1.
The force generated by the rotating puck (usually called the centrifugal force) has to be sufficient to balance the gravitational forces on the suspended objects.
Centrifugal force=mv²/r where m is the mass of the puck, v its speed and r the length of the string above the table (radius of the trajectory). This comes to 0.1v²/0.2=v²/2N.
The combined weight of the objects is 0.4×9.80=3.92N.
So v²=7.84 and v=2.8m/s approx. At this (tangential) speed the system will be in equilibrium.
2.
When the sandbag is pierced, the weight starts to vary. Call the rate at which the mass changes M(t), i.e., dM/dt=-s, so we can write M(t)=-st+c where c is a constant. When t=0, M=0.4kg so c=0.4. When t=T the bag is empty so we can write 0.2=0.4-sT, so s=0.2/T and M(t)=0.4-0.2t/T. The gravitational force becomes Mg=(0.4-0.2t/T)g.
Now centrifugal force mv²/r=(0.4-0.2t/T)g (gravitational force) and v²=(0.4-0.2t/T)gr/m. We can put m=0.1 because the puck’s weight is constant, so v²=(4-2t/T)9.8r. But angular momentum has to be conserved and that means I⍵=mr²⍵=mr²v/r=mrv must stay constant for the puck. (I=mr² is the moment of inertia of a point mass m a distance r from the axis of rotation.) The initial angular momentum is mrv=0.1×0.2×2.8=0.056. So rv=constant 0.56, and v=0.56/r. [Note that the rate of change of mass doesn’t need to be constant because we know that the angular momentum (rotating puck) when the sandbag is full is equal to the angular momentum when it’s empty. So we only really need the initial and final angular momenta. Nevertheless, the function M(t) encapsulates in one function how the mass can vary.]
0.3136/r²=(4-2t/T)9.8r; so r³=0.3136/(9.8(4-2t/T)). When t=T, r³=0.3136/19.6=0.016, so r=0.252m=25.2cm. Therefore v=0.56/0.252=2.22m/s.