First slide

Moment of inertia

Question

Two circular discs are of same thickness. The diameter of A is twice that of B. The moment of inertia of A as compared to that of B is

Easy

A

twice as large
B

four times as large
C

eight times as large
D

16 times as large

Solution

$Mass of disc \propto area M \propto R^{2} ⇒ M_{A} = 4 M_{B} (as R_{A} = 2 R_{B})$
$\frac{I_{A}}{I_{B}} = \frac{\frac{1}{2} M_{A} R_{A}^{2}}{\frac{1}{2} M_{B} R_{B}^{2}} = \frac{4 M_{B} \times {(2 R_{B})}^{2}}{M_{B} \times R_{B}^{2}} = 16$

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