Answer:
The final angular speed is 0.223 rad/s
Explanation:
By the conservation of angular moment:
ĪL=0
Lā=Lā
Lā is the initial angular moment
Lā is the final angular moment
Lā is given by:
[tex]L_1=L_{door} + L_{mud}[/tex]
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:
[tex]L_1= L_{mud}= mvr[/tex]
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
Lā is given by:
[tex]L_2= (I_{door} + I_{mud}) \omega_f[/tex]
Ļf is the final angular speed
The moment of inertia of the door can be considered as a rectangular plate:
[tex]I_{door}=\frac{1}{3}MW^2[/tex]
M is the mass of the door
W is the width of the door
The moment of inertia of the mud is:
[tex]I_{mud}=mr^2[/tex]
Hence,
[tex]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}}[/tex]
[tex]\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}[/tex]
[tex]\omega_f=0.223 \frac{rad}{s}[/tex]
By applying the law of conservation of angular momentum, the final angular speed of the solid wood door and the sticky mud is 0.2230 rad/s.
Given the following data:
To find the final angular speed, we would use the law of conservation of angular momentum:
Applying the law of conservation of angular momentum:
[tex]L_i = L_f[/tex] Ā ....equation 1
First of all, we would determine both the initial and final angular momentum of the solid wood door and the sticky mud.
Note: The initial angular momentum of the the solid wood door would be zero (0) because it is initially open and at rest.
Note: The sticky mud is considered a point of mass.
For initial angular momentum:
The distance of the mud from the center of mass of the the support point to the axis of rotation is the radius = [tex]\frac{1}{2} \times 1=0.5\;meter[/tex]
For a point of mass, angular momentum is the product of linear momentum and radius:
[tex]L_{i} = mvr\\\\L_{i} = 0.500\times 12\times0.5\\\\L_{i} = 3 kgm^2/s[/tex]
For final angular momentum:
[tex]L_f = I_m\omega_f + I_d\omega_f\\\\L_f = (I_m + I_d)\omega_f[/tex]......equation 2.
The moment of inertia of the sticky mud (point of mass) is given by the formula:
[tex]I_d = mr^2\\\\I_d = 0.5 \times 0.5^2\\\\I_d = 0.5 \times 0.25\\\\I_d = 0.125 \;kgm^2[/tex]
The moment of inertia of the solid wood door (rectangular) is given by the formula:
[tex]I_d = \frac{mw^2}{3}\\\\I_d = \frac{40 \times 1^2}{3}\\\\I_d =13.33 \;kg/m^2[/tex]
Substituting the values into eqn. 2, we have:
[tex]L_f = (I_m + I_d)\omega_f\\\\L_f = (0.125 + 13.33)\omega_f\\\\L_f = 13.455\omega_f \;kgm^2/s[/tex]
Substituting the values of [tex]L_i[/tex] and [tex]L_f[/tex] into eqn. 1, we have:
[tex]3 = 13.455\omega_f\\\\\omega_f = \frac{3}{13.455} \\\\\omega_f = 0.2230\; rad/s[/tex]
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