Question

A solid wood door 1.00 mm wide and 2.00 mm high is hinged along one side and has a total mass of 43.0 kgkg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.600 kgkg , traveling perpendicular to the door at 14.0 m/sm/s just before impact.
A) Find the angular speed of the door.B) Does the mud make a significant contribution to the moment of inertia?

Answer 1

Answer:

0.290055248619 rad/s

Explanation:

r = $\dfrac{1}{2}\ m$

m denotes the mass

v = Velocity = 14 m/s

Moment of inertia of the door

$I_d=\dfrac{1}{3}mr^2\\\Rightarrow I_d=\dfrac{1}{3}\times 43\times 1^2\\\Rightarrow I_d=14.33\ kgm^2$

Angular momentum of the mud

$L_m=mvr\\\Rightarrow L_m=0.6\times 14\times \dfrac{1}{2}\\\Rightarrow L_m=4.2\ kgm^2/s$

As the angular momentum is conserved we have

$L_m=L_f\\\Rightarrow 4.2=(I_d+mr^2)\omega\\\Rightarrow \omega=\dfrac{4.2}{I_d+mr^2}\\\Rightarrow \omega=\dfrac{4.2}{14.33+0.6\times 0.5^2}\\\Rightarrow \omega=0.290055248619\ rad/s$

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.