven identical circular planar disks, each of mass ${}^{\prime }{M}^{\prime }$ and radius ${}^{\prime }{R}^{\prime }$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point ${}^{\prime }{O}^{\prime }$ is?

• A. $\frac{73}{2}M{R}^2$
• B. $\frac{181}{2}M{R}^2$
• C. $\frac{19}{2}M{R}^2$
• D. $\frac{55}{2}M{R}^2$

Answer 1

Answer: (D)

$\frac{55}{2}M{R}^2$

Using parallel axis theorem we have,

$I_0=I_{cm}+m{d}^2$

There are 6 circular disk sorrounding a single disc

Hence adding all the Moment of inertia we get

$I_{net}$=$\frac{7M{R}^2}{2}+6(M\times (2R{)}^2)=\frac{55M{R}^2}{2}$