Animated Solution for Physics - Rotational Motion: Consider a badminton racket with length scales as shown in the figure. If the mass of the linear and circular portions of the badminton racket are same ($M$) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end $A$ of the handle will be ........ $M{r}^2$.

Imagine holding a badminton racket. We want to find out how much rotational inertia it possesses when swung around a specific axis near its handle. To make this complex shape manageable, we can break the racket into two standard geometric parts: a uniform rod representing the handle, and a circular ring representing the head of the racket.

The axis of rotation given in the problem does not pass through the center of mass of either the handle or the ring. Whenever we face this situation, we must rely on our trusty tool: the Parallel Axis Theorem. It states that the moment of inertia about any axis is equal to the moment of inertia about a parallel axis passing through the center of mass, plus the mass of the body multiplied by the square of the perpendicular distance between the two axes:

Let's focus on the handle first. It is a uniform rod of mass $M$ and length $6r$. Its center of mass ($C{M}_1$) lies exactly in the middle, at a distance of $3r$ from the end $A$.

Our axis of rotation is located at a distance of $\frac{r}{2}$ from end $A$. Therefore, the distance $d_1$ between the handle's center of mass and our axis is:

Now, applying the parallel axis theorem for the handle. The moment of inertia of a rod about its center is $\frac{M{L}^2}{12}$. Here, $L=6r$:

Next, let's look at the circular ring. Its center of mass ($C{M}_2$) is at its geometric center. Since the handle is $6r$ long and the ring has a radius $r$, its center is located at $6r+r=7r$ from end $A$.

The distance $d_2$ from our axis to the ring's center of mass is:

Here is a crucial trap! The problem states that the axis of rotation is in the plane of the ring. Therefore, the moment of inertia of the ring about its own center of mass is the moment of inertia about its diameter, which is $\frac{M{r}^2}{2}$, not $M{r}^2$! Applying the parallel axis theorem:

We are almost there! The total moment of inertia of the racket is simply the sum of the moments of inertia of the handle and the ring:

Finally, dividing $208$ by $4$ gives us exactly $52$. So, the total moment of inertia is:

The coefficient we were looking for is 52. What a beautiful application of the parallel axis theorem!