Question

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.400 kg , traveling perpendicular to the door at 12.0 m/s just before impact.

A) Find the final angular speed of the door

B) Does the mud make a significant contribution to the moment of inertia? Yes or no?

Answer 1

the door is struck at its centre so you know it is struck at
r = 0.5m from the hinge.

at the instant just before impact, the mud has momentum
p = 0.4kg x 12m/s = 4.8 kg.m/s and it sticks to the door, imparting all of its momentum to the door and causing it to swing shut.

at the instant after impact, the muddy door now has a combined mass m = 45kg + 0.4kg = 45.4kg and a momentum of 4.8 kgm/s.

from the relation p = mv, the (linear, tangential) velocity v can be calculated at this point on the door's circular path:

v = p/m = 4.8 kgm/s / 45.4kg = 0.1057 m/s.

from here the angular velocity can be found, ie. the angle per second swept out by the swinging door.

using the relationships for centripetal acceleration that are fairly easy to prove but i cant be arsed,
a =(v^2)/r = (w^2)r.

this leads to v = rw

ie. the angular velocity w = v/r
= 0.1057m/s / 0.5m = 0.2114 s^-1.

The units are correct, 1/seconds, because an angle has no units - it is essentially a length divided by a length so don't let that confuse you.

the first answerer might be a bit mixed up because the angular velocity is the same for all parts of the door regardless of their distance from the hinge.
the linear (tangential to the circular path at a given moment) velocity will increase with increasing distance from the hinge though.

and there will be no significant contribution to the moment of inertia by mud.