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Question

Physics Question on rotational motion

Radius of gyration of disc of mass 50 g and radius 2.5 cm about an axis passing through its center of gravity and perpendicular to the plane is

A

6.54 cm

B

3.64 cm

C

1.77 cm

D

0.88 cm

Answer

1.77 cm

Explanation

Solution

Here mass M = 50 g and radius R = 2.5 cm Required moment of inertia of the disc is given by I=MR22=MK2I=\frac{MR^2}{2}=MK^2 so,K2=R22orK=R2=2.52=2.522K^2=\frac{R^2}{2}or \, K=\frac{R}{\sqrt{2}}=\frac{2.5}{\sqrt{2}}=\frac{2.5\sqrt{2}}{2} = 1.767= 1.77 cm