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 44.0 kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.300 kg , traveling perpendicular to the door at 14.0 m/s just before impact.

1- Find the final angular speed of the door.

2- Does the mud make a significant contribution to the moment of inertia?

Answer 1

1. Given,

Length, r = 1 m

Height, h = 2 m

Total mass, M = 44 kg

The mass of sticky mud, m = 0.3 kg

Speed, v = 14 m/s

The moment of inertia of the door, I = (1/3) * Mr²

= (1/3) * 44 * 1²

= 14.67 kg. m²

The angular momentum of the mud, L = m*v* (r/2)

= 0.3 * 14 * (1/2)²

= 2.1 Kg m² /s

Angular momentum, L = [ I + m(r/2)² ]

2.1 = [14.67 + 0.3 * (0.5)²]

= 0.142 rad/s

2.

No, the mud does not make a significant contribution to the moment of inertia, because the mass of a door is much larger than the mass of a mud.