(i) Your mistake is when you calculated the moment of inertia about corner A. It should be $\frac{20}{48}mc^2$.

The MI about the centre of the plate is $\frac{1}{12}m[c^2+(\frac{c}{2})^2]=\frac{5}{48}mc^2$. The distance $r$ from the corner A to the centre is given by $r^2=(\frac{c}{2})^2+(\frac{c}{4})^2=\frac{5}{16}c^2=\frac{15}{48}c^2$. So the MI about the corner A is $\frac{5}{48}mc^2+mr^2=\frac{20}{48}mc^2$.

When pin B is released the weight of the plate (plus any centripetal force) is borne by pin A. At the instant of release there is no rotation so no centripetal force. The reaction at A is $mg$ upwards. The force at the CM is also $mg$ but downwards, so there is a torque on the plate equal to $mg\frac{c}{2}$.

After correcting the mistake regarding MI, your calculation gives the correct result of $\frac{6g}{5c}$.

(iii) In this case the spring at A does not have time to extend, so when B is cut the force supplied by A is still $\frac12mg$ upwards. The force at the CM is $mg$. The net force of $\frac12mg$ accelerates the CM downwards. The net torque is $\frac12mg\frac{c}{2}$. Using the CM as the centre of rotation we have

$I\alpha = \tau$

$\frac{5}{48}mc^2 \alpha = (\frac12 mg)(\frac12 c)$

$\alpha = \frac{12g}{5c}=2.4\frac{g}{c}$

which agrees with the given answer.

(ii) It is not clear to me how to deal with this case. The wire is inextensible, so the tension adjusts instantly from $\frac12mg$ to $mg$ as in case (i).

The difference between the 3 cases is that in (i) corner A cannot move in any direction, in (ii) it can only move in an arc, initially horizontally, and in (iii) it can only move downwards. (I think we are expected to assume that the spring resists bending much more than it resists extension/compression.)

(I have moved some other comments into my Answer.)

edited Feb 6, 2017 by sammy gerbil