Question 538: Centre of Mass and Rotational Motion

Question

Two loops ' $A$ ' and ' $B$ ' of radii ' $R_1$ ' and ' $R_2$ ' are made from uniform wire. If moment of inertia of ' A ' is ' $\mathrm{I}_{\mathrm{A}}$ ' and that ' B ' is ' $\mathrm{I}_{\mathrm{B}}$ ', then $\mathrm{R}_2 / \mathrm{R}_1$ is $\left[\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=27\right]$

Accepted Answer

The moment of inertia of a loop made from uniform wire about its center is given by:
$$ I = mR^2 $$
where $ m $ is the mass of the wire, and $ R $ is the radius of the loop.
Assuming that the wire is uniform, both loops $ A $ and $ B $ will have the same linear mass density. This means the mass of a loop is proportional to its circumference:
$$ m = \lambda \cdot 2\pi R $$
where $ \lambda $ is the linear mass density.
Therefore, the moments of inertia for loops $ A $ and $ B $ are:
$$ I_A = (\lambda \cdot 2\pi R_1)R_1^2 = \lambda \cdot 2\pi R_1^3 $$
$$ I_B = (\lambda \cdot 2\pi R_2)R_2^2 = \lambda \cdot 2\pi R_2^3 $$
Given that $ \frac{I_A}{I_B} = 27 $, we have:
$$ \frac{\lambda \cdot 2\pi R_1^3}{\lambda \cdot 2\pi R_2^3} = 27 $$
This simplifies to:
$$ \frac{R_1^3}{R_2^3} = 27 $$
Taking the cube root of both sides:
$$ \frac{R_1}{R_2} = 3 $$
Thus, the ratio $ R_2 / R_1 $ is:
$$ \frac{R_2}{R_1} = \frac{1}{3} $$
Hence, the correct option is:
Option C
$1 : 3$

Answer

$1: 6$

Answer

$1: 4$

Answer

$1: 3$

Answer

$1: 2$

Centre of Mass and Rotational Motion

Two loops ' A ' and ' B ' of radii ' R_1 ' and ' R_2 ' are made from uniform wire. If moment of inertia of ' A ' is ' \mathrm{I}_{\mathrm{A}} ' and that ' B ' is ' \mathrm{I}_{\mathrm{B}} ', then \mathrm{R}_2 / \mathrm{R}_1 is \left[\frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=27\right]

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