Asolid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500 kg, traveling perpendicular to the door at 12.0 m> s just before impact. find the final angular speed

The final angular speed is 0.223 rad/s

Explanation:

By the conservation of angular moment:

ΔL=0

L₁=L₂

L₁ is the initial angular moment

L₂ is the final angular moment

L₁ is given by:

As the door is at rest its angular moment is zero and the angular moment of mud can be considered as a point object, then:

where

r is the distance from the support point to the axis of rotation (the mud hits at the center of the door; r=0.5 m)

v is the speed

m is the mass of the mud

L₂ is given by:

ωf is the final angular speed

The moment of inertia of the door can be considered as a rectangular plate:

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

Hence,

what is the question because that is not a question.

1/0.14nm = 10^10/1.4 a lot