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Chapter 1: Q44E (page 332)

A solid wood door 1.00mwide and 2.00mhigh is hinged along one side and has a total mass of 40.0kg. Initially open and at rest, the door is truck at its center by a handful of sticky mud with mass 0.500kg, travelling perpendicular to the door at 12.0m/sjust before impact. Find the final angular speed of the door. Does the mud make a significant contribution to the moment of inertia?

Short Answer

Expert verified

Thus, the final angular speed is0.875rev/s

Step by step solution

01

Conservation of angular momentum.

Apply the law of conservation of angular momentum to a system whose moment of inertia changes gives:

Ii蝇i=If蝇f=constant

02

Initial momentum of the wood.

The initial momentum of the wood is solved as:

Linitial=Lcoor+Lmud=0+mVr=0+(0.5)(12)(0.5)=3

03

Moment of inertia of the door.

The total moment of inertia with arms rapped around the body is solved as:

Idoor=13mR2=13脳40脳(0.5)2=103=3.3

04

Moment of inertia of the mud.

Imud=mR2=0.5脳(0.5)2=0.125

05

Use conservation of angular momentum.

Use the conservation of angular momentum to obtain angular speed:

Linitital=Lfinal3=(3.3+0.125)蝇f蝇f=33.425=0.875


Hence, the final angular speed is0.875rev/s

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