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 _________ $$Mr^2$$.

We need to determine the total moment of inertia of the badminton racket about an axis perpendicular to the handle, lying in the plane of the ring, and located at a distance of

$$\frac{r}{2}$$

from the free end A of the handle.

1. Identify the Given Parameters

From the problem statement , the badminton racket is split into two geometric components:

• The Handle (Linear Rod): Mass = $$M$$, Length = $$6r$$
• The Head (Circular Ring): Mass = $$M$$, Radius = $$r$$ (Diameter = $$2r$$)

The target axis of rotation is positioned at a distance of $$\frac{r}{2}$$ from end A of the handle.

2. Calculate the Moment of Inertia for the Handle

The handle is treated as a uniform thin rod. We determine its moment of inertia using the parallel axis theorem:

• Center of Mass Location:
  The center of mass of the rod lies at its midpoint, which is at a distance of $$3r$$ from end A.

  $$I_{\text{cm, handle}} = \frac{1}{12} M L^2 = \frac{1}{12} M (6r)^2 = 3Mr^2$$

• Parallel Axis Distance Shift:
  The distance from the rod's center of mass to the target axis (at $$\frac{r}{2}$$ from A) is:

  $$d_1 = 3r - \frac{r}{2} = \frac{5r}{2}$$

Applying the Parallel Axis Theorem ($$I = I_{\text{cm}} + Md^2$$):

$$I_{\text{handle}} = 3Mr^2 + M\left(\frac{5r}{2}\right)^2 = 3Mr^2 + \frac{25}{4}Mr^2 = \frac{37}{4}Mr^2$$

3. Calculate the Moment of Inertia for the Ring Head

The head of the racket is treated as a uniform circular ring. The rotation axis lies along the coplanar diameter plane of the ring:

• Center of Mass Axis:
  The moment of inertia of a thin ring about its coplanar diameter axis passing through its center is:

  $$I_{\text{cm, ring}} = \frac{1}{2}Mr^2$$

• Parallel Axis Distance Shift:
  The center of the ring is located at the far end of the handle, a distance of $$6r + r = 7r$$ from end A. The shift distance to the target axis is:

  $$d_2 = 7r - \frac{r}{2} = \frac{13r}{2}$$

Applying the Parallel Axis Theorem ($$I = I_{\text{cm}} + Md^2$$):

$$I_{\text{ring}} = \frac{1}{2}Mr^2 + M\left(\frac{13r}{2}\right)^2 = \frac{2}{4}Mr^2 + \frac{169}{4}Mr^2 = \frac{171}{4}Mr^2$$

4. Determine Total Moment of Inertia

The total moment of inertia of the entire composite racket system is the scalar sum of both individual components:

$$I_{\text{total}} = I_{\text{handle}} + I_{\text{ring}}$$

$$I_{\text{total}} = \frac{37}{4}Mr^2 + \frac{171}{4}Mr^2 = \frac{208}{4}Mr^2 = 52Mr^2$$

Conclusion

The multiplier value for the final total moment of inertia observed for the badminton racket is 52.