Question 498: Centre of Mass and Rotational Motion

Question

Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is ' $I$ '. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' $\mathrm{I}_1$ '. If $\mathrm{I}_1=\mathrm{xI}$, then the value of ' x ' is

Accepted Answer

Step 1: Moment of inertia of thin uniform rod about perpendicular axis through its centre
Length of rod = $L$, mass = $M$.
For a thin rod about a perpendicular axis through its centre (i.e., an axis normal to rod at centre):
$ I = \frac{1}{12} M L^2 $
Step 2: Bend the same rod into the shape of a ring
Now the rod of length $L$ is bent into a ring (circle).
Radius of the ring:
$ 2 \pi R = L \quad \Rightarrow \quad R = \frac{L}{2\pi} $
Mass = $M$.
Step 3: Moment of inertia of ring about its diameter
For a thin ring of mass $M$, radius $R$:
About axis through center and perpendicular to plane: $MR^2$.
About diameter of ring (axis in plane, through centre): $ 	frac{1}{2}MR^2$.
$ I_1 = \frac{1}{2}MR^2 $
Step 4: Express $I_1$ in terms of given $I$
From Step 1: $ I = \frac{1}{12}ML^2$.
From Step 2: $ R = \frac{L}{2\pi} $.
So:
$ I_1 = \frac{1}{2}M \left(\frac{L}{2\pi}ight)^2 = \frac{1}{2}M \cdot \frac{L^2}{4 \pi^2} = \frac{M L^2}{8 \pi^2} $
Step 5: Ratio
$ \frac{I_1}{I} = \frac{	frac{ML^2}{8\pi^2}}{	frac{1}{12}ML^2} = \frac{12}{8\pi^2} = \frac{3}{2\pi^2} $
Final Answer:
$ x = \frac{3}{2 \pi^2} $
Ã¢ÂÂ Correct option:
Option B

Answer

$\frac{2 \pi^2}{3}$

Answer

$\frac{3}{2 \pi^2}$

Answer

$\frac{3 \pi^2}{4}$

Answer

$\frac{4}{3 \pi^2}$

Centre of Mass and Rotational Motion

Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is ' I '. If the same rod is bent in the form of ring, its moment of inertia about the diameter is ' \mathrm{I}_1 '. If \mathrm{I}_1=\mathrm{xI}, then the value of ' x ' is

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