The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is :
Answer & explanation
Correct answer: option 4
I1 = MR2 / 2
k1 = \(\sqrt{\frac{I_1}{M}}\) = \(\frac{R}{sqrt{2}}\)
I2 = MR2 / 4
k2 = \(\sqrt{\frac{I_2}{M}}\) = \(\frac{R}{2}\)
k1 : k2 = \(\sqrt{2}\) : 1