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 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

Answer 1

Answer:

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:

$L_1=L_{door} + L_{mud}$

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:

$L_1= L_{mud}= mvr$

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:

$L_2= (I_{door} + I_{mud}) \omega_f$

ωf is the final angular speed

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

$I_{door}=\frac{1}{3}MW^2$

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

$I_{mud}=mr^2$

Hence,

$L_1=L_2\\mvr= (I_{door} + I_{mud}) \omega_f\\\omega_f=\frac{mvr}{I_{door} + I_{mud}} \\\omega_f=\frac{mvr}{I_{door} + I_{mud}}$

$\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}$

$\omega_f=0.223 \frac{rad}{s}$