The plate stops due to friction. The time taken is given by \( t = \omega / \alpha \), where \( \alpha = \tau_{friction} / I \). So, \( t \propto I / \tau_{friction} \).

Case 1 (Corner): Using parallel axis theorem, \( I_1 = I_{cm} + m(a/\sqrt{2})^2 = ma^2/6 + ma^2/2 = 2ma^2/3 \).
Case 2 (Center): \( I_2 = ma^2/6 \).
Ratio \( I_1 / I_2 = 4 \).

The frictional torque is \( \tau = \int r (\mu g dm) \). This depends on the average distance of mass elements from the axis.

The calculation of torque for a square plate is complex, but scaling arguments lead to the relation: \[ \frac{t_2}{t_1} = \frac{I_2 / \tau_2}{I_1 / \tau_1} = \frac{1}{2} \]

Thus, the time taken for the center-axis rotation to stop is half the time taken for the corner-axis rotation.