A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.600 kg, traveling perpendicular to the door at 12.0 m/s just before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

(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