First slide

Moment of intertia

Question

The radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Then its radius of gyration about a parallel axis through its centre of mass will be

Easy

A

80 cm
B

8 cm
C

0.8 cm
D

80 m

Solution

According to theorem of parallel axes
$\begin{array}{l} I = I_{CM} + {Ma}^{2} \\ or I_{CM} = {MK}^{2} - {Ma}^{2} = M (K^{2} - a^{2}) \\ = M (100 - 36) = M \times 64 = M \times (8)^{2} \\ ∴ K = 8 cm \end{array}$

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