Answer:
0.290055248619 rad/s
Explanation:
r = [tex]\dfrac{1}{2}\ m[/tex]
m denotes the mass
v = Velocity = 14 m/s
Moment of inertia of the door
[tex]I_d=\dfrac{1}{3}mr^2\\\Rightarrow I_d=\dfrac{1}{3}\times 43\times 1^2\\\Rightarrow I_d=14.33\ kgm^2[/tex]
Angular momentum of the mud
[tex]L_m=mvr\\\Rightarrow L_m=0.6\times 14\times \dfrac{1}{2}\\\Rightarrow L_m=4.2\ kgm^2/s[/tex]
As the angular momentum is conserved we have
[tex]L_m=L_f\\\Rightarrow 4.2=(I_d+mr^2)\omega\\\Rightarrow \omega=\dfrac{4.2}{I_d+mr^2}\\\Rightarrow \omega=\dfrac{4.2}{14.33+0.6\times 0.5^2}\\\Rightarrow \omega=0.290055248619\ rad/s[/tex]
The angular speed of the door is 0.290055248619 rad/s
The mud has a mass 0.6 kg it will not have a significant on the door which is 43 kg. The door is significantly heavier than the mud.
