Question 483: Centre of Mass and Rotational Motion

Question

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' $I_1$ '. The same rod is bent into a ring and its moment of inertia about the diameter is ' $I_2$ '. Then $I_1 / I_2$ is

Accepted Answer

Step 1: Moment of inertia of the straight rod
The moment of inertia for a rod of mass $ M $ and length $ L $, about its center and perpendicular to its length, is given by:
$$ I_1 = \frac{ML^2}{12} $$
Step 2: Bending the rod into a ring
When you bend the rod into a circle (a ring), the length of the rod becomes the circumference of the ring.
$$ L = 2\pi r $$
This means the radius $ r $ of the ring is:
$$ r = \frac{L}{2\pi} $$
Step 3: Moment of inertia of the ring about its diameter
The moment of inertia of a ring of mass $ M $ and radius $ r $ about its diameter is:
$$ I_2 = \frac{Mr^2}{2} $$
Now, substitute the value of $ r $ from above:
$$ I_2 = \frac{M}{2} \left( \frac{L}{2\pi} ight)^2 = \frac{M}{2} \cdot \frac{L^2}{4\pi^2} = \frac{ML^2}{8\pi^2} $$
Step 4: Ratio of the two moments of inertia
To find $ \frac{I_1}{I_2} $, divide the two results:
$$ \frac{I_1}{I_2} = \frac{\frac{ML^2}{12}}{ \frac{ML^2}{8\pi^2} } = \frac{ML^2}{12} 	imes \frac{8\pi^2}{ML^2} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3} $$

Answer

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

Answer

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

Answer

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

Answer

$\frac{\pi^2}{9}$

Centre of Mass and Rotational Motion

Moment of inertia of the rod about an axis passing through the centre and perpendicular to its length is ' I_1 '. The same rod is bent into a ring and its moment of inertia about the diameter is ' I_2 '. Then I_1 / I_2 is

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