Question
Seven identical circular planar disks, each of mass $$M$$ and radius $$R$$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $$P$$ is:
Seven identical circular planar disks, each of mass $$M$$ and radius $$R$$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $$P$$ is:
A.
$$\frac{{19}}{2}M{R^2}$$
B.
$$\frac{{55}}{2}M{R^2}$$
C.
$$\frac{{73}}{2}M{R^2}$$
D.
$$\frac{{181}}{2}M{R^2}$$
Answer :
$$\frac{{181}}{2}M{R^2}$$
Solution :
Using parallel axes theorem, moment of inertia about $$‘O'$$
$$\eqalign{ & {I_o} = {I_{cm}} + m{d^2} \cr & = \frac{{7M{R^2}}}{2} + 6\left( {M \times {{\left( {2R} \right)}^2}} \right) = \frac{{55M{R^2}}}{2} \cr} $$
Again, moment of inertia about point $$P,\,{I_p} = {I_o} + m{d^2}$$
$$ = \frac{{55M{R^2}}}{2} + 7M{\left( {3R} \right)^2} = \frac{{181}}{2}M{R^2}$$
Using parallel axes theorem, moment of inertia about $$‘O'$$
$$\eqalign{ & {I_o} = {I_{cm}} + m{d^2} \cr & = \frac{{7M{R^2}}}{2} + 6\left( {M \times {{\left( {2R} \right)}^2}} \right) = \frac{{55M{R^2}}}{2} \cr} $$
Again, moment of inertia about point $$P,\,{I_p} = {I_o} + m{d^2}$$
$$ = \frac{{55M{R^2}}}{2} + 7M{\left( {3R} \right)^2} = \frac{{181}}{2}M{R^2}$$
