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Step-by-Step Solution
Step 1: Separate the racket into two parts
The badminton racket can be modeled as consisting of two distinct parts:
A straight (linear) portion (the handle) of mass M and length r .
A circular ring portion (the head) of mass M and radius r .
We treat each part separately and then add their moments of inertia about the chosen axis.
Step 2: Identify the axis of rotation
We need the moment of inertia about an axis:
Perpendicular to the handle.
In the plane of the ring.
Located at a distance \tfrac{r}{2} from the end A of the handle.
Since the linear portion (handle) extends from x=0 to x=r, the axis at x=\tfrac{r}{2} passes through the midpoint of the rod.
Step 3: Moment of inertia of the linear (rod) portion
For a uniform rod of mass M and length r, the moment of inertia about its center (midpoint) for an axis perpendicular to its length is:
I_{\text{rod about midpoint}} \;=\; \frac{1}{12} \, M \, r^2.
Because our axis is exactly at the midpoint ( x = \tfrac{r}{2} ), no additional parallel axis shift is needed for the rod.
Step 4: Moment of inertia of the circular (ring) portion
1. Moment of inertia of a ring about any diameter (in its plane) through its center is
I_{\text{ring center, in-plane}} \;=\; \frac{1}{2} \, M \, r^2.
2. Parallel axis shift: The center of the ring is at x = r, so the displacement of the chosen axis from the ring's center is
d \;=\; r - \frac{r}{2} \;=\; \frac{r}{2}.
Using the parallel axis theorem:
I_{\text{ring about given axis}}
\;=\;
I_{\text{ring center, in-plane}}
\;+\;
M \, d^2
\;=\;
\frac{1}{2} \, M \, r^2
\;+\;
M \,\Bigl(\frac{r}{2}\Bigr)^2
\;=\;
\frac{1}{2} \, M \, r^2 \;+\; \frac{1}{4} \, M \, r^2
\;=\;
\frac{3}{4} \, M \, r^2.
Step 5: Total moment of inertia of the racket
Summing the contributions from the rod and the ring:
I_{\text{total}}
\;=\;
I_{\text{rod}} \;+\; I_{\text{ring}}
\;=\;
\frac{1}{12} \, M \, r^2
\;+\;
\frac{3}{4} \, M \, r^2.
Convert to a common denominator:
\frac{1}{12} \, M \, r^2
\;=\;
\frac{1}{12} \, M \, r^2,
\quad
\frac{3}{4} \, M \, r^2
\;=\;
\frac{9}{12} \, M \, r^2.
Hence,
I_{\text{total}}
\;=\;
\frac{1}{12} \, M \, r^2 \;+\; \frac{9}{12} \, M \, r^2
\;=\;
\frac{10}{12} \, M \, r^2
\;=\;
\frac{5}{6} \, M \, r^2.
Final Answer
The moment of inertia of the badminton racket about the specified axis is
\frac{5}{6} \; M \; r^2.
