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) *
Mr ^{2}*

*= (1/3) * 44 * 1 ^{2}*

*= 14.67 kg. m ^{2}*

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

*= 0.3 * 14 * (1/2) ^{2}*

*= 2.1 Kg m ^{2} /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.*

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