The radius of gyration \( k \) of the plate about an axis perpendicular to the plane is given by: \[ k = \sqrt{\frac{I_x + I_y}{m}} \] Where:
\( I_x = 0.2 \, \text{kg} \, \text{m}^2 \) is the moment of inertia about the x-axis,
\( I_y = 0.3 \, \text{kg} \, \text{m}^2 \) is the moment of inertia about the y-axis,
\( m = 2 \, \text{kg} \) is the mass of the plate. Substituting the values: \[ k = \sqrt{\frac{0.2 + 0.3}{2}} = \sqrt{\frac{0.5}{2}} = \sqrt{0.25} = 0.5 \, \text{m} = 50 \, \text{cm} \]
The correct answer is (A) : 50 cm.
A uniform circular disc of radius \( R \) and mass \( M \) is rotating about an axis perpendicular to its plane and passing through its center. A small circular part of radius \( R/2 \) is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.