Answer:
(A) This question invloves the concept of conservation of angular momentum. It is an inelastic or "sticky" collision. So, momentum is conserved and kinetic energy is lost. The final angular momentum will include both masses since the mud sticks to the door.
L=angular momentum = I x w (moment of inertia times angular velocity) = r x p = r x mv (radius times momentum which is mass times velocity)
L(initial)=L(final)
I of the door: The moment of inertia for a thin rectangular plate rotating about an axis on its edge is (1/3)ma2 where a is the distance from the axis or in this case the width of the door.
L(mud)= r x m x v = (.5) x (.7) x (13) = 4.55 kgm2/s = L(initial)
4.55= Ifwf = ((1/3) x mf x a2) x wf = ((1/3) x (mmud + mdoor) x (1)2) x wf =
4.55 = ((1/3) x (43.7) x 1) x wf
wf = .312 rad/sec
(B) No: since the door is so much larger than the mud 43>>>0.7