question_answer

A tennis racket can be idealized as a uniform ring of mass M and radius R, attached to a uniform rod also of mass M and length L. The rod and the ring are coplanar and the line of the rod passes through the centre of the ring. The moment of inertia of the object (racket) about an axis through the centre of the ring and perpendicular to its plane is

A)  \[\frac{1}{2}M(6{{R}^{2}}+{{L}^{2}})\]

B)  \[\frac{1}{12}M(18{{R}^{2}}+{{L}^{2}})\]

C)  \[\frac{1}{3}M(6{{R}^{2}}+{{L}^{2}}+3LR)\]

D)  None of the above

Correct Answer: C

Solution :

Moment of inertia of the racket above the given axis \[I={{I}_{ring}}+{{I}_{rod}}\] \[=M{{R}^{2}}+\frac{1}{2}M{{I}^{2}}+\frac{1}{4}M{{I}^{2}}+M{{R}^{2}}+MIx\] \[=\frac{24M{{R}^{2}}+4M{{I}^{2}}+12MIR}{12}\] \[=\frac{1}{3}M(6{{R}^{2}}+{{I}^{2}}+3IR)\]