First slide

Moment of inertia

Question

A circular disc X of radius R is made from an iron plate of thickness t, and another disc Y of radius 4R is made from an iron plate of thickness t/4. Then the relation between the moment of inertia l_X and I_Y is

Moderate

A

$I_{Y} = 64 I_{X}$
B

$I_{Y} = 32 I_{X}$
C

$I_{Y} = 16 I_{X}$
D

$I_{Y} = I_{X}$

Solution

Moment of Inertia of disc $I = \frac{1}{2} M R^{2}$
$= \frac{1}{2} (π R^{2} t ρ) R^{2} = \frac{1}{2} π t ρ R^{4}$
$\begin{array}{l} [As M = V \times ρ = π R^{2} t ρ where t = thickness, ρ = density] \\ ∴ \frac{I_{y}}{I_{x}} = \frac{t_{y}}{t_{x}} {(\frac{R_{y}}{R_{x}})}^{4} [If ρ = constant] \\ \Rightarrow \frac{I_{y}}{I_{x}} = \frac{1}{4} (4)^{4} = 64 [Given R_{y} = 4 R_{x}, t_{y} = \frac{t_{x}}{4}] \\ \Rightarrow I_{y} = 64 I_{x} \end{array}$

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