= Solution
The <moment of inertia> of a uniform thin ring about its perpendicular central axis is $M_aa^2$. Its two in-plane diameters are equivalent by <rotational symmetry>, so the <perpendicular axis theorem> gives $M_aa^2/2$ about either diameter. The central rod contribution about an axis perpendicular to its length is $M_ll^2/12$.
The requested axis is parallel to these central axes and passes through the whole racquet's <centre of mass>. Applying the <parallel axis theorem> separately gives
$$
\boxed{I_{\mathrm{CM}}=\frac12M_aa^2+M_aX_{\mathrm{CM}}^2
+\frac1{12}M_ll^2+M_l\left(a+\frac l2-X_{\mathrm{CM}}\right)^2.}
$$
An equivalent reduced expression is
$$
I_{\mathrm{CM}}=\frac12M_aa^2+\frac1{12}M_ll^2
+\frac{M_aM_l}{M_a+M_l}\left(a+\frac l2\right)^2.
$$
Both refer to the in-plane axis perpendicular to the handle; the ring contribution would be different for an axis normal to the racquet face.
Back to article page
