First slide

Moment of inertia

Question

Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

Moderate

A

2
B

$2 \sqrt{2}$
C

$\sqrt{2}$
D

4

Solution

$\begin{array}{l} Moment of Inertia , I ∝ {mR}^{2} \\ \Rightarrow I \propto (σ π R^{2} t) R^{2} [Thickness and material is same] \\ \Rightarrow I \propto R^{4} \\ ∴ \frac{R_{2}}{R_{1}} = {(\frac{I_{2}}{I_{1}})}^{1 / 4} = (16)^{1 / 4} = 2 \\ \Rightarrow α = 2 \end{array}$

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