Question

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 _________ \(Mr^2\). 
 

Hint:

When an axis is "in the plane" of a circular ring, its moment of inertia is \(\frac{1}{2}Mr^2\). If it were "perpendicular to the plane", it would be \(Mr^2\).

Answer 1

Correct Answer: 52

Step 1: Understanding the Concept:
The total moment of inertia (\(I\)) of the racket is the sum of the moments of inertia of the handle (a thin rod) and the head (a thin ring) about the specified axis. We will use the Parallel Axis Theorem.
Step 2: Key Formula or Approach:
1. Moment of inertia of a rod about its COM: \(I_{rod} = \frac{1}{12}ML^2\).
2. Moment of inertia of a ring about its diameter: \(I_{ring} = \frac{1}{2}Mr^2\).
3. Parallel Axis Theorem: \(I = I_{cm} + Md^2\).
Step 3: Detailed Explanation:
Let the end A be at \(x = 0\). The axis is at \(x = 0.5r\).

1. Handle (Rod):
Mass = \(M\), Length \(L = 6r\).
COM of handle is at \(x = 3r\).
Distance from axis to handle COM: \(d_h = 3r - 0.5r = 2.5r\).
\[ I_h = \frac{1}{12}M(6r)^2 + M(2.5r)^2 \]
\[ I_h = \frac{36Mr^2}{12} + 6.25Mr^2 = 3Mr^2 + 6.25Mr^2 = 9.25Mr^2 \]

2. Head (Ring):
Mass = \(M\), Radius = \(r\).
COM of ring is at \(x = 6r + r = 7r\).
Distance from axis to ring COM: \(d_r = 7r - 0.5r = 6.5r\).
The axis is in the plane of the ring, so we use the MOI about a diameter.
\[ I_r = \frac{1}{2}Mr^2 + M(6.5r)^2 \]
\[ I_r = 0.5Mr^2 + 42.25Mr^2 = 42.75Mr^2 \]

3. Total Moment of Inertia:
\[ I_{total} = I_h + I_r = 9.25Mr^2 + 42.75Mr^2 = 52Mr^2 \]
Step 4: Final Answer:
The moment of inertia will be 52 \(Mr^2\).