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, r \over 2 distance from the end A of the handle will be ................ Mr2.

Rotational Motion PYQ — JEE Main 2021 · 26 Aug · Shift 1 · Q70 with Solution

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Correct answer: 52

Let the racket consist of:

a linear handle of length $4r$ and mass $M$
a circular ring of radius $r$ and mass $M$

The required axis is perpendicular to the handle and in the plane of the ring, passing through a point on the handle at a distance $\dfrac r2$ from end $A$ .

We compute the moment of inertia of both parts about this axis and add them.

1. Geometry from the figure

The usual scale shown implies:

handle length $=4r$
ring radius $=r$
the ring is attached at the far end of the handle, so the center of the ring lies at distance $4r+r=5r$ from end $A$ along the handle axis.

The given axis passes through a point at distance $\dfrac r2$ from $A$ . Hence the distance of the axis from the center of the ring is $d=5r-\frac r2=\frac{9r}{2}.$

2. Moment of inertia of the handle

The handle is a uniform thin rod of length $4r$ , mass $M$ .

Its center is at distance $2r$ from end $A$ . Therefore distance of axis from the rod’s center is $a=2r-\frac r2=\frac{3r}{2}.$

For a rod about an axis perpendicular to its length through its center, $I_{\text{cm}}=\frac{1}{12}M(4r)^2=\frac{16}{12}Mr^2=\frac{4}{3}Mr^2.$

By parallel axis theorem,

=\frac{4}{3}Mr^2+M\left(\frac{3r}{2}\right)^2.$$ So, $$I_{\text{handle}}=\frac{4}{3}Mr^2+\frac{9}{4}Mr^2 =\left(\frac{16+27}{12}\right)Mr^2 =\frac{43}{12}Mr^2.$$ --- ## 3. Moment of inertia of the circular ring The ring is a thin circular hoop of radius $r$, mass $M$. The required axis is **in the plane of the ring**. For a hoop, the MOI about any diameter is $$I_{\text{diameter}}=\frac{1}{2}Mr^2$$ (using perpendicular axis theorem since $I_z=Mr^2$ for the axis perpendicular to plane through center, and symmetry gives $I_x=I_y$). Now shift this axis from the center of the hoop to the given axis by distance $d=\dfrac{9r}{2}$. Hence, $$I_{\text{ring}}=\frac{1}{2}Mr^2+M\left(\frac{9r}{2}\right)^2 =\frac{1}{2}Mr^2+\frac{81}{4}Mr^2.$$ Thus, $$I_{\text{ring}}=\left(\frac{2+81}{4}\right)Mr^2=\frac{83}{4}Mr^2.$$ --- ## 4. Total moment of inertia $$I=I_{\text{handle}}+I_{\text{ring}}$$ $$I=\frac{43}{12}Mr^2+\frac{83}{4}Mr^2$$ $$I=\frac{43}{12}Mr^2+\frac{249}{12}Mr^2$$ $$I=\frac{292}{12}Mr^2=\frac{73}{3}Mr^2.$$ So the coefficient of $Mr^2$ is $$\boxed{\frac{73}{3}}.$$ --- ## 5. Comparison with stored answer Stored correct answer: $52$ My derived answer is $\dfrac{73}{3}\approx 24.33$, which does **not** match $52$. This suggests either: 1. the figure contains additional length information not visible in the text, or 2. the stored answer is incorrect. Based on the standard interpretation of the racket geometry described above, the answer is $$\boxed{\frac{73}{3}}.$$

More from Rotational Motion

1. Geometry from the figure

The usual scale shown implies:

handle length

=4r

ring radius

=r

the ring is attached at the far end of the handle, so the center of the ring lies at distance

4r+r=5r

from end

A

along the handle axis.

The given axis passes through a point at distance

\dfrac r2

from

A

. Hence the distance of the axis from the center of the ring is

d=5r-\frac r2=\frac{9r}{2}.

2. Moment of inertia of the handle

The handle is a uniform thin rod of length

4r

, mass

M

Its center is at distance

2r

from end

A

. Therefore distance of axis from the rod’s center is

a=2r-\frac r2=\frac{3r}{2}.

For a rod about an axis perpendicular to its length through its center,

I_{\text{cm}}=\frac{1}{12}M(4r)^2=\frac{16}{12}Mr^2=\frac{4}{3}Mr^2.

By parallel axis theorem,

=\frac{4}{3}Mr^2+M\left(\frac{3r}{2}\right)^2.$$ So, $$I_{\text{handle}}=\frac{4}{3}Mr^2+\frac{9}{4}Mr^2 =\left(\frac{16+27}{12}\right)Mr^2 =\frac{43}{12}Mr^2.$$ --- ## 3. Moment of inertia of the circular ring The ring is a thin circular hoop of radius $r$, mass $M$. The required axis is **in the plane of the ring**. For a hoop, the MOI about any diameter is $$I_{\text{diameter}}=\frac{1}{2}Mr^2$$ (using perpendicular axis theorem since $I_z=Mr^2$ for the axis perpendicular to plane through center, and symmetry gives $I_x=I_y$). Now shift this axis from the center of the hoop to the given axis by distance $d=\dfrac{9r}{2}$. Hence, $$I_{\text{ring}}=\frac{1}{2}Mr^2+M\left(\frac{9r}{2}\right)^2 =\frac{1}{2}Mr^2+\frac{81}{4}Mr^2.$$ Thus, $$I_{\text{ring}}=\left(\frac{2+81}{4}\right)Mr^2=\frac{83}{4}Mr^2.$$ --- ## 4. Total moment of inertia $$I=I_{\text{handle}}+I_{\text{ring}}$$ $$I=\frac{43}{12}Mr^2+\frac{83}{4}Mr^2$$ $$I=\frac{43}{12}Mr^2+\frac{249}{12}Mr^2$$ $$I=\frac{292}{12}Mr^2=\frac{73}{3}Mr^2.$$ So the coefficient of $Mr^2$ is $$\boxed{\frac{73}{3}}.$$ --- ## 5. Comparison with stored answer Stored correct answer: $52$ My derived answer is $\dfrac{73}{3}\approx 24.33$, which does **not** match $52$. This suggests either: 1. the figure contains additional length information not visible in the text, or 2. the stored answer is incorrect. Based on the standard interpretation of the racket geometry described above, the answer is $$\boxed{\frac{73}{3}}.$$

Rotational Motion question

Rotational Motion question

1. Geometry from the figure

2. Moment of inertia of the handle

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Rotational Motion question

Rotational Motion question

1. Geometry from the figure

2. Moment of inertia of the handle

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