Question

A thin uniform rectangular plate of mass 2 kg is placed in X-Y plane as shown in figure. The moment of inertial about x-axis is Ix = 0.2 kg m² and the moment of inertia about y-axis is Iy = 0.3 kg m². The radius of gyration of the plate about the axis passing through O and perpendicular to the plane of the plate is

• A. 50 cm
• B. 5 cm
• C. 38.7 cm
• D. 31.6 cm

Answer 1

The Correct Option is A

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.

Answer 2

The radius of gyration \( k \) about an axis is related to the moment of inertia \( I \) by the formula: \[ I = m k^2 \] where \( m \) is the mass of the object, and \( k \) is the radius of gyration about the axis. For the axis passing through the point O and perpendicular to the plane of the plate, the total moment of inertia \( I_z \) can be found using the parallel axis theorem: \[ I_z = I_x + I_y \] Given: - \( I_x = 0.2 \, \text{kg} \, \text{m}^2 \), - \( I_y = 0.3 \, \text{kg} \, \text{m}^2 \), - \( m = 2 \, \text{kg} \). The total moment of inertia is: \[ I_z = 0.2 + 0.3 = 0.5 \, \text{kg} \, \text{m}^2 \] Now, using the formula \( I_z = m k^2 \), we can solve for \( k \): \[ k^2 = \frac{I_z}{m} = \frac{0.5}{2} = 0.25 \] \[ k = \sqrt{0.25} = 0.5 \, \text{m} = 50 \, \text{cm} \] Thus, the radius of gyration is 50 cm.

Therefore, the correct answer is (A).