Answer
Work done will be the same.
$W_{a}=W_{b}=W_{c}=W_{d}$
Work Step by Step
Work done on the disc will be :
$W=\Delta K=K_{f}-K_{i}=\dfrac {1}{2}I\left( w^{2}_{f}-w^{2}_{i}\right) $
$I$ is the moment of inertia of the disc, and $w_{f}$ and $w_{i}$ are final and initial angular speed of disc, respectively.
Plugging in the known values, we obtain:
$a)w_{f}=5\dfrac {rad}{s};w_{i}=-2\dfrac {rad}{s}$ we get $W_{a}=\dfrac {21}{2}I$
$b)w_{f}=5\dfrac {rad}{s};w_{i}=2\dfrac {rad}{s}$ we get $W_{b}=\dfrac {21}{2}I$
$c)w_{f}=-5\dfrac {rad}{s};w_{i}=-2\dfrac {rad}{s}$ we get $W_{c}=\dfrac {21}{2}I$
$d)w_{f}=5\dfrac {rad}{s};w_{i}=2\dfrac {rad}{s}$ we get $W_{d}=\dfrac {21}{2}I$
So, we see that in all 4 conditions, the work done on disc will be same
$W_{a}=W_{b}=W_{c}=W_{d}$
